This module is part of a collection (see http://cnx.org/content/col11294/latest/ ) of modules designed to make physics concepts accessible to blind students. The collection is intended to supplement but not to replace the textbook in an introductory course in high school or college physics.

This module explains various units of force in a format that is accessible to blind students.

In addition to an Internet connection and a browser, you will need the following tools (as a minimum) to work through the exercises in these modules:

The minimum prerequisites for understanding the material in these modules include:

I recommend that you open another copy of this document in a separate browser window and use the following links to easily find and view the figures while you are reading about them.

I recommend that you also study the other lessons in my extensive collection of online programming tutorials. You will find a consolidated index at www.DickBaldwin.com .

The module titled Manual Creation of Tactile Graphics at http://cnx.org/content/m38546/latest/ explained how to create tactile graphics from svg files that I will provide.

If you are going to have an assistant create tactile graphics for this module, you will need to download the file named Phy1150.zip , which contains the svg file for this module. Extract the svg file from the zip file and provide them to your assistant.

Also, if you are going to use tactile graphics, it probably won't be necessary for you to perform the graph board exercises. However, you should still walk through the graph board exercises in your mind because I will often embed important physics concepts in the instructions for doing the graph board exercises.

In each case where I am providing an svg file for the creation of tactile graphics, I will identify the name of the appropriate svg file and display an image of the contents of the file for the benefit of your assistant. As explained at http://cnx.org/content/m38546/latest/ , those images will be mirror images of the actual images so that your assistant can emboss the image from the back of the paper and you can explore it from the front.

I will also display a non-mirror-image version of the image so that your assistant can easily read the text in the image.

Also in those cases, I will provide a table of key-value pairs that explain how the Braille keys in the image relate to text or objects in the image.

One of the more confusing things about physics textbooks can be their treatment of the units in which force is measured and reported. Several different units are often used including:

The first three units in the above list are said to be absolute units of force because they are measured in fundamental units of mass, length, and time. The last three units are tied directly to the gravitational attraction between the earth and other objects.

The newton, which is an SI derived unit, is possibly the most universally accepted unit of force in the year 2011.

Force equals the product of mass and acceleration

You learned in an earlier module that the force that is required to cause a given mass to be accelerated by a given amount is proportional to the product of the mass and the acceleration. If we specify mass and acceleration in consistent units, we can write

f = m * a

where

Many physics books use a system of units called SI units . SI is an abbreviation for a French name, which I am unable to pronounce, and which is probably also not compatible with your screen reader and your Braille display.

I won't attempt to explain much about SI units in this module. I provided some information in an earlier module titled Units and Dimensional Analysis . Also, you can probably find a good explanation in your textbook, and if not, you can Google SI units and find hundreds of web pages that explain the system in varying levels of detail.

Tables of SI units

Most of those references will probably also provide tables for the units, but those tables may be partially incompatible with your screen reader and Braille display due to the extensive use of superscripts. Therefore, I will provide tables that should be accessible in Figure 1 and Figure 2 .

Base units and derived units

When reading about SI units, you will find that they are often divided into base units and derived units. I will put the base units in Figure 1 and some sample derived units in Figure 2 .

Base Quantity               Name        Symbol

length                      meter       m
mass                        kilogram    kg
time                        second      s
electric current            ampere      A
thermodynamic temperature   kelvin      K
amount of substance         mole        mol
luminous intensity          candela     cd

Note that the list of derived units in Figure 2 is only a sampling of different units that can be derived from the base units.

The exponentiation indicator

As is the case throughout these modules, the character "^" that you see used extensively in Figure 2 indicates that the character following the ^ is an exponent. Note also that when the exponent is negative, it is enclosed along with its minus sign in parentheses for clarity.

force            newton                    kg*m/s^2
area             square meter              m^2
volume           cubic meter               m^3
speed, velocity  meter per second          m/s
acceleration     meter per second squared  m/s^2
wave number      reciprocal meter          m^(-1)
mass density     kilogram per cubic meter  kg/m^3
specific volume  cubic meter per kilogram  m^3/kg
current density  ampere per square meter   A/m^2

The newton

The first derived unit listed in Figure 2 is the newton, which is the product of the base unit for mass (kg) and the base unit for acceleration (m/s^2). Thus, the units for the newton are

kg*m/s^2

A newton is a unit of force that causes a mass of one kilogram to be accelerated by one meter per second squared (1 m/s^2).

When making physics calculations, It is extremely important that you understand and keep track of the units that you are using. The Google search box can serve your needs as a scientific calculator if you are careful how you use it. For example, if you enter an expression such as 3+5 in the Google search box and press the Enter key, the result of evaluating that expression will be displayed immediately below the search box.

A physics problem

A typical problem in a physics textbook states that a 2-kg mass is moving in a circular path with a constant angular velocity of 5 radians per second and with a tangential velocity of 3 m/sec. The objective is to find the centripetal force on the mass.

You will learn all that you need to know to solve this problem in future modules. For now, just bear with me and concentrate on the use of the Google search box as a scientific calculator.

As you will learn in a future module, a radian is a dimensionless quantity, and the proper units for an angular velocity of 5 radians per second is simply 5/s. (It also works to spell radians out, but abbreviations for radians may not work in the Google calculator.)

The algebraic solution

Once you work through the algebra for this problem, you will have determined that the answer to the problem is given by the following expression:

centripetal force = 2kg * (5/s) *(3m/s)

Let Google do the work

Enter the following expression into the Google search box and press the Enter key:

2kg * (5/s) *(3m/s)

The following text should appear immediately below the search box.

2 kg * (5 / s) * (3 (m / s)) = 30 newtons

Note that in this case, I used the correct symbols for SI units.

Do it again with the units spelled differently

Enter the following expression into the search box and press the Enter key.

2 kilograms * (5 radians/second) * (3 meters/second)

The following text should appear immediately below the search box.

2 kilograms * ((5 radians) / second) * (3 (meters / second)) = 30 newtons

Do it one more time

Let's do it one more time, this time mixing metric and English units. Enter the following expression into the search box and press the Enter key.

2 kilograms * (5 radians/second) * (9.8425197 feet/second)

The following output should appear below the search box:

2 kilograms * ((5 radians) / second) * (9.8425197 (feet / second)) = 30 newtons

The conclusion

Given an input with units in the correct format, the Google calculator is not only able to deal with those units and perform the calculation correctly, it is also able to properly convert the result to the correct value in derived units (newtons in this case).

When analyzing a physics problem, the safest approach is probably to convert all of the given information into SI units and solve the problem in SI units, converting the result back to some other system of units if required. However, some textbooks and some physics professors may not allow that approach. Therefore, you need to know something about the other units of force that you may encounter.

The values for the next three units of force that I will explain depend on the gravitational attraction between the earth and other objects.

The universal law of gravitation tells us that objects having mass are attracted to one another by a force that is proportional to the product of their masses and inversely proportional to the distance between them.

attractive force = G*m1*m2/d^2

where

It is also true that for large spherical masses like the earth, the effect is as if all of the mass is concentrated at a point located at the center of the sphere.

Objects are attracted to the earth

Therefore, all objects on or near the surface of the earth are attracted toward the center of the earth by an amount given by the above equation .

Neglecting friction such as air resistance, objects near the surface at different points on earth fall with an acceleration somewhere between about 9.78 m/s^2 and 9.82 m/s^2 depending on latitude.

Despite the fact that scales in the grocery stores in the United States have displays that read in pounds (which is a unit of mass), weight is not a measure of mass. Weight is a measure of force.

What exactly is the weight of an object?

The measurement that we normally think of as the weight of an object with a given mass is the force exerted on that object at the surface of the earth by the gravitational attraction between that object and the earth.

The weight of that same object on the surface of the moon would be the force exerted on that object by the gravitational attraction between that object and the moon.

Weighing a package of hamburger

Assume that you have a package that contains a one-pound mass of hamburger. You put it on a scale on the surface of the earth and the display reads 1 pound.

Now assume that you take the same scale and the same package of hamburger to the surface of the moon and place the package on the scale. The display would no longer read 1 pound.

The weight indicated by the scale would be different because the gravitational attraction between the mass of the hamburger and the mass of the moon would be less than the gravitational attraction between the mass of the hamburger and the mass of the earth.

According to the calculator at http://www.moonconnection.com/moon_gravity.phtml , the scale would read 0.2 pounds on the moon.

What does the output of the scale really mean?

When you see a scale with a display that reads pound, it should read pound-force instead. If it reads kilogram, it should read kilogram-force instead. Pound-force, kilogram-force, and gram-force are units that are tied directly to the gravitational attraction between the earth and other objects.

Weightlessness

When astronauts go into space and speak of being weightless, they probably aren't completely weightless. However, their weight is probably so low that it seems to them to be zero.

The reduction in an astronaut's weight occurs because the distance between the astronaut and any large massive object (such as the earth or the moon) is so great that the gravitational attraction between them is very small.

It will probably help you to keep track of everything if you draw the scenario on your graph board.

Draw a side view of two cubes with different masses on the top of a flat level horizontal friction-free table. Label them Mass C and Mass B. Mass C is on the left and Mass B is on the right.

Mass B is close to the rightmost edge of the table and Mass C is to the left of Mass B.

Label Mass C as 3 kg and label Mass B as 2 kg.

A mass hanging on a cord

Draw a strong but lightweight cord, connected to the right side of Mass B and thread the cord over a very light frictionless pulley that changes the orientation of the cord from horizontal to vertical. The pulley is attached to the right edge of the table.

Draw a triangle-shaped mass connected to the cord that hangs down from the pulley. Label this Mass A and label it as 5 kg.

Two additional cords

Draw a strong but lightweight cord connecting Mass B to Mass C.

Draw another strong but lightweight cord connecting Mass C to a vertical wall on the left side of Mass C. That cord prevents any of the masses from moving and keeps the entire system in equilibrium with the 5-kg mass suspended from the cord that is threaded over the pulley.

Everything should be lined up

The pulley, both masses, and all three horizontal segments of cord above the tabletop should be in a straight line. The attachment heights of the cords and the top of the pulley should be such that all of the cords above the tabletop are horizontal. In other words, there should be no vertical components in any of the forces that act on the cords above the tabletop.

Label the tension in each cord

Label the tension in the cord between Mass B and Mass A as P. Label the tension in the cord between Mass C and Mass B as Q Label the tension in the cord between the wall and Mass C as R.

Tactile graphics

The svg file that is required to create tactile graphics for this exercise is named Phy1150a1.svg. You should have downloaded that file earlier. This file contains an image that represents the instructions given above .

Figure 3 shows the mirror image that is contained in that file for the benefit of your assistant who will create the tactile graphic for this exercise.

Figure 3: Mirror image of the image from the file named Phy1150a1.svg.

Mirror image of the image from the file named Phy1150a1.svg.

Figure 4 shows a non-mirror-image version of the same image.

Figure 4: Non-mirror image of the image from the file named Phy1150a1.svg.

Non-mirror image of the image from the file named Phy1150a1.svg.

Figure 5 shows the key-value pairs that go with the image in the file named Phy1150a1.svg

m: R
n: Q
o: Friction free pulley
p: Friction free table
q: P
r: Mass A 5 kg
s: Mass C 3 kg
t: Mass B 2 kg
u: A static scenario
v: File: Phy1150a1.svg

The question

What are the tensions P, Q, and R in the individual cord segments when the system is in equilibrium and nothing is moving?

The answers

Mass A is in equilibrium because the upward force exerted by the cord attached to the top of Mass A is equal to the downward force of gravity that is exerted on Mass A.

Therefore, when the system is in equilibrium, the tension P is equal to the weight of Mass A, which is equal to the product of its mass and the acceleration of gravity.

P = 5*kg*9.81*m/s^2 = 49.05 newtons

The effect of the pulley

A single-wheel, frictionless, lightweight pulley as described here changes the direction of the cord, but does not change the tension P in the cord. The tension in the cord is the same on both sides of the pulley.

Support from the table

Mass B and Mass C are both in vertical equilibrium because their weight is being supported by the table.

Horizontal equilibrium

Mass B is in horizontal equilibrium because the force exerted on one side of the mass by tension Q is equal to the force exerted on the other side of the mass by tension P, which is 49.05 newtons.

Mass C is in horizontal equilibrium because the force exerted on one side of the mass by tension R is equal to the force exerted on the other side of the mass by tension Q, which as explained above, is 49.05 newtons.

The wall exerts a force that is equal in magnitude and opposite in direction to tension R in the cord that attaches Mass C to the wall.

Tensions are all the same

Therefore, the tensions in all three cords are the same and the force exerted by the wall supports the weight of Mass A hanging by the cord on the other end of the chain.

P = Q = R = 49.05 newtons

Now assume that someone cuts the cord that attaches Mass A to Mass B. What happens to the tension in each cord? Which, if any of the masses move, and if they move, what is their acceleration?

We probably don't need to do any calculations to answer these questions. Life experience tells us that the tension in each cord immediately goes to zero when the cord holding up the weight is cut.

Movement

Mass B and Mass C remain in equilibrium with no horizontal forces acting on them and their weights being supported by the table. They don't move.

Mass A immediately begins a free fall toward the floor with an acceleration that is equal to the acceleration of gravity at 9.81 m/s^2. A short segment of cord trails out behind Mass A like an unopened parachute.

Now assume that we start over with the original static scenario and someone cuts the cord that attaches Mass A to Mass B. What happens to the tension in each cord? Which, if any of the masses move, and if they move, what is their acceleration?

This situation is a little more complicated and will probably require some calculations to sort out.

It seem obvious that the tensions labeled Q and R immediately go to zero, but the tension labeled P does not go to zero.

Movement

Mass C remains in equilibrium with no horizontal forces acting on it and its weight being supported by the table. It does not move.

However, Mass A starts falling toward the floor, dragging Mass B horizontally towards the pulley.

The driving force

The only force causing Mass A and Mass B to move is the weight of Mass A (49.05 newtons), which has not changed. However, that force is now trying to move a total of 7 kg instead of only 5 kg as in the first dynamic scenario above.

A force of 49.05 newtons is not sufficient to cause a mass of 7 kg to accelerate at 9.81 m/s^2. Instead, the acceleration of each mass is proportional to the force and inversely proportional to the total mass.

a = f/m = (49.05*newtons)/(7*kg)

Entering the rightmost expression into the Google search box and pressing Enter tells us that

a = 7.0 m / s^2

Thus, the acceleration of Mass A and Mass B is 7 m/s^2, a little less than the acceleration of gravity.

What is the value of tension P?

Tension P is exerting a horizontal force on the right side of Mass B that is causing that mass to accelerate at 7.0 m / s^2. The force required to achieve that acceleration on a mass of 2 kg is

P = m*a = 2 kg*7.0 m / s^2

Once again using the Google calculator, we learn that

P = 14 newtons

Thus, although the tension at P did not go to zero when the cord was cut at Q, the resulting tension in the cord at P was substantially reduced relative to the tension at P while the system was in equilibrium.

Now assume that we start over with the original static scenario and someone cuts the cord that attaches Mass C to the wall. What happens to the tension in each cord? Which, if any of the masses move, and if they move, what is their acceleration?

This situation is considerably more complicated and will definitely require some calculations. The tension labeled R goes to zero immediately, but the tensions labeled P and Q do not go to zero.

Movement

None of the masses remain in equilibrium. Mass A starts falling toward the floor, dragging Mass B and Mass C horizontally toward the pulley.

The driving force

Once again, the only force causing all three masses to move is the weight of Mass A (49.05 newtons), which has not changed. However, that force is now trying to move a total of 10 kg instead of only 5 kg or 7 kg as in the two scenarios described above.

A force of 49.05 newtons is not sufficient to cause a mass of 10 kg to accelerate at 9.81 m/s^2. Instead, the acceleration of each mass is proportional to the force and inversely proportional to the total mass.

a = f/m = (49.05*newtons)/(10*kg)

Entering the rightmost expression into the Google search box and pressing Enter tells us that

a = 4.9 m / s^2

Half the acceleration of gravity

Note that this is half the acceleration of gravity. This makes sense, because the force attributable to gravitational attraction acting on a mass of 5 kg is being applied to 10 kg of mass. It follows, therefore, that the acceleration that is achieved will be only half the acceleration of gravity.

What is the value of tension P?

Tension P is exerting a horizontal force on the right side of Mass B that is causing that mass and Mass C to accelerate at 4.9 m / s^2. The force required to achieve that acceleration on a mass of 5 kg is

P = m*a = 5*kg*4.9 m / s^2

Once again using the Google calculator, we learn that

P = 24.5 newtons

Mass B and Mass C together represent 50-percent of the total mass, and tension P is 50-percent of the force applied to the total mass.

What is the value of tension Q?

Tension Q is exerting a horizontal force on the right side of Mass C that is causing Mass C to accelerate at 4.9 m / s^2. The force required to achieve that acceleration on a mass of 3 kg is

Q = m*a = 3*kg*4.9 m / s^2

Once again using the Google calculator, we learn that

Q = 14.7 newtons

Mass C is 30-percent of the total mass, and tension Q is 30-percent of the force applied to the total mass.