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Introduction

In Grade 10 we studied motion but not what caused the motion. In this chapter we will learn that a net force is needed to cause motion. We recall what a force is and learn about how force and motion are related. We are introduced to two new concepts, momentum and impulse, and we learn more about turning forces and the force of gravity.

Force

What is a force?

A force is anything that can cause a change to objects. Forces can:

  • change the shape of an object
  • accelerate or stop an object
  • change the direction of a moving object.

A force can be classified as either a contact force or a non-contact force.

A contact force must touch or be in contact with an object to cause a change. Examples of contact forces are:

  • the force that is used to push or pull things, like on a door to open or close it
  • the force that a sculptor uses to turn clay into a pot
  • the force of the wind to turn a windmill

Write down 5 examples (excluding those given above) of contact forces that you see on your way to school.

A non-contact force does not have to touch an object to cause a change. Examples of non-contact forces are:

  • the force due to gravity, like the Earth pulling the Moon towards itself
  • the force due to electricity, like a proton and an electron attracting each other
  • the force due to magnetism, like a magnet pulling a paper clip towards itself

The unit of force is the newton (symbol N). This unit is named after Sir Isaac Newton who first defined force. Force is a vector quantity and has a magnitude and a direction. We use the abbreviation F for force.

Note: Interesting Fact :

There is a popular story that while Sir Isaac Newton was sitting under an apple tree, an apple fell on his head, and he suddenly thought of the Universal Law of Gravitation. Coincidently, the weight of a small apple is approximately 1 N.

Note: Interesting Fact :

Force was first described by Archimedes of Syracuse (circa 287 BC - 212 BC). Archimedes was a Greek mathematician, astronomer, philosopher, physicist and engineer. He was killed by a Roman soldier during the sack of the city, despite orders from the Roman general, Marcellus, that he was not to be harmed.

This chapter will often refer to the resultant force acting on an object. The resultant force is simply the vector sum of all the forces acting on the object. It is very important to remember that all the forces must be acting on the same object. The resultant force is the force that has the same effect as all the other forces added together.

Examples of Forces in Physics

Most of Physics revolves around the study of forces. Although there are many different forces, all are handled in the same way. All forces in Physics can be put into one of four groups. These are gravitational forces, electromagnetic forces, strong nuclear forces and weak nuclear forces. You will mostly come across gravitational or electromagnetic forces at school.

Gravitational Forces

Gravity is the attractive force between two objects due to the mass of the objects. When you throw a ball in the air, its mass and the Earth's mass attract each other, which leads to a force between them. The ball falls back towards the Earth, and the Earth accelerates towards the ball. The movement of the Earth towards the ball is, however, so small that you couldn't possibly measure it.

Electromagnetic Forces

Almost all of the forces that we experience in everyday life are electromagnetic in origin. They have this unusual name because long ago people thought that electric forces and magnetic forces were different things. After much work and experimentation, it has been realised that they are actually different manifestations of the same underlying theory.

Electric or Electrostatic Forces

If we have objects carrying electrical charge, which are not moving, then we are dealing with electrostatic forces (Coulomb's Law). This force is actually much stronger than gravity. This may seem strange, since gravity is obviously very powerful, and holding a balloon to the wall seems to be the most impressive thing electrostatic forces have done, but if we think about it: for gravity to be detectable, we need to have a very large mass nearby. But a balloon rubbed against someone's hair can stick to a wall with a force so strong that it overcomes the force of gravity between the entire Earth and the balloon—with just the charges in the balloon and the wall!

Magnetic Forces

The magnetic force is a different manifestation of the electromagnetic force. It stems from the interaction between moving charges as opposed to the fixed charges involved in Coulomb's Law. Examples of the magnetic force in action include magnets, compasses, car engines and computer data storage. Magnets are also used in the wrecking industry to pick up cars and move them around sites.

Friction

According to Newton's First Law (we will discuss this later in the chapter) an object moving without a force acting on it will keep on moving. Then why does a box sliding on a table come to a stop? The answer is friction. Friction arises where two surfaces are in contact and moving relative to eachother as a result of the interaction between the molecules of the two contact surfaces—for instance the interactions between the molecules on the bottom of the box with molecules on the top of the table. This interaction is electromagnetic in origin, hence friction is just another view of the electromagnetic force. Later in this chapter we will discuss frictional forces a little more.

Drag Forces

This is the force an object experiences while travelling through a medium like an aeroplane flying through air. When something travels through the air it needs to displace air as it travels and because of this, the air exerts a force on the object. This becomes an important force when you move fast and a lot of thought is taken to try and reduce the amount of drag force a sports car or an aeroplane experiences. The drag force is very useful for parachutists. They jump from high altitudes and if there was no drag force, then they would continue accelerating all the way to the ground. Parachutes are wide because the more surface area you have, the greater the drag force and hence the slower you hit the ground.

Systems and External Forces

The concepts of systems and forces external to such systems are very important in Physics. A system is any collection of objects. If one draws an imaginary box around such a system then an external force is one that is applied by an object or person outside the box. Imagine for example a car pulling two trailers.

Figure 1
Figure 1 (PG11C11_001.png)

If we draw a box around the two trailers they can be considered a closed system or unit. When we look at the forces on this closed system the following forces will apply (we assume drag forces are absent):

  • The force of the car pulling the unit (trailer A and B)
  • The force of friction between the wheels of the trailers and the road (opposite to the direction of motion)
  • The force of the Earth pulling downwards on the system (gravity)
  • The force of the road pushing upwards on the system

These forces are called external forces to the system.

The following forces will not apply:

  • The force of A pulling B
  • The force of B pulling A
  • The force of friction between the wheels of the car and the road (opposite to the direction of motion)

We can also draw a box around trailer A or B, in which case the forces will be different.

Figure 2
Figure 2 (PG11C11_002.png)

If we consider trailer A as a system, the following external forces will apply:

  • The force of the car pulling on A (towards the right)
  • The force of B pulling on A (towards the left)
  • The force of the Earth pulling downwards on the trailer (gravity)
  • The force of the road pushing upwards on the trailer
  • The force of friction between the wheels of A and the road (opposite to the direction of motion)

Force Diagrams

If we look at the example above and draw a force diagram of all the forces acting on the two-trailer-unit, the diagram would look like this:

Figure 3
Figure 3 (PG11C11_003.png)

It is important to keep the following in mind when you draw force diagrams:

  • Make your drawing large and clear.
  • You must use arrows and the direction of the arrow will show the direction of the force.
  • The length of the arrow will indicate the size of the force, in other words, the longer arrows in the diagram (F11 for example) indicates a bigger force than a shorter arrow (Fff). Arrows of the same length indicate forces of equal size (FNN and Fgg). Use “little lines” like in maths to show this.
  • Draw neat lines using a ruler. The arrows must touch the system or object.
  • All arrows must have labels. Use letters with a key on the side if you do not have enough space on your drawing.
  • The labels must indicate what is applying the force (the force of the car?) on what the force is applied (?on the trailer?) and in which direction (to the right)
  • If the values of the forces are known, these values can be added to the diagram or key.

Exercise 1: Force diagrams

Draw a labeled force diagram to indicate all the forces acting on trailer A in the example above.

Solution
  1. Step 1. Draw a large diagram of the ?picture? from your question :
    Figure 4
    Figure 4 (PG11C11_004.png)
  2. Step 2. Add all the forces :
    Figure 5
    Figure 5 (PG11C11_005.png)
  3. Step 3. Add the labels :
    Figure 6
    Figure 6 (PG11C11_006.png)

Free Body Diagrams

In a free-body diagram, the object of interest is drawn as a dot and all the forces acting on it are drawn as arrows pointing away from the dot. A free body diagram for the two-trailer-system will therefore look like this:

Figure 7
Figure 7 (PG11C11_007.png)

Exercise 2: Free body diagram

Draw a free body diagram of all the forces acting on trailer A in the example above.

Solution
  1. Step 1. Draw a dot to indicate the object :
    Figure 8
    Figure 8 (PG11C11_008.png)
  2. Step 2. Draw arrows to indicate all the forces acting on the object :
    Figure 9
    Figure 9 (PG11C11_009.png)
  3. Step 3. Label the forces :
    Figure 10
    Figure 10 (PG11C11_010.png)

Finding the Resultant Force

The easiest way to determine a resultant force is to draw a free body diagram. Remember from Chapter (Reference) that we use the length of the arrow to indicate the vector's magnitude and the direction of the arrow to show which direction it acts in.

After we have done this, we have a diagram of vectors and we simply find the sum of the vectors to get the resultant force.

Figure 11: (a) Force diagram of 2 forces acting on a box. (b) Free body diagram of the box.
Figure 11 (PG11C11_011.png)

For example, two people push on a box from opposite sides with forces of 4 N and 6 N respectively as shown in Figure 11(a). The free body diagram in Figure 11(b) shows the object represented by a dot and the two forces are represented by arrows with their tails on the dot.

As you can see, the arrows point in opposite directions and have different lengths. The resultant force is 2 N to the left. This result can be obtained algebraically too, since the two forces act along the same line. First, as in motion in one direction, choose a frame of reference. Secondly, add the two vectors taking their directions into account.

For the example, assume that the positive direction is to the right, then:

F R = ( + 4 N ) + ( - 6 N ) = - 2 N = 2 N to the left F R = ( + 4 N ) + ( - 6 N ) = - 2 N = 2 N to the left
(1)

Remember that a negative answer means that the force acts in the opposite direction to the one that you chose to be positive. You can choose the positive direction to be any way you want, but once you have chosen it you must keep it.

As you work with more force diagrams in which the forces exactly balance, you may notice that you get a zero answer (e.g. 0 N). This simply means that the forces are balanced and that the object will not accelerate.

Once a force diagram has been drawn the techniques of vector addition introduced in Chapter (Reference) can be used. Depending on the situation you might choose to use a graphical technique such as the tail-to-head method or the parallelogram method, or else an algebraic approach to determine the resultant. Since force is a vector quantity all of these methods apply.

Exercise 3: Finding the resultant force

A car (mass 1200 kg) applies a force of 2000 N on a trailer (mass 250 kg). A constant frictional force of 200 N is acting on the trailer, and a constant frictional force of 300 N is acting on the car.

  1. Draw a force diagram of all the forces acting on the car.
  2. Draw a free body diagram of all the horizontal forces acting on the trailer.
  3. Use the force diagram to determine the resultant force on the trailer.
Solution
  1. Step 1. Draw the force diagram for the car. :

    The question asks us to draw all the forces on the car. This means that we must include horizontal and vertical forces.

    Figure 12
    Figure 12 (PG11C11_012.png)
  2. Step 2. Draw the free body diagram for the trailer. :

    The question only asks for horizontal forces. We will therefore not include the force of the Earth on the trailer, or the force of the road on the trailer as these forces are in a vertical direction.

    Figure 13
    Figure 13 (PG11C11_013.png)
  3. Step 3. Determine the resultant force on the trailer. :

    To find the resultant force we need to add all the horizontal forces together. We do not add vertical forces as the movement of the car and trailer will be in a horizontal direction, and not up or down. FRR = 2000 + (-200) = 1800 N to the right.

run demo

Figure 14
Figure 14 (forces-and-motion-screenshot.png)

Exercise

  1. A force acts on an object. Name three effects that the force can have on the object.
  2. Identify each of the following forces as contact or non-contact forces.
    1. The force between the north pole of a magnet and a paper clip.
    2. The force required to open the door of a taxi.
    3. The force required to stop a soccer ball.
    4. The force causing a ball, dropped from a height of 2 m, to fall to the floor.
  3. A book of mass 2 kg is lying on a table. Draw a labeled force diagram indicating all the forces on the book.
  4. A boy pushes a shopping trolley (mass 15 kg) with a constant force of 75 N. A constant frictional force of 20 N is present.
    1. Draw a labeled force diagram to identify all the forces acting on the shopping trolley.
    2. Draw a free body diagram of all the horizontal forces acting on the trolley.
    3. Determine the resultant force on the trolley.
  5. A donkey (mass 250 kg) is trying to pull a cart (mass 80 kg) with a force of 400 N. The rope between the donkey and the cart makes an angle of 30 with the cart. The cart does not move.
    1. Draw a free body diagram of all the forces acting on the donkey.
    2. Draw a force diagram of all the forces acting on the cart.
    3. Find the magnitude and direction of the frictional force preventing the cart from moving.

Newton's Laws

In grade 10 you learned about motion, but did not look at how things start to move. You have also learned about forces. In this section we will look at the effect of forces on objects and how we can make things move.

Newton's First Law

Sir Isaac Newton was a scientist who lived in England (1642-1727) who was interested in the motion of objects under various conditions. He suggested that a stationary object will remain stationary unless a force acts on it and that a moving object will continue moving unless a force slows it down, speeds it up or changes its direction of motion. From this he formulated what is known as Newton's First Law of Motion:

Definition 1: Newton's First Law of Motion

An object will remain in a state of rest or continue traveling at constant velocity, unless acted upon by an unbalanced (net) force.

Let us consider the following situations:

An ice skater pushes herself away from the side of the ice rink and skates across the ice. She will continue to move in a straight line across the ice unless something stops her. Objects are also like that. If we kick a soccer ball across a soccer field, according to Newton's First Law, the soccer ball should keep on moving forever! However, in real life this does not happen. Is Newton's Law wrong? Not really. Newton's First Law applies to situations where there aren't any external forces present. This means that friction is not present. In the case of the ice skater, the friction between the skates and the ice is very little and she will continue moving for quite a distance. In the case of the soccer ball, air resistance (friction between the air and the ball) and friction between the grass and the ball is present and this will slow the ball down.

Figure 15
Figure 15 (PG11C11_014.png)

Figure 16
Figure 16 (PG11C11_015.png)

Figure 17
Khan academy video on newtons laws - 1

Newton's First Law in action

We experience Newton's First Law in every day life. Let us look at the following examples:

Rockets:

A spaceship is launched into space. The force of the exploding gases pushes the rocket through the air into space. Once it is in space, the engines are switched off and it will keep on moving at a constant velocity. If the astronauts want to change the direction of the spaceship they need to fire an engine. This will then apply a force on the rocket and it will change its direction.

Figure 18: Newton's First Law and rockets
Figure 18 (PG11C11_016.png)

Seat belts:

We wear seat belts in cars. This is to protect us when the car is involved in an accident. If a car is traveling at 120 km··hr-1-1, the passengers in the car is also traveling at 120 km··hr-1-1. When the car suddenly stops a force is exerted on the car (making it slow down), but not on the passengers. The passengers will carry on moving forward at 120 km··hr-1-1according to Newton I. If they are wearing seat belts, the seat belts will stop them by exerting a force on them and so prevent them from getting hurt.

Figure 19
Figure 19 (PG11C11_017.png)

Exercise 4: Newton's First Law in action

Why do passengers get thrown to the side when the car they are driving in goes around a corner?

Solution
  1. Step 1. What happens before the car turns :

    Before the car starts turning both the passengers and the car are traveling at the same velocity. (picture A)

  2. Step 2. What happens while the car turns :

    The driver turns the wheels of the car, which then exert a force on the car and the car turns. This force acts on the car but not the passengers, hence (by Newton's First Law) the passengers continue moving with the same original velocity. (picture B)

  3. Step 3. Why passengers get thrown to the side? :

    If the passengers are wearing seat belts they will exert a force on the passengers until the passengers' velocity is the same as that of the car (picture C). Without a seat belt the passenger may hit the side of the car.

    Figure 20
    Figure 20 (PG11C11_018.png)

Newton's Second Law of Motion

According to Newton I, things 'like to keep on doing what they are doing'. In other words, if an object is moving, it tends to continue moving (in a straight line and at the same speed) and if an object is stationary, it tends to remain stationary. So how do objects start moving?

Let us look at the example of a 10 kg box on a rough table. If we push lightly on the box as indicated in the diagram, the box won't move. Let's say we applied a force of 100 N, yet the box remains stationary. At this point a frictional force of 100 N is acting on the box, preventing the box from moving. If we increase the force, let's say to 150 N and the box almost starts to move, the frictional force is 150 N. To be able to move the box, we need to push hard enough to overcome the friction and then move the box. If we therefore apply a force of 200 N remembering that a frictional force of 150 N is present, the 'first' 150 N will be used to overcome or 'cancel' the friction and the other 50 N will be used to move (accelerate) the block. In order to accelerate an object we must have a resultant force acting on the block.

Figure 21
Figure 21 (PG11C11_019.png)

Now, what do you think will happen if we pushed harder, lets say 300 N? Or, what do you think will happen if the mass of the block was more, say 20 kg, or what if it was less? Let us investigate how the motion of an object is affected by mass and force.

Investigation : Newton's Second Law of Motion

Aim:

To investigate the relation between the acceleration of objects and the application of a constant resultant force.

Method:

Figure 22
Figure 22 (PG11C11_020.png)
  1. A constant force of 20 N, acting at an angle of 60 to the horizontal, is applied to a dynamics trolley.
  2. Ticker tape attached to the trolley runs through a ticker timer of frequency 20 Hz as the trolley is moving on the frictionless surface.
  3. The above procedure is repeated 4 times, each time using the same force, but varying the mass of the trolley as follows:
    • Case 1: 6,25 kg
    • Case 2: 3,57 kg
    • Case 3: 2,27 kg
    • Case 4: 1,67 kg
  4. Shown below are sections of the four ticker tapes obtained. The tapes are marked with the letters A, B, C, D, etc. A is the first dot, B is the second dot and so on. The distance between each dot is also shown.
Figure 23
Figure 23 (PG11C11_021.png)

Instructions:

  1. Use each tape to calculate the instantaneous velocity (in m··s-1-1) of the trolley at points B and F (remember to convert the distances to m first!). Use these velocities to calculate the trolley's acceleration in each case.
  2. Tabulate the mass and corresponding acceleration values as calculated in each case. Ensure that each column and row in your table is appropriately labeled.
  3. Draw a graph of acceleration vs. mass, using a scale of 1 cm = 1 m··s-2-2on the y-axis and 1 cm = 1 kg on the x-axis.
  4. Use your graph to read off the acceleration of the trolley if its mass is 5 kg.
  5. Write down a conclusion for the experiment.

You will have noted in the investigation above that the heavier the trolley is, the slower it moved. The acceleration is inversely proportional to the mass. In mathematical terms:

a 1 m a 1 m
(2)

In a similar investigation where the mass is kept constant, but the applied force is varied, you will find that the bigger the force is, the faster the object will move. The acceleration of the trolley is therefore directly proportional to the resultant force. In mathematical terms:

a F a F
(3)

Rearranging the above equations, we get aaFmFm OR F=maF=ma

Newton formulated his second law as follows:

Definition 2: Newton's Second Law of Motion

If a resultant force acts on a body, it will cause the body to accelerate in the direction of the resultant force. The acceleration of the body will be directly proportional to the resultant force and inversely proportional to the mass of the body. The mathematical representation is:

F = m a . F = m a .
(4)

Figure 24
Khan academy video on newtons laws - 2

Applying Newton's Second Law

Newton's Second Law can be applied to a variety of situations. We will look at the main types of examples that you need to study.

Exercise 5: Newton II - Box on a surface 1

A 10 kg box is placed on a table. A horizontal force of 32 N is applied to the box. A frictional force of 7 N is present between the surface and the box.

  1. Draw a force diagram indicating all the horizontal forces acting on the box.
  2. Calculate the acceleration of the box.
Figure 25
Figure 25 (PG11C11_022.png)
Solution
  1. Step 1. Identify the horizontal forces and draw a force diagram :

    We only look at the forces acting in a horizontal direction (left-right) and not vertical (up-down) forces. The applied force and the force of friction will be included. The force of gravity, which is a vertical force, will not be included.

    Figure 26
    Figure 26 (PG11C11_023.png)
  2. Step 2. Calculate the acceleration of the box :

    We have been given:

    Applied force F11 = 32 N

    Frictional force Fff = - 7 N

    Mass m = 10 kg

    To calculate the acceleration of the box we will be using the equation FR=maFR=ma. Therefore:

    F R = m a F 1 + F f = ( 10 ) ( a ) 32 - 7 = 10 a 25 = 10 a a = 2 , 5 m · s - 1 towards the left F R = m a F 1 + F f = ( 10 ) ( a ) 32 - 7 = 10 a 25 = 10 a a = 2 , 5 m · s - 1 towards the left
    (5)
Exercise 6: Newton II - box on surface 2

Two crates, 10 kg and 15 kg respectively, are connected with a thick rope according to the diagram. A force of 500 N is applied. The boxes move with an acceleration of 2 m··s-2-2. One third of the total frictional force is acting on the 10 kg block and two thirds on the 15 kg block. Calculate:

  1. the magnitude and direction of the frictional force present.
  2. the magnitude of the tension in the rope at T.
Figure 27: Two crates on a surface
Figure 27 (PG11C11_024.png)
Solution
  1. Step 1. Draw a force diagram :

    Always draw a force diagram although the question might not ask for it. The acceleration of the whole system is given, therefore a force diagram of the whole system will be drawn. Because the two crates are seen as a unit, the force diagram will look like this:

    Figure 28: Force diagram for two crates on a surface
    Figure 28 (PG11C11_025.png)
  2. Step 2. Calculate the frictional force :

    To find the frictional force we will apply Newton's Second Law. We are given the mass (10 + 15 kg) and the acceleration (2 m··s-2-2). Choose the direction of motion to be the positive direction (to the right is positive).

    F R = m a F applied + F f = m a 500 + F f = ( 10 + 15 ) ( 2 ) F f = 50 - 500 F f = - 450 N F R = m a F applied + F f = m a 500 + F f = ( 10 + 15 ) ( 2 ) F f = 50 - 500 F f = - 450 N
    (6)

    The frictional force is 450 N opposite to the direction of motion (to the left).

  3. Step 3. Find the tension in the rope :

    To find the tension in the rope we need to look at one of the two crates on their own. Let's choose the 10 kg crate. Firstly, we need to draw a force diagram:

    Figure 29: Force diagram of 10 kg crate
    Figure 29 (PG11C11_026.png)

    The frictional force on the 10 kg block is one third of the total, therefore:

    Ff=13×Ff=13× 450

    F f = 150 N F f = 150 N

    If we apply Newton's Second Law:

    F R = m a T + F f = ( 10 ) ( 2 ) T + ( - 150 ) = 20 T = 170 N F R = m a T + F f = ( 10 ) ( 2 ) T + ( - 150 ) = 20 T = 170 N
    (7)

    Note: If we had used the same principle and applied it to 15 kg crate, our calculations would have been the following:

    F R = m a F applied + T + F f = ( 15 ) ( 2 ) 500 + T + ( - 300 ) = 30 T = - 170 N F R = m a F applied + T + F f = ( 15 ) ( 2 ) 500 + T + ( - 300 ) = 30 T = - 170 N
    (8)

    The negative answer here means that the force is in the direction opposite to the motion, in other words to the left, which is correct. However, the question asks for the magnitude of the force and your answer will be quoted as 170 N.

Exercise 7: Newton II - Man pulling a box

A man is pulling a 20 kg box with a rope that makes an angle of 60 with the horizontal. If he applies a force of 150 N and a frictional force of 15 N is present, calculate the acceleration of the box.

Figure 30: Man pulling a box
Figure 30 (PG11C11_027.png)
Solution
  1. Step 1. Draw a force diagram :

    The motion is horizontal and therefore we will only consider the forces in a horizontal direction. Remember that vertical forces do not influence horizontal motion and vice versa.

    Figure 31: Force diagram
    Figure 31 (PG11C11_028.png)
  2. Step 2. Calculate the horizontal component of the applied force :

    The applied force is acting at an angle of 60 to the horizontal. We can only consider forces that are parallel to the motion. The horizontal component of the applied force needs to be calculated before we can continue:

    F x = 150 cos 60 F x = 75 N F x = 150 cos 60 F x = 75 N
    (9)
  3. Step 3. Calculate the acceleration :

    To find the acceleration we apply Newton's Second Law:

    F R = m a F x + F f = ( 20 ) ( a ) 75 + ( - 15 ) = 20 a a = 3 m · s - 2 to the right F R = m a F x + F f = ( 20 ) ( a ) 75 + ( - 15 ) = 20 a a = 3 m · s - 2 to the right
    (10)
Exercise 8: Newton II - Truck and trailor

A 2000 kg truck pulls a 500 kg trailer with a constant acceleration. The engine of the truck produces a thrust of 10 000 N. Ignore the effect of friction.

  1. Calculate the acceleration of the truck.
  2. Calculate the tension in the tow bar T between the truck and the trailer, if the tow bar makes an angle of 25 with the horizontal.
Figure 32: Truck pulling a trailer
Figure 32 (PG11C11_029.png)
Solution
  1. Step 1. Draw a force diagram :

    Draw a force diagram indicating all the horizontal forces on the system as a whole:

    Figure 33: Force diagram for truck pulling a trailer
    Figure 33 (PG11C11_030.png)
  2. Step 2. Find the acceleration of the system :

    In the absence of friction, the only force that causes the system to accelerate is the thrust of the engine. If we now apply Newton's Second Law:

    F R = m a 10 000 = ( 500 + 2000 ) a a = 4 m · s - 2 to the right F R = m a 10 000 = ( 500 + 2000 ) a a = 4 m · s - 2 to the right
    (11)
  3. Step 3. Find the horizontal component of T :

    We are asked to find the tension in the tow bar, but because the tow bar is acting at an angle, we need to find the horizontal component first. We will find the horizontal component in terms of T and then use it in the next step to find T.

    Figure 34
    Figure 34 (PG11C11_031.png)

    The horizontal component is T cos 25.

  4. Step 4. Find the tension in the tow bar :

    To find T, we will apply Newton's Second Law:

    F R = m a F - T cos 25 = m a 10 000 - T cos 25 = ( 2000 ) ( 4 ) T cos 25 = 2000 T = 2206 , 76 N F R = m a F - T cos 25 = m a 10 000 - T cos 25 = ( 2000 ) ( 4 ) T cos 25 = 2000 T = 2206 , 76 N
    (12)

Object on an inclined plane

When we place an object on a slope the force of gravity (Fgg) acts straight down and not perpendicular to the slope. Due to gravity pulling straight down, the object will tend to slide down the slope with a force equal to the horizontal component of the force of gravity (Fgg sin θθ). The object will 'stick' to the slope due to the frictional force between the object and the surface. As you increase the angle of the slope, the horizontal component will also increase until the frictional force is overcome and the object starts to slide down the slope.

The force of gravity will also tend to push an object 'into' the slope. The vertical component of this force is equal to the vertical component of the force of gravity (Fgg cos θθ). There is no movement in this direction as this force is balanced by the slope pushing up against the object. This “pushing force” is called the normal force (N) and is equal to the force required to make the component of the resultant force perpendicularly into the plane zero, Fgg cos θθ in this case, but opposite in direction.

Tip:
Do not use the abbreviation W for weight as it is used to abbreviate 'work'. Rather use the force of gravity Fgg for weight.
Figure 35
Figure 35 (PG11C11_032.png)

Figure 36
Figure 36 (ramp-forces-and-motion-screenshot.png)
run demo

Exercise 9: Newton II - Box on inclined plane

A body of mass M is at rest on an inclined plane.

Figure 37
Figure 37 (PG11C11_033.png)

What is the magnitude of the frictional force acting on the body?

  1. Mg
  2. Mg cos θθ
  3. Mg sin θθ
  4. Mg tan θθ
Solution
  1. Step 1. Analyse the situation :

    The question asks us to identify the frictional force. The body is said to be at rest on the plane, which means that it is not moving and therefore there is no resultant force. The frictional force must therefore be balanced by the force F up the inclined plane.

  2. Step 2. Choose the correct answer :

    The frictional force is equal to the component of the weight (Mg) parallel to the surface, which is equal to Mg sin θθ.

Exercise 10: Newton II - Object on a slope

A force T = 312 N is required to keep a body at rest on a frictionless inclined plane which makes an angle of 35 with the horizontal. The forces acting on the body are shown. Calculate the magnitudes of forces P and R, giving your answers to three significant figures.

Figure 38
Figure 38 (PG11C11_034.png)
Solution
  1. Step 1. Find the magnitude of P :

    We are usually asked to find the magnitude of T, but in this case T is given and we are asked to find P. We can use the same equation. T is the force that balances the component of P parallel to the plane (Pxx) and therefore it has the same magnitude.

    T = P sin θ 312 = P sin 35 P = 544 N T = P sin θ 312 = P sin 35 P = 544 N
    (13)
  2. Step 2. Find the magnitude of R :

    R can also be determined with the use of trigonometric ratios. The tan or cos ratio can be used. We recommend that you use the tan ratio because it does not involve using the value for P (for in case you made a mistake in calculating P).

    tan 55 = R T tan 55 = R 312 R = tan 55 × 312 R = 445 , 6 N R = 446 N tan 55 = R T tan 55 = R 312 R = tan 55 × 312 R = 445 , 6 N R = 446 N
    (14)

    Note that the question asks that the answers be given to 3 significant figures. We therefore round 445,6 N up to 446 N.

Lifts and rockets

So far we have looked at objects being pulled or pushed across a surface, in other words motion parallel to the surface the object rests on. Here we only considered forces parallel to the surface, but we can also lift objects up or let them fall. This is vertical motion where only vertical forces are being considered.

Let us consider a 500 kg lift, with no passengers, hanging on a cable. The purpose of the cable is to pull the lift upwards so that it can reach the next floor or to let go a little so that it can move downwards to the floor below. We will look at five possible stages during the motion of the lift.

Stage 1:

The 500 kg lift is stationary at the second floor of a tall building.

Because the lift is stationary (not moving) there is no resultant force acting on the lift. This means that the upward forces must be balanced by the downward forces. The only force acting down is the force of gravity which is equal to (500 x 9,8 = 4900 N) in this case. The cable must therefore pull upwards with a force of 4900 N to keep the lift stationary at this point.

Stage 2:

The lift moves upwards at an acceleration of 1 m··s-2-2.

If the lift is accelerating, it means that there is a resultant force in the direction of the motion. This means that the force acting upwards is now greater than the force of gravity Fgg (down). To find the magnitude of the force applied by the cable (Fcc) we can do the following calculation: (Remember to choose a direction as positive. We have chosen upwards as positive.)

F R = m a F c + F g = m a F c + ( - 4900 ) = ( 500 ) ( 1 ) F c = 5400 N upwards F R = m a F c + F g = m a F c + ( - 4900 ) = ( 500 ) ( 1 ) F c = 5400 N upwards
(15)

The answer makes sense as we need a bigger force upwards to cancel the effect of gravity as well as make the lift go faster.

Stage 3:

The lift moves at a constant velocity.

When the lift moves at a constant velocity, it means that all the forces are balanced and that there is no resultant force. The acceleration is zero, therefore FRR = 0. The force acting upwards is equal to the force acting downwards, therefore Fcc = 4900 N.

Stage 4:

The lift slows down at a rate of 2m··s-2-2.

As the lift is now slowing down there is a resultant force downwards. This means that the force acting downwards is greater than the force acting upwards. To find the magnitude of the force applied by the cable (Fcc) we can do the following calculation: Again we have chosen upwards as positive, which means that the acceleration will be a negative number.

F R = m a F c + F g = m a F c + ( - 4900 ) = ( 500 ) ( - 2 ) F c = 3900 N upwards F R = m a F c + F g = m a F c + ( - 4900 ) = ( 500 ) ( - 2 ) F c = 3900 N upwards
(16)

This makes sense as we need a smaller force upwards to ensure that the resultant force is downward. The force of gravity is now greater than the upward pull of the cable and the lift will slow down.

Stage 5:

The cable snaps.

When the cable snaps, the force that used to be acting upwards is no longer present. The only force that is present would be the force of gravity. The lift will freefall and its acceleration can be calculated as follows:

F R = m a F c + F g = m a 0 + ( - 4900 ) = ( 500 ) ( a ) a = - 9 , 8 m · s - 2 a = 9 , 8 m · s - 2 downwards F R = m a F c + F g = m a 0 + ( - 4900 ) = ( 500 ) ( a ) a = - 9 , 8 m · s - 2 a = 9 , 8 m · s - 2 downwards
(17)

Rockets

As with lifts, rockets are also examples of objects in vertical motion. The force of gravity pulls the rocket down while the thrust of the engine pushes the rocket upwards. The force that the engine exerts must overcome the force of gravity so that the rocket can accelerate upwards. The worked example below looks at the application of Newton's Second Law in launching a rocket.

Exercise 11: Newton II - rocket

A rocket is launched vertically upwards into the sky at an acceleration of 20 m··s-2-2. If the mass of the rocket is 5000 kg, calculate the magnitude and direction of the thrust of the rocket's engines.

Solution
  1. Step 1. Analyse what is given and what is asked :

    We have the following:

    m = 5000 kg

    a = 20 m··s-2-2

    Fgg = 5000 x 9,8 = 49000 N

    We are asked to find the thrust of the rocket engine F11.

  2. Step 2. Find the thrust of the engine :

    We will apply Newton's Second Law:

    F R = m a F 1 + F g = m a F 1 + ( - 49000 ) = ( 5000 ) ( 20 ) F 1 = 149 000 N upwards F R = m a F 1 + F g = m a F 1 + ( - 49000 ) = ( 5000 ) ( 20 ) F 1 = 149 000 N upwards
    (18)
Exercise 12: Rockets

How do rockets accelerate in space?

Figure 39
Figure 39 (PG11C11_035.png)

Solution
  1. Step 1. What happens to the rocket fuel? :

    Gas explodes inside the rocket.

  2. Step 2. What effect does this have? :

    This exploding gas exerts a force on each side of the rocket (as shown in the picture below of the explosion chamber inside the rocket).

    Figure 40
    Figure 40 (PG11C11_036.png)

  3. Step 3. What are the forces? :

    Due to the symmetry of the situation, all the forces exerted on the rocket are balanced by forces on the opposite side, except for the force opposite the open side. This force on the upper surface is unbalanced.

  4. Step 4. Think about the resultant force :

    This is therefore the resultant force acting on the rocket and it makes the rocket accelerate forwards.

Exercise 13: Newton II - lifts

A lift, mass 250 kg, is initially at rest on the ground floor of a tall building. Passengers with an unknown total mass, m, climb into the lift. The lift accelerates upwards at 1,6 m··s-2-2. The cable supporting the lift exerts a constant upward force of 7700 N. Use g = 10 m··s-2-2.

  1. Draw a labeled force diagram indicating all the forces acting on the lift while it accelerates upwards.
  2. What is the maximum mass, m, of the passengers the lift can carry in order to achieve a constant upward acceleration of 1,6 m··s-2-2.
Solution
  1. Step 1. Draw a force diagram. :
    Figure 41
    Figure 41 (PG11C11_037.png)
  2. Step 2. Find the mass, m. :

    Let us look at the lift with its passengers as a unit. The mass of this unit will be (250 + m) kg and the force of the Earth pulling downwards (Fgg) will be (250 + m) x 10 m.s-2-2. If we apply Newton's Second Law to the situation we get:

    F n e t = m a F C - F g = m a 7700 - ( 250 + m ) ( 10 ) = ( 250 + m ) ( 1 , 6 ) 7700 - 2500 - 10 m = 400 + 1 , 6 m 4800 = 11 , 6 m m = 413 , 79 kg F n e t = m a F C - F g = m a 7700 - ( 250 + m ) ( 10 ) = ( 250 + m ) ( 1 , 6 ) 7700 - 2500 - 10 m = 400 + 1 , 6 m 4800 = 11 , 6 m m = 413 , 79 kg
    (19)

Exercise

  1. A tug is capable of pulling a ship with a force of 100 kN. If two such tugs are pulling on one ship, they can produce any force ranging from a minimum of 0 N to a maximum of 200 kN. Give a detailed explanation of how this is possible. Use diagrams to support your result.
  2. A car of mass 850 kg accelerates at 2 m··s-2-2. Calculate the magnitude of the resultant force that is causing the acceleration.
  3. Find the force needed to accelerate a 3 kg object at 4 m··s-2-2.
  4. Calculate the acceleration of an object of mass 1000 kg accelerated by a force of 100 N.
  5. An object of mass 7 kg is accelerating at 2,5 m··s-2-2. What resultant force acts on it?
  6. Find the mass of an object if a force of 40 N gives it an acceleration of 2 m··s-2-2.
  7. Find the acceleration of a body of mass 1 000 kg that has a 150 N force acting on it.
  8. Find the mass of an object which is accelerated at 2 m··s-2-2 by a force of 40 N.
  9. Determine the acceleration of a mass of 24 kg when a force of 6 N acts on it. What is the acceleration if the force were doubled and the mass was halved?
  10. A mass of 8 kg is accelerating at 5 m··s-2-2.
    1. Determine the resultant force that is causing the acceleration.
    2. What acceleration would be produced if we doubled the force and reduced the mass by half?
  11. A motorcycle of mass 100 kg is accelerated by a resultant force of 500  N. If the motorcycle starts from rest:
    1. What is its acceleration?
    2. How fast will it be travelling after 20 s?
    3. How long will it take to reach a speed of 35 m··s-1-1?
    4. How far will it travel from its starting point in 15 s?
  12. A force acting on a trolley on a frictionless horizontal plane causes an acceleration of magnitude 6 m··s-2-2. Determine the mass of the trolley.
  13. A force of 200 N, acting at 60 to the horizontal, accelerates a block of mass 50 kg along a horizontal plane as shown.
    Figure 42
    Figure 42 (PG11C11_038.png)
    1. Calculate the component of the 200 N force that accelerates the block horizontally.
    2. If the acceleration of the block is 1,5 m··s-2-2, calculate the magnitude of the frictional force on the block.
    3. Calculate the vertical force exerted by the block on the plane.
  14. A toy rocket of mass 0,5 kg is supported vertically by placing it in a bottle. The rocket is then ignited. Calculate the force that is required to accelerate the rocket vertically upwards at 8 m··s-2-2.
  15. A constant force of 70 N is applied vertically to a block of mass 5 kg as shown. Calculate the acceleration of the block.
    Figure 43
    Figure 43 (PG11C11_039.png)
  16. A stationary block of mass 3kg is on top of a plane inclined at 35 to the horizontal.
    Figure 44
    Figure 44 (PG11C11_040.png)
    1. Draw a force diagram (not to scale). Include the weight (Fgg) of the block as well as the components of the weight that are perpendicular and parallel to the inclined plane.
    2. Determine the values of the weight's perpendicular and parallel components (Fgxgx and Fgygy).
    3. Determine the magnitude and direction of the frictional force between the block and plane.
  17. A student of mass 70 kg investigates the motion of a lift. While he stands in the lift on a bathroom scale (calibrated in newton), he notes three stages of his journey.
    1. For 2 s immediately after the lift starts, the scale reads 574 N.
    2. For a further 6 s it reads 700 N.
    3. For the final 2 s it reads 854 N.
    Answer the following questions:
    1. Is the motion of the lift upward or downward? Give a reason for your answer.
    2. Write down the magnitude and the direction of the resultant force acting on the student for each of the stages I, II and III.
    3. Calculate the magnitude of the acceleration of the lift during the first 2s.
  18. A car of mass 800 kg accelerates along a level road at 4 m··s-2-2. A frictional force of 700 N opposes its motion. What force is produced by the car's engine?
  19. Two objects, with masses of 1 kg and 2 kg respectively, are placed on a smooth surface and connected with a piece of string. A horizontal force of 6 N is applied with the help of a spring balance to the 1 kg object. Ignoring friction, what will the force acting on the 2 kg mass, as measured by a second spring balance, be?
    Figure 45
    Figure 45 (PG11C11_041.png)
  20. A rocket of mass 200 kg has a resultant force of 4000 N upwards on it.
    1. What is its acceleration in space, where it has no weight?
    2. What is its acceleration on the Earth, where it has weight?
    3. What driving force does the rocket engine need to exert on the back of the rocket in space?
    4. What driving force does the rocket engine need to exert on the back of the rocket on the Earth?
  21. A car going at 20 m··s-1-1accelerates uniformly and comes to a stop in a distance of 20 m.
    1. What is its acceleration?
    2. If the car is 1000 kg how much force do the brakes exert?

Newton's Third Law of Motion

Newton's Third Law of Motion deals with the interaction between pairs of objects. For example, if you hold a book up against a wall you are exerting a force on the book (to keep it there) and the book is exerting a force back at you (to keep you from falling through the book). This may sound strange, but if the book was not pushing back at you, your hand would push through the book! These two forces (the force of the hand on the book (F11) and the force of the book on the hand (F22)) are called an action-reaction pair of forces. They have the same magnitude, but act in opposite directions and act on different objects (the one force is onto the book and the other is onto your hand).

There is another action-reaction pair of forces present in this situation. The book is pushing against the wall (action force) and the wall is pushing back at the book (reaction). The force of the book on the wall (F33) and the force of the wall on the book (F44) are shown in the diagram.

Figure 46: Newton's action-reaction pairs
Figure 46 (PG11C11_042.png)
Definition 3: Newton's Third Law of Motion

If body A exerts a force on body B, then body B exerts a force of equal magnitude on body A, but in the opposite direction.

Figure 47
Khan academy video on newtons laws - 3

Newton's action-reaction pairs can be found everywhere in life where two objects interact with one another. The following worked examples will illustrate this:

Exercise 14: Newton III - seat belt

Dineo is seated in the passenger seat of a car with the seat belt on. The car suddenly stops and he moves forwards until the seat belt stops him. Draw a labeled force diagram identifying two action-reaction pairs in this situation.

Figure 48
Figure 48 (PG11C11_043.png)
Solution
  1. Step 1. Draw a force diagram :

    Start by drawing the picture. You will be using arrows to indicate the forces so make your picture large enough so that detailed labels can also be added. The picture needs to be accurate, but not artistic! Use stickmen if you have to.

  2. Step 2. Label the diagram :

    Take one pair at a time and label them carefully. If there is not enough space on the drawing, then use a key on the side.

    Figure 49
    Figure 49 (PG11C11_044.png)

Exercise 15: Newton III - forces in a lift

Tammy travels from the ground floor to the fifth floor of a hotel in a lift. Which ONE of the following statements is TRUE about the force exerted by the floor of the lift on Tammy's feet?

  1. It is greater than the magnitude of Tammy's weight.
  2. It is equal in magnitude to the force Tammy's feet exert on the floor.
  3. It is equal to what it would be in a stationary lift.
  4. It is greater than what it would be in a stationary lift.
Solution
  1. Step 1. Analyse the situation :

    This is a Newton's Third Law question and not Newton II. We need to focus on the action-reaction pairs of forces and not the motion of the lift. The following diagram will show the action-reaction pairs that are present when a person is standing on a scale in a lift.

    Figure 50: Newton's action-reaction pairs in a lift
    Figure 50 (PG11C11_045.png)

    In this question statements are made about the force of the floor (lift) on Tammy's feet. This force corresponds to F22 in our diagram. The reaction force that pairs up with this one is F11, which is the force that Tammy's feet exerts on the floor of the lift. The magnitude of these two forces are the same, but they act in opposite directions.

  2. Step 2. Choose the correct answer :

    It is important to analyse the question first, before looking at the answers as the answers might confuse you. Make sure that you understand the situation and know what is asked before you look at the options.

    The correct answer is B.

Exercise 16: Newton III - book and wall

Tumi presses a book against a vertical wall as shown in the sketch.

  1. Draw a labelled force diagram indicating all the forces acting on the book.
  2. State, in words, Newton's Third Law of Motion.
  3. Name the action-reaction pairs of forces acting in the horizontal plane.
Figure 51
Figure 51 (PG11C11_046.png)
Solution
  1. Step 1. Draw a force diagram :

    A force diagram will look like this:

    Figure 52
    Figure 52 (PG11C11_047.png)

    Note that we had to draw all the forces acting on the book and not the action-reaction pairs. None of the forces drawn are action-reaction pairs, because they all act on the same object (the book). When you label forces, be as specific as possible, including the direction of the force and both objects involved, for example, do not say gravity (which is an incomplete answer) but rather say 'Downward (direction) gravitational force of the Earth (object) on the book (object)'.

  2. Step 2. State Newton's Third Law :

    If body A exerts a force onto body B, then body B will exert a force equal in magnitude, but opposite in direction, onto body A.

  3. Step 3. Name the action-reaction pairs :

    The question only asks for action-reaction forces in the horizontal plane. Therefore:

    Pair 1: Action: Applied force of the girl on the book; Reaction: The force of the book on the girl.

    Pair 2: Action: Force of the book on the wall; Reaction: Force of the wall on the book.

    Note that a Newton III pair will always involve the same combination of words, like 'book on wall' and 'wall on book'. The objects are 'swapped around' in naming the pairs.

Experiment : Balloon Rocket

Aim:

In this experiment for the entire class, you will use a balloon rocket to investigate Newton's Third Law. A fishing line will be used as a track and a plastic straw taped to the balloon will help attach the balloon to the track.

Apparatus:

You will need the following items for this experiment:

  1. balloons (one for each team)
  2. plastic straws (one for each team)
  3. tape (cellophane or masking)
  4. fishing line, 10 meters in length
  5. a stopwatch - optional (a cell phone can also be used)
  6. a measuring tape - optional

Method:

  1. Divide into groups of at least five.
  2. Attach one end of the fishing line to the blackboard with tape. Have one teammate hold the other end of the fishing line so that it is taut and roughly horizontal. The line must be held steady and must not be moved up or down during the experiment.
  3. Have one teammate blow up a balloon and hold it shut with his or her fingers. Have another teammate tape the straw along the side of the balloon. Thread the fishing line through the straw and hold the balloon at the far end of the line.
  4. Let go of the rocket and observe how the rocket moves forward.
  5. Optionally, the rockets of each group can be timed to determine a winner of the fastest rocket.
    1. Assign one teammate to time the event. The balloon should be let go when the time keeper yells “Go!” Observe how your rocket moves toward the blackboard.
    2. Have another teammate stand right next to the blackboard and yell “Stop!” when the rocket hits its target. If the balloon does not make it all the way to the blackboard, “Stop!” should be called when the balloon stops moving. The timekeeper should record the flight time.
    3. Measure the exact distance the rocket traveled. Calculate the average speed at which the balloon traveled. To do this, divide the distance traveled by the time the balloon was “in flight.” Fill in your results for Trial 1 in the Table below.
    4. Each team should conduct two more races and complete the sections in the Table for Trials 2 and 3. Then calculate the average speed for the three trials to determine your team's race entry time.
    Results:
    Table 1
     Distance (m)Time (s)Speed (m··s-1-1)
    Trial 1   
    Trial 2   
    Trial 3   
      Average: 
    Conclusions: The winner of this race is the team with the fastest average balloon speed.

While doing the experiment, you should think about,

  1. What made your rocket move?
  2. How is Newton's Third Law of Motion demonstrated by this activity?
  3. Draw pictures using labeled arrows to show the forces acting on the inside of the balloon before it was released and after it was released.

Exercise

  1. A fly hits the front windscreen of a moving car. Compared to the magnitude of the force the fly exerts on the windscreen, the magnitude of the force the windscreen exerts on the fly during the collision, is ...
    1. zero.
    2. smaller, but not zero.
    3. bigger.
    4. the same.
    Figure 53
    Figure 53 (PG11C11_048.png)
    Figure 54
    Figure 54 (PG11C11_049.png)
  2. Which of the following pairs of forces correctly illustrates Newton's Third Law?
    Figure 55
    Figure 55 (PG11C11_050.png)

Different types of forces

Tension

Tension is the magnitude of the force that exists in objects like ropes, chains and struts that are providing support. For example, there are tension forces in the ropes supporting a child's swing hanging from a tree.

Contact and non-contact forces

In this chapter we have come across a number of different types of forces, for example a push or a pull, tension in a string, frictional forces and the normal force. These are all examples of contact forces where there is a physical point of contact between applying the force and the object. Non-contact forces are forces that act over a distance, for example magnetic forces, electrostatic forces and gravitational forces.

When an object is placed on a surface, two types of surface forces can be identified. Friction is a force that acts between the surface and the object and is parallel to the surface. The normal force is a force that acts between the object and the surface and is perpendicular to the surface.

The normal force

A 5 kg box is placed on a rough surface and a 10 N force is applied at an angle of 36,9 to the horizontal. The box does not move. The normal force (N or FNN) is the force between the box and the surface acting in the vertical direction. If this force is not present the box would fall through the surface because the force of gravity pulls it downwards. The normal force therefore acts upwards. We can calculate the normal force by considering all the forces in the vertical direction. All the forces in the vertical direction must add up to zero because there is no movement in the vertical direction.

N + F y + F g = 0 N + 6 + ( - 49 ) = 0 N = 43 N upwards N + F y + F g = 0 N + 6 + ( - 49 ) = 0 N = 43 N upwards
(20)
Figure 56: Friction and the normal force
Figure 56 (PG11C11_051.png)

The most interesting and illustrative normal force question, that is often asked, has to do with a scale in a lift. Using Newton's third law we can solve these problems quite easily.

When you stand on a scale to measure your weight you are pulled down by gravity. There is no acceleration downwards because there is a reaction force we call the normal force acting upwards on you. This is the force that the scale would measure. If the gravitational force were less then the reading on the scale would be less.

Exercise 17: Normal Forces 1

A man with a mass of 100 kg stands on a scale (measuring newtons). What is the reading on the scale?

Solution
  1. Step 1. Identify what information is given and what is asked for :

    We are given the mass of the man. We know the gravitational acceleration that acts on him is 9,8 = m··s-2-2.

  2. Step 2. Decide what equation to use to solve the problem :

    The scale measures the normal force on the man. This is the force that balances gravity. We can use Newton's laws to solve the problem:

    F r = F g + F N F r = F g + F N
    (21)

    where FrFr is the resultant force on the man.

  3. Step 3. Firstly we determine the force on the man due to gravity :
    F g = m g = ( 100 kg ) ( 9 , 8 m · s - 2 ) = 980 kg · m · s - 2 = 980 N downwards F g = m g = ( 100 kg ) ( 9 , 8 m · s - 2 ) = 980 kg · m · s - 2 = 980 N downwards
    (22)
  4. Step 4. Now determine the normal force acting upwards on the man :

    We now know the gravitational force downwards. We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is 0 - so Fr=0Fr=0.

    F r = F g + F N 0 = - 980 N + F N F N = 980 N upwards F r = F g + F N 0 = - 980 N + F N F N = 980 N upwards
    (23)
  5. Step 5. Quote the final answer :

    The normal force is 980 N upwards. It exactly balances the gravitational force downwards so there is no net force and no acceleration on the man. The reading on the scale is 980 N.

Now we are going to add things to exactly the same problem to show how things change slightly. We will now move to a lift moving at constant velocity. Remember if velocity is constant then acceleration is zero.

Exercise 18: Normal Forces 2

A man with a mass of 100 kg stands on a scale (measuring newtons) inside a lift that is moving downwards at a constant velocity of 2 m··s-1-1. What is the reading on the scale?

Solution
  1. Step 1. Identify what information is given and what is asked for :

    We are given the mass of the man and the acceleration of the lift. We know the gravitational acceleration that acts on him.

  2. Step 2. Decide which equation to use to solve the problem :

    Once again we can use Newton's laws. We know that the sum of all the forces must equal the resultant force, FrFr.

    F r = F g + F N F r = F g + F N
    (24)
  3. Step 3. Determine the force due to gravity :
    F g = m g = ( 100 kg ) ( 9 , 8 m · s - 2 ) = 980 kg · m · s - 2 = 980 N downwards F g = m g = ( 100 kg ) ( 9 , 8 m · s - 2 ) = 980 kg · m · s - 2 = 980 N downwards
    (25)
  4. Step 4. Now determine the normal force acting upwards on the man :

    The scale measures this normal force, so once we have determined it we will know the reading on the scale. Because the lift is moving at constant velocity the overall resultant acceleration of the man on the scale is 0. If we write out the equation:

    F r = F g + F N m a = F g + F N ( 100 ) ( 0 ) = - 980 N + F N F N = 980 N upwards F r = F g + F N m a = F g + F N ( 100 ) ( 0 ) = - 980 N + F N F N = 980 N upwards
    (26)
  5. Step 5. Quote the final answer :

    The normal force is 980 N upwards. It exactly balances the gravitational force downwards so there is no net force and no acceleration on the man. The reading on the scale is 980 N.

In the previous two examples we got exactly the same result because the net acceleration on the man was zero! If the lift is accelerating downwards things are slightly different and now we will get a more interesting answer!

Exercise 19: Normal Forces 3

A man with a mass of 100 kg stands on a scale (measuring newtons) inside a lift that is accelerating downwards at 2 m··s-2-2. What is the reading on the scale?

Solution
  1. Step 1. Identify what information is given and what is asked for :

    We are given the mass of the man and his resultant acceleration - this is just the acceleration of the lift. We know the gravitational acceleration also acts on him.

  2. Step 2. Decide which equation to use to solve the problem :

    Once again we can use Newton's laws. We know that the sum of all the forces must equal the resultant force, FrFr.

    F r = F g + F N F r = F g + F N
    (27)
  3. Step 3. Determine the force due to gravity, FgFg :
    F g = m g = ( 100 kg ) ( 9 , 8 m · s - 2 ) = 980 kg · m · s - 2 = 980 N downwards F g = m g = ( 100 kg ) ( 9 , 8 m · s - 2 ) = 980 kg · m · s - 2 = 980 N downwards
    (28)
  4. Step 4. Determine the resultant force, FrFr :

    The resultant force can be calculated by applying Newton's Second Law:

    F r = m a F r = ( 100 kg ) ( - 2 m · s - 2 ) = - 200 N = 200 N down F r = m a F r = ( 100 kg ) ( - 2 m · s - 2 ) = - 200 N = 200 N down
    (29)
  5. Step 5. Determine the normal force, FNFN :

    The sum of all the vertical forces is equal to the resultant force, therefore

    F r = F g + F N - 200 N = - 980 N + F N F N = 780 N upwards F r = F g + F N - 200 N = - 980 N + F N F N = 780 N upwards
    (30)
  6. Step 6. Quote the final answer :

    The normal force is 780 N upwards. It balances the gravitational force downwards just enough so that the man only accelerates downwards at 2 m··s-2-2. The reading on the scale is 780 N.

Exercise 20: Normal Forces 4

A man with a mass of 100 kg stands on a scale (measuring newtons) inside a lift that is accelerating upwards at 4 m··s-2-2. What is the reading on the scale?

Solution
  1. Step 1. Identify what information is given and what is asked for :

    We are given the mass of the man and his resultant acceleration - this is just the acceleration of the lift. We know the gravitational acceleration also acts on him.

  2. Step 2. Decide which equation to use to solve the problem :

    Once again we can use Newton's laws. We know that the sum of all the forces must equal the resultant force, FrFr.

    F r = F g + F N F r = F g + F N
    (31)
  3. Step 3. Determine the force due to gravity, FgFg :
    F g = m g = ( 100 kg ) ( 9 , 8 m · s - 2 ) = 980 kg · m · s - 2 = 980 N downwards F g = m g = ( 100 kg ) ( 9 , 8 m · s - 2 ) = 980 kg · m · s - 2 = 980 N downwards
    (32)
  4. Step 4. Determine the resultant force, FrFr :

    The resultant force can be calculated by applying Newton's Second Law:

    F r = m a F r = ( 100 kg ) ( 4 m · s - 2 ) = 400 N upwards F r = m a F r = ( 100 kg ) ( 4 m · s - 2 ) = 400 N upwards
    (33)
  5. Step 5. Determine the normal force, FNFN :

    The sum of all the vertical forces is equal to the resultant force, therefore

    F r = F g + F N 400 N = - 980 N + F N F N = 1380 N upwards F r = F g + F N 400 N = - 980 N + F N F N = 1380 N upwards
    (34)
  6. Step 6. Quote the final answer :

    The normal force is 1380 N upwards. It balances the gravitational force downwards and then in addition applies sufficient force to accelerate the man upwards at 4m··s-2-2. The reading on the scale is 1380 N.

Friction forces

When the surface of one object slides over the surface of another, each body exerts a frictional force on the other. For example if a book slides across a table, the table exerts a frictional force onto the book and the book exerts a frictional force onto the table (Newton's Third Law). Frictional forces act parallel to surfaces.

A force is not always powerful enough to make an object move, for example a small applied force might not be able to move a heavy crate. The frictional force opposing the motion of the crate is equal to the applied force but acting in the opposite direction. This frictional force is called static friction. When we increase the applied force (push harder), the frictional force will also increase until the applied force overcomes it. This frictional force can vary from zero (when no other forces are present and the object is stationary) to a maximum that depends on the surfaces. When the applied force is greater than the maximum frictional force, the crate will move. Once the object moves, the frictional force will decrease and remain at that level, which is also dependent on the surfaces, while the objects are moving. This is called kinetic friction. In both cases the maximum frictional force is related to the normal force and can be calculated as follows:

For static friction: Fffμsμs N

Where μsμs = the coefficient of static friction

and N = normal force

For kinetic friction: Fff = μkμk N

Where μkμk = the coefficient of kinetic friction

and N = normal force

Remember that static friction is present when the object is not moving and kinetic friction while the object is moving. For example when you drive at constant velocity in a car on a tar road you have to keep the accelerator pushed in slightly to overcome the friction between the tar road and the wheels of the car. However, while moving at a constant velocity the wheels of the car are rolling, so this is not a case of two surfaces “rubbing” against eachother and we are in fact looking at static friction. If you should break hard, causing the car to skid to a halt, we would be dealing with two surfaces rubbing against eachother and hence kinetic friction. The higher the value for the coefficient of friction, the more 'sticky' the surface is and the lower the value, the more 'slippery' the surface is.

The frictional force (Fff) acts in the horizontal direction and can be calculated in a similar way to the normal for as long as there is no movement. If we use the same example as in Figure 56 and we choose to the rightward direction as positive,

F f + F x = 0 F f + ( + 8 ) = 0 F f = - 8 F f = 8 N to the left F f + F x = 0 F f + ( + 8 ) = 0 F f = - 8 F f = 8 N to the left
(35)
Exercise 21: Forces on a slope

A 50 kg crate is placed on a slope that makes an angle of 30 with the horizontal. The box does not slide down the slope. Calculate the magnitude and direction of the frictional force and the normal force present in this situation.

Solution
  1. Step 1. Draw a force diagram :

    Draw a force diagram and fill in all the details on the diagram. This makes it easier to understand the problem.

    Figure 57: Friction and the normal forces on a slope
    Figure 57 (PG11C11_052.png)
  2. Step 2. Calculate the normal force :

    The normal force acts perpendicular to the surface (and not vertically upwards). It's magnitude is equal to the component of the weight perpendicular to the slope. Therefore:

    N = F g c o s 30 N = 490 c o s 30 N = 224 N perpendicular to the surface N = F g c o s 30 N = 490 c o s 30 N = 224 N perpendicular to the surface
    (36)
  3. Step 3. Calculate the frictional force :

    The frictional force acts parallel to the sloped surface. It's magnitude is equal to the component of the weight parallel to the slope. Therefore:

    F f = F g s i n 30 F f = 490 s i n 30 F f = 245 N up the slope F f = F g s i n 30 F f = 490 s i n 30 F f = 245 N up the slope
    (37)

We often think about friction in a negative way but very often friction is useful without us realizing it. If there was no friction and you tried to prop a ladder up against a wall, it would simply slide to the ground. Rock climbers use friction to maintain their grip on cliffs. The brakes of cars would be useless if it wasn't for friction!

Exercise 22: Coefficients of friction

A block of wood weighing 32 N is placed on a rough, flat inclline and a rope is tied to it. The tension in the rope can be increased to 8 N before the block starts to slide. A force of 4 N will keep the block moving at constant speed once it has been set in motion. Determine the coefficients of static and kinetic friction.

Solution
  1. Step 1. Analyse the question and determine what is asked :

    The weight of the block is given (32 N) and two situations are identified: One where the block is not moving (applied force is 8 N), and one where the block is moving (applied force is 4 N).

    We are asked to find the coefficient for static friction μsμs and kinetic friction μkμk.

  2. Step 2. Find the coefficient of static friction :
    F f = μ s N 8 = μ s ( 32 ) μ s = 0 , 25 F f = μ s N 8 = μ s ( 32 ) μ s = 0 , 25
    (38)

    Note that the coefficient of friction does not have a unit as it shows a ratio. The value for the coefficient of friction friction can have any value up to a maximum of 0,25. When a force less than 8 N is applied, the coefficient of friction will be less than 0,25.

  3. Step 3. Find the coefficient of kinetic friction :

    The coefficient of kinetic friction is sometimes also called the coefficient of dynamic friction. Here we look at when the block is moving:

    F f = μ k N 4 = μ k ( 32 ) μ k = 0 , 125 F f = μ k N 4 = μ k ( 32 ) μ k = 0 , 125
    (39)

Exercise

  1. A 12 kg box is placed on a rough surface. A force of 20 N applied at an angle of 30 to the horizontal cannot move the box. Calculate the magnitude and direction of the normal and friction forces.
  2. A 100 kg crate is placed on a slope that makes an angle of 45 with the horizontal. The box does not slide down the slope. Calculate the magnitude and direction of the frictional force and the normal force present in this situation.
  3. What force T at an angle of 30 above the horizontal, is required to drag a block weighing 20 N to the right at constant speed, if the coefficient of kinetic friction between the block and the surface is 0,20?
  4. A block weighing 20 N rests on a horizontal surface. The coefficient of static friction between the block and the surface is 0,40 and the coefficient of dynamic friction is 0,20.
    1. What is the magnitude of the frictional force exerted on the block while the block is at rest?
    2. What will the magnitude of the frictional force be if a horizontal force of 5 N is exerted on the block?
    3. What is the minimum force required to start the block moving?
    4. What is the minimum force required to keep the block in motion once it has been started?
    5. If the horizontal force is 10 N, determine the frictional force.
  5. A stationary block of mass 3kg is on top of a plane inclined at 3535 to the horizontal.
    Figure 58
    Figure 58 (PG11C11_053.png)
    1. Draw a force diagram (not to scale). Include the weight of the block as well as the components of the weight that are perpendicular and parallel to the inclined plane.
    2. Determine the values of the weight's perpendicular and parallel components.
    3. There exists a frictional force between the block and plane. Determine this force (magnitude and direction).
  6. A lady injured her back when she slipped and fell in a supermarket. She holds the owner of the supermarket accountable for her medical expenses. The owner claims that the floor covering was not wet and meets the accepted standards. He therefore cannot accept responsibility. The matter eventually ends up in court. Before passing judgement, the judge approaches you, a science student, to determine whether the coefficient of static friction of the floor is a minimum of 0,5 as required. He provides you with a tile from the floor, as well as one of the shoes the lady was wearing on the day of the incident.
    1. Write down an expression for the coefficient of static friction.
    2. Plan an investigation that you will perform to assist the judge in his judgement. Follow the steps outlined below to ensure that your plan meets the requirements.
      1. Formulate an investigation question.
      2. Apparatus: List all the other apparatus, except the tile and the shoe, that you will need.
      3. A stepwise method: How will you perform the investigation? Include a relevant, labelled free body-diagram.
      4. Results: What will you record?
      5. Conclusion: How will you interpret the results to draw a conclusion?

Forces in equilibrium

At the beginning of this chapter it was mentioned that resultant forces cause objects to accelerate in a straight line. If an object is stationary or moving at constant velocity then either,

  • no forces are acting on the object, or
  • the forces acting on that object are exactly balanced.

In other words, for stationary objects or objects moving with constant velocity, the resultant force acting on the object is zero. Additionally, if there is a perpendicular moment of force, then the object will rotate. You will learn more about moments of force later in this chapter.

Therefore, in order for an object not to move or to be in equilibrium, the sum of the forces (resultant force) must be zero and the sum of the moments of force must be zero.

Definition 4: Equilibrium

An object in equilibrium has both the sum of the forces acting on it and the sum of the moments of the forces equal to zero.

If a resultant force acts on an object then that object can be brought into equilibrium by applying an additional force that exactly balances this resultant. Such a force is called the equilibrant and is equal in magnitude but opposite in direction to the original resultant force acting on the object.

Definition 5: Equilibrant

The equilibrant of any number of forces is the single force required to produce equilibrium, and is equal in magnitude but opposite in direction to the resultant force.

Figure 59
Figure 59 (PG11C11_054.png)

In the figure the resultant of F1F1 and F2F2 is shown. The equilibrant of F1F1 and F2F2 is then the vector opposite in direction to this resultant with the same magnitude (i.e. F3F3).

  • F1F1, F2F2 and F3F3 are in equilibrium
  • F3F3 is the equilibrant of F1F1 and F2F2
  • F1F1 and F2F2 are kept in equilibrium by F3F3

As an example of an object in equilibrium, consider an object held stationary by two ropes in the arrangement below:

Figure 60
Figure 60 (PG11C11_055.png)

Let us draw a free body diagram for the object. In the free body diagram the object is drawn as a dot and all forces acting on the object are drawn in the correct directions starting from that dot. In this case, three forces are acting on the object.

Figure 61
Figure 61 (PG11C11_056.png)

Each rope exerts a force on the object in the direction of the rope away from the object. These tension forces are represented by T1T1 and T2T2. Since the object has mass, it is attracted towards the centre of the Earth. This weight is represented in the force diagram as FgFg.

Since the object is stationary, the resultant force acting on the object is zero. In other words the three force vectors drawn tail-to-head form a closed triangle:

Figure 62
Figure 62 (PG11C11_057.png)

Exercise 23: Equilibrium

A car engine of weight 2000 N is lifted by means of a chain and pulley system. The engine is initially suspended by the chain, hanging stationary. Then, the engine is pulled sideways by a mechanic, using a rope. The engine is held in such a position that the chain makes an angle of 3030 with the vertical. In the questions that follow, the masses of the chain and the rope can be ignored.

Figure 63
Figure 63 (PG11C11_058.png)

  1. Draw a free body representing the forces acting on the engine in the initial situation.
  2. Determine the tension in the chain initially.
  3. Draw a free body diagram representing the forces acting on the engine in the final situation.
  4. Determine the magnitude of the applied force and the tension in the chain in the final situations.
Solution
  1. Step 1. Initial free body diagram for the engine :

    There are only two forces acting on the engine initially: the tension in the chain, TchainTchain and the weight of the engine, FgFg.

    Figure 64
    Figure 64 (PG11C11_059.png)

  2. Step 2. Determine the tension in the chain :

    The engine is initially stationary, which means that the resultant force on the engine is zero. There are also no moments of force. Thus the tension in the chain exactly balances the weight of the engine. The tension in the chain is:

    T c h a i n = F g = 2000 N T c h a i n = F g = 2000 N
    (40)
  3. Step 3. Final free body diagram for the engine :

    There are three forces acting on the engine finally: The tension in the chain, the applied force and the weight of the engine.

    Figure 65
    Figure 65 (PG11C11_060.png)

  4. Step 4. Calculate the magnitude of the applied force and the tension in the chain in the final situation :

    Since no method was specified let us calculate the magnitudes algebraically. Since the triangle formed by the three forces is a right-angle triangle this is easily done:

    F a p p l i e d F g = tan 30 F a p p l i e d = ( 2000 N ) tan 30 = 1 155 N F a p p l i e d F g = tan 30 F a p p l i e d = ( 2000 N ) tan 30 = 1 155 N
    (41)

    and

    T c h a i n F g = 1 cos 30 T c h a i n = 2000 N cos 30 = 2 309 N T c h a i n F g = 1 cos 30 T c h a i n = 2000 N cos 30 = 2 309 N
    (42)

Exercise

  1. The diagram shows an object of weight W, attached to a string. A horizontal force F is applied to the object so that the string makes an angle of θθ with the vertical when the object is at rest. The force exerted by the string is T. Which one of the following expressions is incorrect?
    Figure 66
    Figure 66 (PG11C11_061.png)
    1. A: F + T + W = 0
    2. B: W = T cos θθ
    3. C: tan θθ = FWFW
    4. D: W = T sin θθ
  2. The point Q is in equilibrium due to three forces F1F1, F2F2 and F3F3 acting on it. Which of the statements about these forces is INCORRECT?
    1. A: The sum of the forces F1F1, F2F2 and F3F3 is zero.
    2. B: The three forces all lie in the same plane.
    3. C: The resultant force of F1F1 and F3F3 is F2F2.
    4. D: The sum of the components of the forces in any direction is zero.
    Figure 67
    Figure 67 (PG11C11_062.png)
  3. A point is acted on by two forces in equilibrium. The forces
    1. A: have equal magnitudes and directions.
    2. B: have equal magnitudes but opposite directions.
    3. C: act perpendicular to each other.
    4. D: act in the same direction.
  4. A point in equilibrium is acted on by three forces. Force F1F1 has components 15 N due south and 13 N due west. What are the components of force F2F2?
    Figure 68
    Figure 68 (PG11C11_063.png)
    1. A: 13 N due north and 20 due west
    2. B: 13 N due north and 13 N due west
    3. C: 15 N due north and 7 N due west
    4. D: 15 N due north and 13 N due east
    1. Define the term 'equilibrant'.
    2. Two tugs, one with a pull of 2500 N and the other with a pull of 3 000 N are used to tow an oil drilling platform. The angle between the two cables is 30 . Determine, either by scale diagram or by calculation (a clearly labelled rough sketch must be given), the equilibrant of the two forces.
  5. A 10 kg block is held motionless by a force F on a frictionless plane, which is inclined at an angle of 50 to the horizontal, as shown below:
    Figure 69
    Figure 69 (PG11C11_064.png)
    1. Draw a force diagram (not a triangle) indicating all the forces acting on the block.
    2. Calculate the magnitude of force F. Include a labelled diagram showing a triangle of forces in your answer.
  6. A rope of negligible mass is strung between two vertical struts. A mass M of weight W hangs from the rope through a hook fixed at point Y
    1. Draw a vector diagram, plotted head to tail, of the forces acting at point Y. Label each force and show the size of each angle.
    2. Where will the tension be greatest? Part P or Q? Motivate your answer.
    3. When the tension in the rope is greater than 600N it will break. What is the maximum mass that the above set up can support?
    Figure 70
    Figure 70 (PG11C11_065.png)
  7. An object of weight ww is supported by two cables attached to the ceiling and wall as shown. The tensions in the two cables are T1T1 and T2T2 respectively. Tension T1=1200T1=1200 N. Determine the tension T2T2 and weight ww of the object by accurate construction and measurement or calculation.
    Figure 71
    Figure 71 (PG11C11_066.png)
  8. A rope is tied at two points which are 70 cm apart from each other, on the same horizontal line. The total length of rope is 1 m, and the maximum tension it can withstand is 1000 N. Find the largest mass (m), in kg, that can be carried at the midpoint of the rope, without breaking the rope. Include a labelled diagram showing the triangle of forces in your answer.
    Figure 72
    Figure 72 (PG11C11_067.png)

Forces between Masses

In Grade 10, you saw that gravitational fields exert forces on masses in the field. A field is a region of space in which an object experiences a force. The strength of a field is defined by a field strength. For example, the gravitational field strength, gg, on or near the surface of the Earth has a value that is approximately 9,8 m··s-2-2.

The force exerted by a field of strength gg on an object of mass mm is given by:

F = m · g F = m · g
(43)

This can be re-written in terms of gg as:

g = F m g = F m
(44)

This means that gg can be understood to be a measure of force exerted per unit mass.

The force defined in Equation 43 is known as weight.

Objects in a gravitational field exert forces on each other without touching. The gravitational force is an example of a non-contact force.

Gravity is a force and therefore must be described by a vector - so remember thta gravity has both magnitude and direction.

Newton's Law of Universal Gravitation

Definition 6: Newton's Law of Universal Gravitation

Every point mass attracts every other point mass by a force directed along the line connecting the two. This force is proportional to the product of the masses and inversely proportional to the square of the distance between them.

The magnitude of the attractive gravitational force between the two point masses, FF is given by:

F = G m 1 m 2 r 2 F = G m 1 m 2 r 2
(45)

where: GG is the gravitational constant, m1m1 is the mass of the first point mass, m2m2 is the mass of the second point mass and rr is the distance between the two point masses.

Assuming SI units, FF is measured in newtons (N), m1m1 and m2m2 in kilograms (kg), rr in meters (m), and the constant GG is approximately equal to 6,67×10-11N·m2·kg-26,67×10-11N·m2·kg-2. Remember that this is a force of attraction.

For example, consider a man of mass 80 kg standing 10 m from a woman with a mass of 65 kg. The attractive gravitational force between them would be:

F = G m 1 m 2 r 2 = ( 6 , 67 × 10 - 11 N · m 2 · kg - 2 ) ( ( 80 kg ) ( 65 kg ) ( 10 m ) 2 ) = 3 , 47 × 10 - 9 N F = G m 1 m 2 r 2 = ( 6 , 67 × 10 - 11 N · m 2 · kg - 2 ) ( ( 80 kg ) ( 65 kg ) ( 10 m ) 2 ) = 3 , 47 × 10 - 9 N
(46)

If the man and woman move to 1 m apart, then the force is:

F = G m 1 m 2 r 2 = ( 6 , 67 × 10 - 11 N · m 2 · kg - 2 ) ( ( 80 kg ) ( 65 kg ) ( 1 m ) 2 ) = 3 , 47 × 10 - 7 N F = G m 1 m 2 r 2 = ( 6 , 67 × 10 - 11 N · m 2 · kg - 2 ) ( ( 80 kg ) ( 65 kg ) ( 1 m ) 2 ) = 3 , 47 × 10 - 7 N
(47)

As you can see, these forces are very small.

Now consider the gravitational force between the Earth and the Moon. The mass of the Earth is 5,98×10245,98×1024 kg, the mass of the Moon is 7,35×10227,35×1022 kg and the Earth and Moon are 3,8×1083,8×108 m apart. The gravitational force between the Earth and Moon is:

F = G m 1 m 2 r 2 = ( 6 , 67 × 10 - 11 N · m 2 · kg - 2 ) ( ( 5 , 98 × 10 24 kg ) ( 7 , 35 × 10 22 kg ) ( 0 , 38 × 10 9 m ) 2 ) = 2 , 03 × 10 20 N F = G m 1 m 2 r 2 = ( 6 , 67 × 10 - 11 N · m 2 · kg - 2 ) ( ( 5 , 98 × 10 24 kg ) ( 7 , 35 × 10 22 kg ) ( 0 , 38 × 10 9 m ) 2 ) = 2 , 03 × 10 20 N
(48)

From this example you can see that the force is very large.

These two examples demonstrate that the greater the masses, the greater the force between them. The 1/r21/r2 factor tells us that the distance between the two bodies plays a role as well. The closer two bodies are, the stronger the gravitational force between them is. We feel the gravitational attraction of the Earth most at the surface since that is the closest we can get to it, but if we were in outer-space, we would barely feel the effect of the Earth's gravity!

Remember that

F = m · a F = m · a
(49)

which means that every object on Earth feels the same gravitational acceleration! That means whether you drop a pen or a book (from the same height), they will both take the same length of time to hit the ground... in fact they will be head to head for the entire fall if you drop them at the same time. We can show this easily by using the two equations above (Equations Equation 45 and Equation 49). The force between the Earth (which has the mass meme) and an object of mass momo is

F = G m o m e r 2 F = G m o m e r 2
(50)

and the acceleration of an object of mass momo (in terms of the force acting on it) is

a o = F m o a o = F m o
(51)

So we substitute equation Equation 50 into Equation Equation 51, and we find that

a o = G m e r 2 a o = G m e r 2
(52)

Since it doesn't matter what momo is, this tells us that the acceleration on a body (due to the Earth's gravity) does not depend on the mass of the body. Thus all objects experience the same gravitational acceleration. The force on different bodies will be different but the acceleration will be the same. Due to the fact that this acceleration caused by gravity is the same on all objects we label it differently, instead of using aa we use gg which we call the gravitational acceleration.

Comparative Problems

Comparative problems involve calculation of something in terms of something else that we know. For example, if you weigh 490 N on Earth and the gravitational acceleration on Venus is 0,903 that of the gravitational acceleration on the Earth, then you would weigh 0,903 x 490 N = 442,5 N on Venus.

Principles for answering comparative problems

  • Write out equations and calculate all quantities for the given situation
  • Write out all relationships between variable from first and second case
  • Write out second case
  • Substitute all first case variables into second case
  • Write second case in terms of first case
Exercise 24: Comparative Problem 1

A man has a mass of 70 kg. The planet Zirgon is the same size as the Earth but has twice the mass of the Earth. What would the man weigh on Zirgon, if the gravitational acceleration on Earth is 9,8 m··s-2-2?

Solution
  1. Step 1. Determine what information has been given :

    The following has been provided:

    • the mass of the man, mm
    • the mass of the planet Zirgon (mZmZ) in terms of the mass of the Earth (mEmE), mZ=2mEmZ=2mE
    • the radius of the planet Zirgon (rZrZ) in terms of the radius of the Earth (rErE), rZ=rErZ=rE
  2. Step 2. Determine how to approach the problem :

    We are required to determine the man's weight on Zirgon (wZwZ). We can do this by using:

    w = m g = G m 1 · m 2 r 2 w = m g = G m 1 · m 2 r 2
    (53)

    to calculate the weight of the man on Earth and then use this value to determine the weight of the man on Zirgon.

  3. Step 3. Situation on Earth :
    w E = m g E = G m E · m r E 2 = ( 70 kg ) ( 9 , 8 m · s - 2 ) = 686 N w E = m g E = G m E · m r E 2 = ( 70 kg ) ( 9 , 8 m · s - 2 ) = 686 N
    (54)
  4. Step 4. Situation on Zirgon in terms of situation on Earth :

    Write the equation for the gravitational force on Zirgon and then substitute the values for mZmZ and rZrZ, in terms of the values for the Earth.

    w Z = m g Z = G m Z · m r Z 2 = G 2 m E · m r E 2 = 2 ( G m E · m r E 2 ) = 2 w E = 2 ( 686 N ) = 1 372 N w Z = m g Z = G m Z · m r Z 2 = G 2 m E · m r E 2 = 2 ( G m E · m r E 2 ) = 2 w E = 2 ( 686 N ) = 1 372 N
    (55)
  5. Step 5. Quote the final answer :

    The man weighs 1 372 N on Zirgon.

Exercise 25: Comparative Problem 2

A man has a mass of 70 kg. On the planet Beeble how much will he weigh if Beeble has mass half of that of the Earth and a radius one quarter that of the Earth. Gravitational acceleration on Earth is 9,8 m··s-2-2.

Solution
  1. Step 1. Determine what information has been given :

    The following has been provided:

    • the mass of the man on Earth, mm
    • the mass of the planet Beeble (mBmB) in terms of the mass of the Earth (mEmE), mB=12mEmB=12mE
    • the radius of the planet Beeble (rBrB) in terms of the radius of the Earth (rErE), rB=14rErB=14rE
  2. Step 2. Determine how to approach the problem :

    We are required to determine the man's weight on Beeble (wBwB). We can do this by using:

    w = m g = G m 1 · m 2 r 2 w = m g = G m 1 · m 2 r 2
    (56)

    to calculate the weight of the man on Earth and then use this value to determine the weight of the man on Beeble.

  3. Step 3. Situation on Earth :
    w E = m g E = G m E · m r E 2 = ( 70 kg ) ( 9 , 8 m · s - 2 ) = 686 N w E = m g E = G m E · m r E 2 = ( 70 kg ) ( 9 , 8 m · s - 2 ) = 686 N
    (57)
  4. Step 4. Situation on Beeble in terms of situation on Earth :

    Write the equation for the gravitational force on Beeble and then substitute the values for mBmB and rBrB, in terms of the values for the Earth.

    w B = m g B = G m B · m r B 2 = G 1 2 m E · m ( 1 4 r E ) 2 = 8 ( G m E · m r E 2 ) = 8 w E = 8 ( 686 N ) = 5 488 N w B = m g B = G m B · m r B 2 = G 1 2 m E · m ( 1 4 r E ) 2 = 8 ( G m E · m r E 2 ) = 8 w E = 8 ( 686 N ) = 5 488 N
    (58)
  5. Step 5. Quote the final answer :

    The man weighs 5 488 N on Beeble.

Physics is all about being simple - all we do is look at the world around us and notice how it really works. It is the one thing everyone is qualified to do - we spend most of our time when we are really young experimenting to find out how things work.

The actual force of air resistance is quite complicated. Experiment by moving a book through the air with the face of the book and then the side of the book forward, you will agree that the area of the book makes a difference as to how much you must work in order to move the book at the same speed in both cases. This is why racing cars are slim-lined in design, and not shaped like a big box!

Exercise

  1. Two objects of mass 2m and 3m respectively exert a force F on each other when they are a certain distance apart. What will be the force between two objects situated the same distance apart but having a mass of 5m and 6m respectively?
    1. A: 0,2 F
    2. B: 1,2 F
    3. C: 2,2 F
    4. D: 5 F
  2. As the distance of an object above the surface of the Earth is greatly increased, the weight of the object would
    1. A: increase
    2. B: decrease
    3. C: increase and then suddenly decrease
    4. D: remain the same
  3. A satellite circles around the Earth at a height where the gravitational force is a factor 4 less than at the surface of the Earth. If the Earth's radius is R, then the height of the satellite above the surface is:
    1. A: R
    2. B: 2 R
    3. C: 4 R
    4. D: 16 R
  4. A satellite experiences a force F when at the surface of the Earth. What will be the force on the satellite if it orbits at a height equal to the diameter of the Earth:
    1. A: 1F1F
    2. B: 1212FF
    3. C: 1313FF
    4. D: 1919FF
  5. The weight of a rock lying on surface of the Moon is W. The radius of the Moon is R. On planet Alpha, the same rock has weight 8W. If the radius of planet Alpha is half that of the Moon, and the mass of the Moon is M, then the mass, in kg, of planet Alpha is:
    1. A: M2M2
    2. B: M4M4
    3. C: 2 M
    4. D: 4 M
  6. Consider the symbols of the two physical quantities gg and GG used in Physics.
    1. Name the physical quantities represented by gg and GG.
    2. Derive a formula for calculating gg near the Earth's surface using Newton's Law of Universal Gravitation. M and R represent the mass and radius of the Earth respectively.
  7. Two spheres of mass 800g and 500g respectively are situated so that their centers are 200 cm apart. Calculate the gravitational force between them.
  8. Two spheres of mass 2 kg and 3 kg respectively are situated so that the gravitational force between them is 2,5 x 10-8-8 N. Calculate the distance between them.
  9. Two identical spheres are placed 10 cm apart. A force of 1,6675 x 10-9-9 N exists between them. Find the masses of the spheres.
  10. Halley's comet, of approximate mass 1 x 101515 kg was 1,3 x 1088 km from the Earth, at its point of closest approach during its last sighting in 1986.
    1. Name the force through which the Earth and the comet interact.
    2. Is the magnitude of the force experienced by the comet the same, greater than or less than the force experienced by the Earth? Explain.
    3. Does the acceleration of the comet increase, decrease or remain the same as it moves closer to the Earth? Explain.
    4. If the mass of the Earth is 6 x 102424 kg, calculate the magnitude of the force exerted by the Earth on Halley's comet at its point of closest approach.

Momentum and Impulse

Momentum is a physical quantity which is closely related to forces. Momentum is a property which applies to moving objects.

Definition 7: Momentum

Momentum is the tendency of an object to continue to move in its direction of travel. Momentum is calculated from the product of the mass and velocity of an object.

The momentum (symbol pp) of an object of mass mm moving at velocity vv is:

p = m · v p = m · v
(59)

According to this equation, momentum is related to both the mass and velocity of an object. A small car travelling at the same velocity as a big truck will have a smaller momentum than the truck. The smaller the mass, the smaller the velocity.

A car travelling at 120 km··hr-1-1will have a bigger momentum than the same car travelling at 60 km··hr-1-1. Momentum is also related to velocity; the smaller the velocity, the smaller the momentum.

Different objects can also have the same momentum, for example a car travelling slowly can have the same momentum as a motor cycle travelling relatively fast. We can easily demonstrate this. Consider a car of mass 1 000 kg with a velocity of 8 m··s-1-1(about 30 km··hr-1-1). The momentum of the car is therefore

p = m · v = ( 1000 kg ) ( 8 m · s - 1 ) = 8000 kg · m · s - 1 p = m · v = ( 1000 kg ) ( 8 m · s - 1 ) = 8000 kg · m · s - 1
(60)

Now consider a motor cycle of mass 250 kg travelling at 32 m··s-1-1 (about 115 km··hr-1-1). The momentum of the motor cycle is:

p = m · v = ( 250 kg ) ( 32 m · s - 1 ) = 8000 kg · m · s - 1 p = m · v = ( 250 kg ) ( 32 m · s - 1 ) = 8000 kg · m · s - 1
(61)

Even though the motor cycle is considerably lighter than the car, the fact that the motor cycle is travelling much faster than the car means that the momentum of both vehicles is the same.

From the calculations above, you are able to derive the unit for momentum as kg··m··s-1-1.

Momentum is also vector quantity, because it is the product of a scalar (mm) with a vector (vv).

This means that whenever we calculate the momentum of an object, we need to include the direction of the momentum.

Figure 73
Khan academy video on momentum - 1

Exercise 26: Momentum of a Soccer Ball

A soccer ball of mass 420 g is kicked at 20 m··s-1-1 towards the goal post. Calculate the momentum of the ball.

Solution

  1. Step 1. Identify what information is given and what is asked for :

    The question explicitly gives

    • the mass of the ball, and
    • the velocity of the ball

    The mass of the ball must be converted to SI units.

    420 g = 0 , 42 kg 420 g = 0 , 42 kg
    (62)

    We are asked to calculate the momentum of the ball. From the definition of momentum,

    p = m · v p = m · v
    (63)

    we see that we need the mass and velocity of the ball, which we are given.

  2. Step 2. Do the calculation :

    We calculate the magnitude of the momentum of the ball,

    p = m · v = ( 0 , 42 kg ) ( 20 m · s - 1 ) = 8 , 4 kg · m · s - 1 p = m · v = ( 0 , 42 kg ) ( 20 m · s - 1 ) = 8 , 4 kg · m · s - 1
    (64)
  3. Step 3. Quote the final answer :

    We quote the answer with the direction of motion included, pp = 8,4 kg··m··s-1-1 in the direction of the goal post.

Exercise 27: Momentum of a cricket ball

A cricket ball of mass 160 g is bowled at 40 m··s-1-1 towards a batsman. Calculate the momentum of the cricket ball.

Solution

  1. Step 1. Identify what information is given and what is asked for :

    The question explicitly gives

    • the mass of the ball (mm = 160 g = 0,16 kg), and
    • the velocity of the ball (vv = 40 m··s-1-1)

    To calculate the momentum we will use

    p = m · v p = m · v
    (65)

    .

  2. Step 2. Do the calculation :
    p = m · v = ( 0 , 16 kg ) ( 40 m · s - 1 ) = 6 , 4 kg · m · s - 1 = 6 , 4 kg · m · s - 1 in the direction of the batsman p = m · v = ( 0 , 16 kg ) ( 40 m · s - 1 ) = 6 , 4 kg · m · s - 1 = 6 , 4 kg · m · s - 1 in the direction of the batsman
    (66)

Exercise 28: Momentum of the Moon

The Moon is 384 400 km away from the Earth and orbits the Earth in 27,3 days. If the Moon has a mass of 7,35 x 102222kg, what is the magnitude of its momentum if we assume a circular orbit?

Solution

  1. Step 1. Identify what information is given and what is asked for :

    The question explicitly gives

    • the mass of the Moon (m = 7,35 x 102222 kg)
    • the distance to the Moon (384 400 km = 384 400 000 m = 3,844 x 1088 m)
    • the time for one orbit of the Moon (27,3 days = 27,3 x 24 x 60 x 60 = 2,36 x 1066 s)

    We are asked to calculate only the magnitude of the momentum of the Moon (i.e. we do not need to specify a direction). In order to do this we require the mass and the magnitude of the velocity of the Moon, since

    p = m · v p = m · v
    (67)
  2. Step 2. Find the magnitude of the velocity of the Moon :

    The magnitude of the average velocity is the same as the speed. Therefore:

    s = d Δ t s = d Δ t
    (68)

    We are given the time the Moon takes for one orbit but not how far it travels in that time. However, we can work this out from the distance to the Moon and the fact that the Moon has a circular orbit. Using the equation for the circumference, CC, of a circle in terms of its radius, we can determine the distance travelled by the Moon in one orbit:

    C = 2 π r = 2 π ( 3 , 844 × 10 8 m ) = 2 , 42 × 10 9 m C = 2 π r = 2 π ( 3 , 844 × 10 8 m ) = 2 , 42 × 10 9 m
    (69)

    Combining the distance travelled by the Moon in an orbit and the time taken by the Moon to complete one orbit, we can determine the magnitude of the Moon's velocity or speed,

    s = d Δ t = C T = 2 , 42 × 10 9 m 2 , 36 × 10 6 s = 1 , 02 × 10 3 m · s - 1 . s = d Δ t = C T = 2 , 42 × 10 9 m 2 , 36 × 10 6 s = 1 , 02 × 10 3 m · s - 1 .
    (70)
  3. Step 3. Finally calculate the momentum and quote the answer :

    The magnitude of the Moon's momentum is:

    p = m · v = ( 7 , 35 × 10 22 kg ) ( 1 , 02 × 10 3 m · s - 1 ) = 7 , 50 × 10 25 kg · m · s - 1 p = m · v = ( 7 , 35 × 10 22 kg ) ( 1 , 02 × 10 3 m · s - 1 ) = 7 , 50 × 10 25 kg · m · s - 1
    (71)

Vector Nature of Momentum

As we have said, momentum is a vector quantity. Since momentum is a vector, the techniques of vector addition discussed in Chapter (Reference) must be used to calculate the total momentum of a system.

Exercise 29: Calculating the Total Momentum of a System

Two billiard balls roll towards each other. They each have a mass of 0,3 kg. Ball 1 is moving at v1=1m·s-1v1=1m·s-1 to the right, while ball 2 is moving at v2=0,8m·s-1v2=0,8m·s-1 to the left. Calculate the total momentum of the system.

Solution
  1. Step 1. Identify what information is given and what is asked for :

    The question explicitly gives

    • the mass of each ball,
    • the velocity of ball 1, v1v1, and
    • the velocity of ball 2, v2v2,

    all in the correct units!

    We are asked to calculate the total momentum of the system. In this example our system consists of two balls. To find the total momentum we must determine the momentum of each ball and add them.

    p t o t a l = p 1 + p 2 p t o t a l = p 1 + p 2
    (72)

    Since ball 1 is moving to the right, its momentum is in this direction, while the second ball's momentum is directed towards the left.

    Figure 74
    Figure 74 (PG11C11_068.png)

    Thus, we are required to find the sum of two vectors acting along the same straight line. The algebraic method of vector addition introduced in Chapter (Reference) can thus be used.

  2. Step 2. Choose a frame of reference :

    Let us choose right as the positive direction, then obviously left is negative.

  3. Step 3. Calculate the momentum :

    The total momentum of the system is then the sum of the two momenta taking the directions of the velocities into account. Ball 1 is travelling at 1 m··s-1-1to the right or +1 m··s-1-1. Ball 2 is travelling at 0,8 m··s-1-1to the left or -0,8 m··s-1-1. Thus,

    p t o t a l = m 1 v 1 + m 2 v 2 = ( 0 , 3 kg ) ( + 1 m · s - 1 ) + ( 0 , 3 kg ) ( - 0 , 8 m · s - 1 ) = ( + 0 , 3 kg · m · s - 1 ) + ( - 0 , 24 kg · m · s - 1 ) = + 0 , 06 kg · m · s - 1 = 0 , 06 kg · m · s - 1 to the right p t o t a l = m 1 v 1 + m 2 v 2 = ( 0 , 3 kg ) ( + 1 m · s - 1 ) + ( 0 , 3 kg ) ( - 0 , 8 m · s - 1 ) = ( + 0 , 3 kg · m · s - 1 ) + ( - 0 , 24 kg · m · s - 1 ) = + 0 , 06 kg · m · s - 1 = 0 , 06 kg · m · s - 1 to the right
    (73)

    In the last step the direction was added in words. Since the result in the second last line is positive, the total momentum of the system is in the positive direction (i.e. to the right).

Exercise

    1. The fastest recorded delivery for a cricket ball is 161,3 km··hr-1-1, bowled by Shoaib Akhtar of Pakistan during a match against England in the 2003 Cricket World Cup, held in South Africa. Calculate the ball's momentum if it has a mass of 160 g.
    2. The fastest tennis service by a man is 246,2 km··hr-1-1by Andy Roddick of the United States of America during a match in London in 2004. Calculate the ball's momentum if it has a mass of 58 g.
    3. The fastest server in the women's game is Venus Williams of the United States of America, who recorded a serve of 205 km··hr-1-1during a match in Switzerland in 1998. Calculate the ball's momentum if it has a mass of 58 g.
    4. If you had a choice of facing Shoaib, Andy or Venus and didn't want to get hurt, who would you choose based on the momentum of each ball.
  1. Two golf balls roll towards each other. They each have a mass of 100 g. Ball 1 is moving at v1v1 = 2,4 m··s-1-1to the right, while ball 2 is moving at v2v2 = 3 m··s-1-1to the left. Calculate the total momentum of the system.
  2. Two motor cycles are involved in a head on collision. Motorcycle A has a mass of 200 kg and was travelling at 120 km··hr-1-1south. Motor cycle B has a mass of 250 kg and was travelling north at 100 km··hr-1-1. A and B are about to collide. Calculate the momentum of the system before the collision takes place.

Change in Momentum

Let us consider a tennis ball (mass = 0,1 kg) that is dropped at an initial velocity of 5 m··s-1-1and bounces back at a final velocity of 3 m··s-1-1. As the ball approaches the floor it has a momentum that we call the momentum before the collision. When it moves away from the floor it has a different momentum called the momentum after the collision. The bounce on the floor can be thought of as a collision taking place where the floor exerts a force on the tennis ball to change its momentum.

The momentum before the bounce can be calculated as follows:

Because momentum and velocity are vectors, we have to choose a direction as positive. For this example we choose the initial direction of motion as positive, in other words, downwards is positive.

p i = m · v i = ( 0 , 1 kg ) ( + 5 m · s - 1 ) = 0 , 5 kg · m · s - 1 downwards p i = m · v i = ( 0 , 1 kg ) ( + 5 m · s - 1 ) = 0 , 5 kg · m · s - 1 downwards
(74)

When the tennis ball bounces back it changes direction. The final velocity will thus have a negative value. The momentum after the bounce can be calculated as follows:

p f = m · v f = ( 0 , 1 kg ) ( - 3 m · s - 1 ) = - 0 , 3 kg · m · s - 1 = 0 , 3 kg · m · s - 1 upwards p f = m · v f = ( 0 , 1 kg ) ( - 3 m · s - 1 ) = - 0 , 3 kg · m · s - 1 = 0 , 3 kg · m · s - 1 upwards
(75)

Now let us look at what happens to the momentum of the tennis ball. The momentum changes during this bounce. We can calculate the change in momentum as follows:

Again we have to choose a direction as positive and we will stick to our initial choice as downwards is positive. This means that the final momentum will have a negative number.

Δ p = p f - p i = m · v f - m · v i = ( - 0 , 3 kg ) - ( 0 , 5 m · s - 1 ) = - 0 , 8 kg · m · s - 1 = 0 , 8 kg · m · s - 1 upwards Δ p = p f - p i = m · v f - m · v i = ( - 0 , 3 kg ) - ( 0 , 5 m · s - 1 ) = - 0 , 8 kg · m · s - 1 = 0 , 8 kg · m · s - 1 upwards
(76)

You will notice that this number is bigger than the previous momenta calculated. This is should be the case as the ball needed to be stopped and then given momentum to bounce back.

Exercise 30: Change in Momentum

A rubber ball of mass 0,8 kg is dropped and strikes the floor with an initial velocity of 6 m··s-1-1. It bounces back with a final velocity of 4 m··s-1-1. Calculate the change in the momentum of the rubber ball caused by the floor.

Figure 75
Figure 75 (PG11C11_069.png)
Solution
  1. Step 1. Identify the information given and what is asked :

    The question explicitly gives

    • the ball's mass (m = 0,8 kg),
    • the ball's initial velocity (vii = 6 m··s-1-1), and
    • the ball's final velocity (vff = 4 m··s-1-1)

    all in the correct units.

    We are asked to calculate the change in momentum of the ball,

    Δ p = m v f - m v i Δ p = m v f - m v i
    (77)

    We have everything we need to find ΔpΔp. Since the initial momentum is directed downwards and the final momentum is in the upward direction, we can use the algebraic method of subtraction discussed in the vectors chapter.

  2. Step 2. Choose a frame of reference :

    Let us choose down as the positive direction.

  3. Step 3. Do the calculation and quote the answer :
    Δ p = m v f - m v i = ( 0 , 8 kg ) ( - 4 m · s - 1 ) - ( 0 , 8 kg ) ( + 6 m · s - 1 ) = ( - 3 , 2 kg · m · s - 1 ) - ( 4 , 8 kg · m · s - 1 ) = - 8 = 8 kg · m · s - 1 upwards Δ p = m v f - m v i = ( 0 , 8 kg ) ( - 4 m · s - 1 ) - ( 0 , 8 kg ) ( + 6 m · s - 1 ) = ( - 3 , 2 kg · m · s - 1 ) - ( 4 , 8 kg · m · s - 1 ) = - 8 = 8 kg · m · s - 1 upwards
    (78)

Exercise

  1. Which expression accurately describes the change of momentum of an object?
    1. A: FmFm
    2. B: FtFt
    3. C: F·mF·m
    4. D: F·tF·t
  2. A child drops a ball of mass 100 g. The ball strikes the ground with a velocity of 5 m··s-1-1and rebounds with a velocity of 4 m··s-1-1. Calculate the change of momentum of the ball.
  3. A 700 kg truck is travelling north at a velocity of 40 km··hr-1-1when it is approached by a 500 kg car travelling south at a velocity of 100 km··hr-1-1. Calculate the total momentum of the system.

Newton's Second Law revisited

You have learned about Newton's Second Law of motion earlier in this chapter. Newton's Second Law describes the relationship between the motion of an object and the net force on the object. We said that the motion of an object, and therefore its momentum, can only change when a resultant force is acting on it. We can therefore say that because a net force causes an object to move, it also causes its momentum to change. We can now define Newton's Second Law of motion in terms of momentum.

Definition 8: Newton's Second Law of Motion (N2)

The net or resultant force acting on an object is equal to the rate of change of momentum.

Mathematically, Newton's Second Law can be stated as:

F n e t = Δ p Δ t F n e t = Δ p Δ t
(79)

Impulse

Impulse is the product of the net force and the time interval for which the force acts. Impulse is defined as:

Impulse = F · Δ t Impulse = F · Δ t
(80)

However, from Newton's Second Law, we know that

F = Δ p Δ t F · Δ t = Δ p = Impulse F = Δ p Δ t F · Δ t = Δ p = Impulse
(81)

Therefore,

Impulse = Δ p Impulse = Δ p
(82)

Impulse is equal to the change in momentum of an object. From this equation we see, that for a given change in momentum, FnetΔtFnetΔt is fixed. Thus, if FnetFnet is reduced, ΔtΔt must be increased (i.e. a smaller resultant force must be applied for longer to bring about the same change in momentum). Alternatively if ΔtΔt is reduced (i.e. the resultant force is applied for a shorter period) then the resultant force must be increased to bring about the same change in momentum.

Exercise 31: Impulse and Change in momentum

A 150 N resultant force acts on a 300 kg trailer. Calculate how long it takes this force to change the trailer's velocity from 2 m··s-1-1to 6 m··s-1-1in the same direction. Assume that the forces acts to the right.

Solution
  1. Step 1. Identify what information is given and what is asked for :

    The question explicitly gives

    • the trailer's mass as 300 kg,
    • the trailer's initial velocity as 2 m··s-1-1to the right,
    • the trailer's final velocity as 6 m··s-1-1to the right, and
    • the resultant force acting on the object

    all in the correct units!

    We are asked to calculate the time taken ΔtΔt to accelerate the trailer from the 2 to 6 m··s-1-1. From the Law of Momentum,

    F n e t Δ t = Δ p = m v f - m v i = m ( v f - v i ) . F n e t Δ t = Δ p = m v f - m v i = m ( v f - v i ) .
    (83)

    Thus we have everything we need to find ΔtΔt!

  2. Step 2. Choose a frame of reference :

    Choose right as the positive direction.

  3. Step 3. Do the calculation and quote the final answer :
    F n e t Δ t = m ( v f - v i ) ( + 150 N ) Δ t = ( 300 kg ) ( ( + 6 m · s - 1 ) - ( + 2 m · s - 1 ) ) ( + 150 N ) Δ t = ( 300 kg ) ( + 4 m · s - 1 ) Δ t = ( 300 kg ) ( + 4 m · s - 1 ) + 150 N Δ t = 8 s F n e t Δ t = m ( v f - v i ) ( + 150 N ) Δ t = ( 300 kg ) ( ( + 6 m · s - 1 ) - ( + 2 m · s - 1 ) ) ( + 150 N ) Δ t = ( 300 kg ) ( + 4 m · s - 1 ) Δ t = ( 300 kg ) ( + 4 m · s - 1 ) + 150 N Δ t = 8 s
    (84)

    It takes 8 s for the force to change the object's velocity from 2 m··s-1-1to the right to 6 m··s-1-1to the right.

Exercise 32: Impulsive cricketers!

A cricket ball weighing 156 g is moving at 54 km··hr-1-1towards a batsman. It is hit by the batsman back towards the bowler at 36 km··hr-1-1. Calculate

  1. the ball's impulse, and
  2. the average force exerted by the bat if the ball is in contact with the bat for 0,13 s.
Solution
  1. Step 1. Identify what information is given and what is asked for :

    The question explicitly gives

    • the ball's mass,
    • the ball's initial velocity,
    • the ball's final velocity, and
    • the time of contact between bat and ball

    We are asked to calculate the impulse

    Impulse = Δ p = F n e t Δ t Impulse = Δ p = F n e t Δ t
    (85)

    Since we do not have the force exerted by the bat on the ball (Fnetnet), we have to calculate the impulse from the change in momentum of the ball. Now, since

    Δ p = p f - p i = m v f - m v i , Δ p = p f - p i = m v f - m v i ,
    (86)

    we need the ball's mass, initial velocity and final velocity, which we are given.

  2. Step 2. Convert to S.I. units :

    Firstly let us change units for the mass

    1000 g = 1 kg So , 1 g = 1 1000 kg 156 × 1 g = 156 × 1 1000 kg = 0 , 156 kg 1000 g = 1 kg So , 1 g = 1 1000 kg 156 × 1 g = 156 × 1 1000 kg = 0 , 156 kg
    (87)

    Next we change units for the velocity

    1 km · h - 1 = 1000 m 3 600 s 54 × 1 km · h - 1 = 54 × 1 000 m 3 600 s = 15 m · s - 1 1 km · h - 1 = 1000 m 3 600 s 54 × 1 km · h - 1 = 54 × 1 000 m 3 600 s = 15 m · s - 1
    (88)

    Similarly, 36 km··hr-1-1= 10 m··s-1-1.

  3. Step 3. Choose a frame of reference :

    Let us choose the direction from the batsman to the bowler as the positive direction. Then the initial velocity of the ball is vivi = -15 m··s-1-1, while the final velocity of the ball is vfvf = 10 m··s-1-1.

  4. Step 4. Calculate the momentum :

    Now we calculate the change in momentum,

    p = p f - p i = m v f - m v i = m ( v f - v i ) = ( 0 , 156 kg ) ( ( + 10 m · s - 1 ) - ( - 15 m · s - 1 ) ) = + 3 , 9 kg · m · s - 1 = 3 , 9 kg · m · s - 1 in the direction from batsman to bowler p = p f - p i = m v f - m v i = m ( v f - v i ) = ( 0 , 156 kg ) ( ( + 10 m · s - 1 ) - ( - 15 m · s - 1 ) ) = + 3 , 9 kg · m · s - 1 = 3 , 9 kg · m · s - 1 in the direction from batsman to bowler
    (89)
  5. Step 5. Determine the impulse :

    Finally since impulse is just the change in momentum of the ball,

    Impulse = Δ p = 3 , 9 kg · m · s - 1 in the direction from the batsman to the bowler Impulse = Δ p = 3 , 9 kg · m · s - 1 in the direction from the batsman to the bowler
    (90)
  6. Step 6. Determine the average force exerted by the bat :
    Impulse = F n e t Δ t = Δ p Impulse = F n e t Δ t = Δ p
    (91)

    We are given ΔtΔt and we have calculated the impulse of the ball.

    F n e t Δ t = Impulse F n e t ( 0 , 13 s ) = + 3 , 9 N · s F n e t = + 3 , 9 N · s 0 , 13 s = + 30 N = 30 N in the direction from batsman to bowler F n e t Δ t = Impulse F n e t ( 0 , 13 s ) = + 3 , 9 N · s F n e t = + 3 , 9 N · s 0 , 13 s = + 30 N = 30 N in the direction from batsman to bowler
    (92)

Exercise

  1. Which one of the following is NOT a unit of impulse?
    1. A: N·sN·s
    2. B: kg·m·s-1kg·m·s-1
    3. C: J·m·s-1J·m·s-1
    4. D: J·m-1·sJ·m-1·s
  2. A toy car of mass 1 kg moves eastwards with a speed of 2 m··s-1-1. It collides head-on with a toy train. The train has a mass of 2 kg and is moving at a speed of 1,5 m··s-1-1westwards. The car rebounds (bounces back) at 3,4 m··s-1-1and the train rebounds at 1,2 m··s-1-1.
    1. Calculate the change in momentum for each toy.
    2. Determine the impulse for each toy.
    3. Determine the duration of the collision if the magnitude of the force exerted by each toy is 8 N.
  3. A bullet of mass 20 g strikes a target at 300 m··s-1-1and exits at 200 m··s-1-1. The tip of the bullet takes 0,0001s to pass through the target. Determine:
    1. the change of momentum of the bullet.
    2. the impulse of the bullet.
    3. the magnitude of the force experienced by the bullet.
  4. A bullet of mass 20 g strikes a target at 300 m··s-1-1. Determine under which circumstances the bullet experiences the greatest change in momentum, and hence impulse:
    1. When the bullet exits the target at 200 m··s-1-1.
    2. When the bullet stops in the target.
    3. When the bullet rebounds at 200 m··s-1-1.
  5. A ball with a mass of 200 g strikes a wall at right angles at a velocity of 12 m··s-1-1and rebounds at a velocity of 9 m··s-1-1.
    1. Calculate the change in the momentum of the ball.
    2. What is the impulse of the wall on the ball?
    3. Calculate the magnitude of the force exerted by the wall on the ball if the collision takes 0,02s.
  6. If the ball in the previous problem is replaced with a piece of clay of 200 g which is thrown against the wall with the same velocity, but then sticks to the wall, calculate:
    1. The impulse of the clay on the wall.
    2. The force exerted by the clay on the wall if it is in contact with the wall for 0,5 s before it comes to rest.

Conservation of Momentum

In the absence of an external force acting on a system, momentum is conserved.

Definition 9: Conservation of Linear Momentum

The total linear momentum of an isolated system is constant. An isolated system has no forces acting on it from the outside.

This means that in an isolated system the total momentum before a collision or explosion is equal to the total momentum after the collision or explosion.

Consider a simple collision of two billiard balls. The balls are rolling on a frictionless surface and the system is isolated. So, we can apply conservation of momentum. The first ball has a mass m1m1 and an initial velocity vi1vi1. The second ball has a mass m2m2 and moves towards the first ball with an initial velocity vi2vi2. This situation is shown in Figure 76.

Figure 76: Before the collision.
Figure 76 (PG11C11_070.png)

The total momentum of the system before the collision, pipi is:

p i = m 1 v i 1 + m 2 v i 2 p i = m 1 v i 1 + m 2 v i 2
(93)

After the two balls collide and move away they each have a different momentum. If the first ball has a final velocity of vf1vf1 and the second ball has a final velocity of vf2vf2 then we have the situation shown in Figure 77.

Figure 77: After the collision.
Figure 77 (PG11C11_071.png)

The total momentum of the system after the collision, pfpf is:

p f = m 1 v f 1 + m 2 v f 2 p f = m 1 v f 1 + m 2 v f 2
(94)

This system of two balls is isolated since there are no external forces acting on the balls. Therefore, by the principle of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision. This gives the equation for the conservation of momentum in a collision of two objects,

p i = p f p i = p f
(95)
m 1 v i 1 + m 2 v i 2 = m 1 v f 1 + m 2 v f 2 m 1 v i 1 + m 2 v i 2 = m 1 v f 1 + m 2 v f 2
(96)
where: m 1 m 1 is the mass of object 1 (kg)
m 2 m 2 is the mass of object 2 (kg)
v i 1 v i 1 is the initial velocity of object 1 (m··s-1-1+ direction)
v i 2 v i 2 is the initial velocity of object 2 (m··s-1-1- direction)
v f 1 v f 1 is the final velocity of object 1 (m··s-1-1- direction)
v f 2 v f 2 is the final velocity of object 2 (m··s-1-1+ direction)

This equation is always true - momentum is always conserved in collisions.

Exercise 33: Conservation of Momentum 1

A toy car of mass 1 kg moves westwards with a speed of 2 m··s-1-1. It collides head-on with a toy train. The train has a mass of 1,5 kg and is moving at a speed of 1,5 m··s-1-1eastwards. If the car rebounds at 2,05 m··s-1-1, calculate the velocity of the train.

Solution
  1. Step 1. Draw rough sketch of the situation :
    Figure 78
    Figure 78 (PG11C11_072.png)
  2. Step 2. Choose a frame of reference :

    We will choose to the east as positive.

  3. Step 3. Apply the Law of Conservation of momentum :
    p i = p f m 1 v i 1 + m 2 v i 2 = m 1 v f 1 + m 2 v f 2 ( 1 , 5 kg ) ( + 1 , 5 m · s - 1 ) + ( 2 kg ) ( - 2 m · s - 1 ) = ( 1 , 5 kg ) ( v f 1 ) + ( 2 kg ) ( 2 , 05 m · s - 1 ) 2 , 25 kg · m · s - 1 - 4 kg · m · s - 1 - 4 , 1 kg · m · s - 1 = ( 1 , 5 kg ) v f 1 5 , 85 kg · m · s - 1 = ( 1 , 5 kg ) v f 1 v f 1 = 3 , 9 m · s - 1 eastwards p i = p f m 1 v i 1 + m 2 v i 2 = m 1 v f 1 + m 2 v f 2 ( 1 , 5 kg ) ( + 1 , 5 m · s - 1 ) + ( 2 kg ) ( - 2 m · s - 1 ) = ( 1 , 5 kg ) ( v f 1 ) + ( 2 kg ) ( 2 , 05 m · s - 1 ) 2 , 25 kg · m · s - 1 - 4 kg · m · s - 1 - 4 , 1 kg · m · s - 1 = ( 1 , 5 kg ) v f 1 5 , 85 kg · m · s - 1 = ( 1 , 5 kg ) v f 1 v f 1 = 3 , 9 m · s - 1 eastwards
    (97)

Exercise 34: Conservation of Momentum 2

A helicopter flies at a speed of 275 m··s-1-1. The pilot fires a missile forward out of a gun barrel at a speed of 700 m··s-1-1. The respective masses of the helicopter and the missile are 5000 kg and 50 kg. Calculate the new speed of the helicopter immmediately after the missile had been fired.

Solution
  1. Step 1. Draw rough sketch of the situation :
    Figure 79: helicopter and missile
    Figure 79 (PG11C11_073.png)
  2. Step 2. Analyse the question and list what is given :

    m1m1 = 5000 kg

    m2m2 = 50 kg

    vi1vi1 = vi2vi2 = 275 m··s-1-1

    vf1vf1 = ?

    vf2vf2 = 700 m··s-1-1

  3. Step 3. Apply the Law of Conservation of momentum :

    The helicopter and missile are connected initially and move at the same velocity. We will therefore combine their masses and change the momentum equation as follows:

    p i = p f ( m 1 + m 2 ) v i = m 1 v f 1 + m 2 v f 2 ( 5000 kg + 50 kg ) ( 275 m · s - 1 ) = ( 5000 kg ) ( v f 1 ) + ( 50 kg ) ( 700 m · s - 1 ) 1388750 kg · m · s - 1 - 35000 kg · m · s - 1 = ( 5000 kg ) ( v f 1 ) v f 1 = 270 , 75 m · s - 1 p i = p f ( m 1 + m 2 ) v i = m 1 v f 1 + m 2 v f 2 ( 5000 kg + 50 kg ) ( 275 m · s - 1 ) = ( 5000 kg ) ( v f 1 ) + ( 50 kg ) ( 700 m · s - 1 ) 1388750 kg · m · s - 1 - 35000 kg · m · s - 1 = ( 5000 kg ) ( v f 1 ) v f 1 = 270 , 75 m · s - 1
    (98)

    Note that speed is asked and not velocity, therefore no direction is included in the answer.

Exercise 35: Conservation of Momentum 3

A bullet of mass 50 g travelling horizontally at 500 m··s-1-1strikes a stationary wooden block of mass 2 kg resting on a smooth horizontal surface. The bullet goes through the block and comes out on the other side at 200 m··s-1-1. Calculate the speed of the block after the bullet has come out the other side.

Solution
  1. Step 1. Draw rough sketch of the situation :
    Figure 80
    Figure 80 (PG11C11_074.png)
  2. Step 2. Choose a frame of reference :

    We will choose to the right as positive.

  3. Step 3. Apply the Law of Conservation of momentum :
    p i = p f m 1 v i 1 + m 2 v i 2 = m 1 v f 1 + m 2 v f 2 ( 0 , 05 kg ) ( + 500 m · s - 1 ) + ( 2 kg ) ( 0 m · s - 1 ) = ( 0 , 05 kg ) ( + 200 m · s - 1 ) + ( 2 kg ) ( v f 2 ) 25 + 0 - 10 = 2 v f 2 v f 2 = 7 , 5 m · s - 1 in the same direction as the bullet the same direction as the bullet p i = p f m 1 v i 1 + m 2 v i 2 = m 1 v f 1 + m 2 v f 2 ( 0 , 05 kg ) ( + 500 m · s - 1 ) + ( 2 kg ) ( 0 m · s - 1 ) = ( 0 , 05 kg ) ( + 200 m · s - 1 ) + ( 2 kg ) ( v f 2 ) 25 + 0 - 10 = 2 v f 2 v f 2 = 7 , 5 m · s - 1 in the same direction as the bullet the same direction as the bullet
    (99)

Physics in Action: Impulse

A very important application of impulse is improving safety and reducing injuries. In many cases, an object needs to be brought to rest from a certain initial velocity. This means there is a certain specified change in momentum. If the time during which the momentum changes can be increased then the force that must be applied will be less and so it will cause less damage. This is the principle behind arrestor beds for trucks, airbags, and bending your knees when you jump off a chair and land on the ground.

Air-Bags in Motor Vehicles

Air bags are used in motor vehicles because they are able to reduce the effect of the force experienced by a person during an accident. Air bags extend the time required to stop the momentum of the driver and passenger. During a collision, the motion of the driver and passenger carries them towards the windshield. If they are stopped by a collision with the windshield, it would result in a large force exerted over a short time in order to bring them to a stop. If instead of hitting the windshield, the driver and passenger hit an air bag, then the time of the impact is increased. Increasing the time of the impact results in a decrease in the force.

Padding as Protection During Sports

The same principle explains why wicket keepers in cricket use padded gloves or why there are padded mats in gymnastics. In cricket, when the wicket keeper catches the ball, the padding is slightly compressible, thus reducing the effect of the force on the wicket keepers hands. Similarly, if a gymnast falls, the padding compresses and reduces the effect of the force on the gymnast's body.

Arrestor Beds for Trucks

An arrestor bed is a patch of ground that is softer than the road. Trucks use these when they have to make an emergency stop. When a trucks reaches an arrestor bed the time interval over which the momentum is changed is increased. This decreases the force and causes the truck to slow down.

Follow-Through in Sports

In sports where rackets and bats are used, like tennis, cricket, squash, badminton and baseball, the hitter is often encouraged to follow-through when striking the ball. High speed films of the collisions between bats/rackets and balls have shown that following through increases the time over which the collision between the racket/bat and ball occurs. This increase in the time of the collision causes an increase in the velocity change of the ball. This means that a hitter can cause the ball to leave the racket/bat faster by following through. In these sports, returning the ball with a higher velocity often increases the chances of success.

Crumple Zones in Cars

Another safety application of trying to reduce the force experienced is in crumple zones in cars. When two cars have a collision, two things can happen:

  1. the cars bounce off each other, or
  2. the cars crumple together.

Which situation is more dangerous for the occupants of the cars? When cars bounce off each other, or rebound, there is a larger change in momentum and therefore a larger impulse. A larger impulse means that a greater force is experienced by the occupants of the cars. When cars crumple together, there is a smaller change in momentum and therefore a smaller impulse. The smaller impulse means that the occupants of the cars experience a smaller force. Car manufacturers use this idea and design crumple zones into cars, such that the car has a greater chance of crumpling than rebounding in a collision. Also, when the car crumples, the change in the car's momentum happens over a longer time. Both these effects result in a smaller force on the occupants of the car, thereby increasing their chances of survival.

Egg Throw

This activity demonstrates the effect of impulse and how it is used to improve safety. Have two learners hold up a bed sheet or large piece of fabric. Then toss an egg at the sheet. The egg should not break, because the collision between the egg and the bed sheet lasts over an extended period of time since the bed sheet has some give in it. By increasing the time of the collision, the force of the impact is minimized. Take care to aim at the sheet, because if you miss the sheet, you will definitely break the egg and have to clean up the mess!

Exercise

  1. A canon, mass 500 kg, fires a shell, mass 1 kg, horizontally to the right at 500 m··s-1-1. What is the magnitude and direction of the initial recoil velocity of the canon?
  2. A trolley of mass 1 kg is moving with a speed of 3 m··s-1-1. A block of wood, mass 0,5 kg, is dropped vertically into the trolley. Immediately after the collision, the speed of the trolley and block is 2 m··s-1-1. By way of calculation, show whether momentum is conserved in the collision.
  3. A 7200 kg empty railway truck is stationary. A fertilizer firm loads 10800 kg fertilizer into the truck. A second, identical, empty truck is moving at 10 m··s-1-1when it collides with the loaded truck.
    1. If the empty truck stops completely immediately after the collision, use a conservation law to calculate the velocity of the loaded truck immediately after the collision.
    2. Calculate the distance that the loaded truck moves after collision, if a constant frictional force of 24 kN acts on the truck.
  4. A child drops a squash ball of mass 0,05 kg. The ball strikes the ground with a velocity of 4 m··s-1-1and rebounds with a velocity of 3 m··s-1-1. Does the law of conservation of momentum apply to this situation? Explain.
  5. A bullet of mass 50 g travelling horizontally at 600 m··s-1-1strikes a stationary wooden block of mass 2 kg resting on a smooth horizontal surface. The bullet gets stuck in the block.
    1. Name and state the principle which can be applied to find the speed of the block-and-bullet system after the bullet entered the block.
    2. Calculate the speed of the bullet-and-block system immediately after impact.
    3. If the time of impact was 5 x 10-4-4 seconds, calculate the force that the bullet exerts on the block during impact.

Torque and Levers

Torque

This chapter has dealt with forces and how they lead to motion in a straight line. In this section, we examine how forces lead to rotational motion.

When an object is fixed or supported at one point and a force acts on it a distance away from the support, it tends to make the object turn. The moment of force or torque (symbol, ττ read tau) is defined as the product of the distance from the support or pivot (rr) and the component of force perpendicular to the object, FF.

τ = F · r τ = F · r
(100)

Torque can be seen as a rotational force. The unit of torque is N··m and torque is a vector quantity. Some examples of where torque arises are shown in Figures Figure 81, Figure 82 and Figure 83.

Figure 81: The force exerted on one side of a see-saw causes it to swing.
Figure 81 (PG11C11_075.png)
Figure 82: The force exerted on the edge of a propellor causes the propellor to spin.
Figure 82 (PG11C11_076.png)
Figure 83: The force exerted on a spanner helps to loosen the bolt.
Figure 83 (PG11C11_077.png)

For example in Figure 83, if a force FF of 10 N is applied perpendicularly to the spanner at a distance rr of 0,3 m from the center of the bolt, then the torque applied to the bolt is:

τ = F · r = ( 10 N ) ( 0 , 3 m ) = 3 N · m τ = F · r = ( 10 N ) ( 0 , 3 m ) = 3 N · m
(101)

If the force of 10 N is now applied at a distance of 0,15 m from the centre of the bolt, then the torque is:

τ = F · r = ( 10 N ) ( 0 , 15 m ) = 1 , 5 N · m τ = F · r = ( 10 N ) ( 0 , 15 m ) = 1 , 5 N · m
(102)

This shows that there is less torque when the force is applied closer to the bolt than further away.

Tip:

Loosening a bolt

If you are trying to loosen (or tighten) a bolt, apply the force on the spanner further away from the bolt, as this results in a greater torque to the bolt making it easier to loosen.

Tip:

Any component of a force exerted parallel to an object will not cause the object to turn. Only perpendicular components cause turning.

Tip:

Torques

The direction of a torque is either clockwise or anticlockwise. When torques are added, choose one direction as positive and the opposite direction as negative. If equal clockwise and anticlockwise torques are applied to an object, they will cancel out and there will be no net turning effect.

Exercise 36: Merry-go-round

Several children are playing in the park. One child pushes the merry-go-round with a force of 50 N. The diameter of the merry-go-round is 3,0 m. What torque does the child apply if the force is applied perpendicularly at point A?

Figure 84
Figure 84 (PG11C11_078.png)

Solution
  1. Step 1. Identify what has been given :

    The following has been given:

    • the force applied, FF = 50 N
    • the diameter of the merry-go-round, 2r2r = 3 m, therefore rr = 1,5 m.

    The quantities are in SI units.

  2. Step 2. Decide how to approach the problem :

    We are required to determine the torque applied to the merry-go-round. We can do this by using:

    τ = F · r τ = F · r
    (103)

    We are given FF and we are given the diameter of the merry-go-round. Therefore, rr = 1,5 m.

  3. Step 3. Solve the problem :
    τ = F · r = ( 50 N ) ( 1 , 5 m ) = 75 N · m τ = F · r = ( 50 N ) ( 1 , 5 m ) = 75 N · m
    (104)
  4. Step 4. Write the final answer :

    75 N·mN·m of torque is applied to the merry-go-round.

Exercise 37: Flat tyre

Kevin is helping his dad replace the flat tyre on the car. Kevin has been asked to undo all the wheel nuts. Kevin holds the spanner at the same distance for all nuts, but applies the force at two angles (90 and 60). If Kevin applies a force of 60 N, at a distance of 0,3 m away from the nut, which angle is the best to use? Prove your answer by means of calculations.

Figure 85
Figure 85 (PG11C11_079.png)

Solution
  1. Step 1. Identify what has been given :

    The following has been given:

    • the force applied, FF = 60 N
    • the angles at which the force is applied, θ=90θ=90 and θ=60θ=60
    • the distance from the centre of the nut at which the force is applied, rr = 0,3 m

    The quantities are in SI units.

  2. Step 2. Decide how to approach the problem :

    We are required to determine which angle is more better to use. This means that we must find which angle gives the higher torque. We can use

    τ = F · r τ = F · r
    (105)

    to determine the torque. We are given FF for each situation. F=FsinθF=Fsinθ and we are given θθ. We are also given the distance away from the nut, at which the force is applied.

  3. Step 3. Solve the problem for θ=90θ=90 :

    F = F F = F

    τ = F · r = ( 60 N ) ( 0 , 3 m ) = 18 N · m τ = F · r = ( 60 N ) ( 0 , 3 m ) = 18 N · m
    (106)
  4. Step 4. Solve the problem for θ=60θ=60 :
    τ = F · r = F sin θ · r = ( 60 N ) sin ( θ ) ( 0 , 3 m ) = 15 , 6 N · m τ = F · r = F sin θ · r = ( 60 N ) sin ( θ ) ( 0 , 3 m ) = 15 , 6 N · m
    (107)
  5. Step 5. Write the final answer :

    The torque from the perpendicular force is greater than the torque from the force applied at 6060. Therefore, the best angle is 9090.

Mechanical Advantage and Levers

We can use our knowlegde about the moments of forces (torque) to determine whether situations are balanced. For example two mass pieces are placed on a seesaw as shown in Figure 86. The one mass is 3 kg and the other is 6 kg. The masses are placed at distances of 2 m and 1 m (respectively) from the pivot. By looking at the clockwise and anti-clockwise moments, we can determine whether the seesaw will pivot (move) or not. If the sum of the clockwise and anti-clockwise moments is zero, the seesaw is in equilibrium (i.e. balanced).

Figure 86: The moments of force are balanced.
Figure 86 (PG11C11_080.png)

The clockwise moment can be calculated as follows:

τ 1 = F · r τ 1 = ( 6 kg ) ( 9 , 8 m · s - 2 ) ( 1 m ) τ 1 = 58 , 8 N · m clockwise τ 1 = F · r τ 1 = ( 6 kg ) ( 9 , 8 m · s - 2 ) ( 1 m ) τ 1 = 58 , 8 N · m clockwise
(108)

The anti-clockwise moment can be calculated as follows:

τ 2 = F · r τ 2 = ( 3 kg ) ( 9 , 8