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Vectors: Components (Grade 11)

Module by: Free High School Science Texts Project. E-mail the author

Components of Vectors

In the discussion of vector addition we saw that a number of vectors acting together can be combined to give a single vector (the resultant). In much the same way a single vector can be broken down into a number of vectors which when added give that original vector. These vectors which sum to the original are called components of the original vector. The process of breaking a vector into its components is called resolving into components.

While summing a given set of vectors gives just one answer (the resultant), a single vector can be resolved into infinitely many sets of components. In the diagrams below the same black vector is resolved into different pairs of components. These components are shown as dashed lines. When added together the dashed vectors give the original black vector (i.e. the original vector is the resultant of its components).

Figure 1
Figure 1 (PG11C1_058.png)

In practice it is most useful to resolve a vector into components which are at right angles to one another, usually horizontal and vertical.

Any vector can be resolved into a horizontal and a vertical component. If AA is a vector, then the horizontal component of AA is AxAx and the vertical component is AyAy.

Figure 2
Figure 2 (PG11C1_059.png)

Exercise 1: Resolving a vector into components

A motorist undergoes a displacement of 250 km in a direction 30 north of east. Resolve this displacement into components in the directions north (xNxN) and east (xExE).

Solution

  1. Step 1. Draw a rough sketch of the original vector :

    Figure 3
    Figure 3 (PG11C1_060.png)

  2. Step 2. Determine the vector component :

    Next we resolve the displacement into its components north and east. Since these directions are perpendicular to one another, the components form a right-angled triangle with the original displacement as its hypotenuse.

    Figure 4
    Figure 4 (PG11C1_061.png)

    Notice how the two components acting together give the original vector as their resultant.

  3. Step 3. Determine the lengths of the component vectors :

    Now we can use trigonometry to calculate the magnitudes of the components of the original displacement:

    x N = ( 250 ) ( sin 30 ) = 125 km x N = ( 250 ) ( sin 30 ) = 125 km
    (1)

    and

    x E = ( 250 ) ( cos 30 ) = 216 , 5 km x E = ( 250 ) ( cos 30 ) = 216 , 5 km
    (2)

    Remember xNxN and xExE are the magnitudes of the components – they are in the directions north and east respectively.

Block on an incline

As a further example of components let us consider a block of mass mm placed on a frictionless surface inclined at some angle θθ to the horizontal. The block will obviously slide down the incline, but what causes this motion?

The forces acting on the block are its weight mgmg and the normal force NN exerted by the surface on the object. These two forces are shown in the diagram below.

Figure 5
Figure 5 (PG11C1_062.png)

Now the object's weight can be resolved into components parallel and perpendicular to the inclined surface. These components are shown as dashed arrows in the diagram above and are at right angles to each other. The components have been drawn acting from the same point. Applying the parallelogram method, the two components of the block's weight sum to the weight vector.

To find the components in terms of the weight we can use trigonometry:

F g = m g sin θ F g = m g cos θ F g = m g sin θ F g = m g cos θ
(3)

The component of the weight perpendicular to the slope FgFg exactly balances the normal force NN exerted by the surface. The parallel component, however, FgFg is unbalanced and causes the block to slide down the slope.

Worked example

Exercise 2: Block on an incline plane

Determine the force needed to keep a 10 kg block from sliding down a frictionless slope. The slope makes an angle of 30 with the horizontal.

Solution
  1. Step 1. Draw a diagram of the situation :

    Figure 6
    Figure 6 (PG11C1_063.png)

    The force that will keep the block from sliding is equal to the parallel component of the weight, but its direction is up the slope.

  2. Step 2. Calculate FgFg :
    F g = m g sin θ = ( 10 ) ( 9 , 8 ) ( sin 30 ) = 49 N F g = m g sin θ = ( 10 ) ( 9 , 8 ) ( sin 30 ) = 49 N
    (4)
  3. Step 3. Write final answer :

    The force is 49 N up the slope.

Vector addition using components

Components can also be used to find the resultant of vectors. This technique can be applied to both graphical and algebraic methods of finding the resultant. The method is simple: make a rough sketch of the problem, find the horizontal and vertical components of each vector, find the sum of all horizontal components and the sum of all the vertical components and then use them to find the resultant.

Consider the two vectors, AA and BB, in Figure 7, together with their resultant, RR.

Figure 7: An example of two vectors being added to give a resultant
Figure 7 (PG11C1_064.png)

Each vector in Figure 7 can be broken down into one component in the xx-direction (horizontal) and one in the yy-direction (vertical). These components are two vectors which when added give you the original vector as the resultant. This is shown in Figure 8 where we can see that:

A = A x + A y B = B x + B y R = R x + R y A = A x + A y B = B x + B y R = R x + R y
(5)
But , R x = A x + B x and R y = A y + B y But , R x = A x + B x and R y = A y + B y
(6)

In summary, addition of the xx components of the two original vectors gives the xx component of the resultant. The same applies to the yy components. So if we just added all the components together we would get the same answer! This is another important property of vectors.

Figure 8: Adding vectors using components.
Figure 8 (PG11C1_065.png)

Exercise 3: Adding Vectors Using Components

If in Figure 8, A=5,385m·s-1A=5,385m·s-1 at an angle of 21.8 to the horizontal and B=5m·s-1B=5m·s-1 at an angle of 53,13 to the horizontal, find RR.

Solution
  1. Step 1. Decide how to tackle the problem :

    The first thing we must realise is that the order that we add the vectors does not matter. Therefore, we can work through the vectors to be added in any order.

  2. Step 2. Resolve AA into components :

    We find the components of AA by using known trigonometric ratios. First we find the magnitude of the vertical component, AyAy:

    sin θ = A y A sin 21 , 8 = A y 5 , 385 A y = ( 5 , 385 ) ( sin 21 , 8 ) = 2 m · s - 1 sin θ = A y A sin 21 , 8 = A y 5 , 385 A y = ( 5 , 385 ) ( sin 21 , 8 ) = 2 m · s - 1
    (7)

    Secondly we find the magnitude of the horizontal component, AxAx:

    cos θ = A x A cos 21 . 8 = A x 5 , 385 A x = ( 5 , 385 ) ( cos 21 , 8 ) = 5 m · s - 1 cos θ = A x A cos 21 . 8 = A x 5 , 385 A x = ( 5 , 385 ) ( cos 21 , 8 ) = 5 m · s - 1
    (8)

    Figure 9
    Figure 9 (PG11C1_066.png)

    The components give the sides of the right angle triangle, for which the original vector, AA, is the hypotenuse.

  3. Step 3. Resolve BB into components :

    We find the components of BB by using known trigonometric ratios. First we find the magnitude of the vertical component, ByBy:

    sin θ = B y B sin 53 , 13 = B y 5 B y = ( 5 ) ( sin 53 , 13 ) = 4 m · s - 1 sin θ = B y B sin 53 , 13 = B y 5 B y = ( 5 ) ( sin 53 , 13 ) = 4 m · s - 1
    (9)

    Secondly we find the magnitude of the horizontal component, BxBx:

    cos θ = B x B cos 21 , 8 = B x 5 , 385 B x = ( 5 , 385 ) ( cos 53 , 13 ) = 5 m · s - 1 cos θ = B x B cos 21 , 8 = B x 5 , 385 B x = ( 5 , 385 ) ( cos 53 , 13 ) = 5 m · s - 1
    (10)

    Figure 10
    Figure 10 (PG11C1_067.png)

  4. Step 4. Determine the components of the resultant vector :

    Now we have all the components. If we add all the horizontal components then we will have the xx-component of the resultant vector, RxRx. Similarly, we add all the vertical components then we will have the yy-component of the resultant vector, RyRy.

    R x = A x + B x = 5 m · s - 1 + 3 m · s - 1 = 8 m · s - 1 R x = A x + B x = 5 m · s - 1 + 3 m · s - 1 = 8 m · s - 1
    (11)

    Therefore, RxRx is 8 m to the right.

    R y = A y + B y = 2 m · s - 1 + 4 m · s - 1 = 6 m · s - 1 R y = A y + B y = 2 m · s - 1 + 4 m · s - 1 = 6 m · s - 1
    (12)

    Therefore, RyRy is 6 m up.

  5. Step 5. Determine the magnitude and direction of the resultant vector :

    Now that we have the components of the resultant, we can use the Theorem of Pythagoras to determine the magnitude of the resultant, RR.

    R 2 = ( R x ) 2 + ( R y ) 2 R 2 = ( 6 ) 2 + ( 8 ) 2 R 2 = 100 R = 10 m · s - 1 R 2 = ( R x ) 2 + ( R y ) 2 R 2 = ( 6 ) 2 + ( 8 ) 2 R 2 = 100 R = 10 m · s - 1
    (13)

    Figure 11
    Figure 11 (PG11C1_068.png)

    The magnitude of the resultant, RR is 10 m. So all we have to do is calculate its direction. We can specify the direction as the angle the vectors makes with a known direction. To do this you only need to visualise the vector as starting at the origin of a coordinate system. We have drawn this explicitly below and the angle we will calculate is labeled αα.

    Using our known trigonometric ratios we can calculate the value of αα;

    tan α = 6 m · s - 1 8 m · s - 1 α = tan - 1 6 m · s - 1 8 m · s - 1 α = 36 , 8 . tan α = 6 m · s - 1 8 m · s - 1 α = tan - 1 6 m · s - 1 8 m · s - 1 α = 36 , 8 .
    (14)
  6. Step 6. Quote the final answer :

    RR is 10 m at an angle of 36,836,8 to the positive xx-axis.

Adding and Subtracting Components of Vectors

  1. Harold walks to school by walking 600 m Northeast and then 500 m N 40o W. Determine his resultant displacement by means of addition of components of vectors.
  2. A dove flies from her nest, looking for food for her chick. She flies at a velocity of 2 m··s-1-1 on a bearing of 135oo in a wind with a velocity of 1,2 m··s-1-1 on a bearing of 230oo. Calculate her resultant velocity by adding the horizontal and vertical components of vectors.

Vector Multiplication

Vectors are special, they are more than just numbers. This means that multiplying vectors is not necessarily the same as just multiplying their magnitudes. There are two different types of multiplication defined for vectors. You can find the dot product of two vectors or the cross product.

The dot product is most similar to regular multiplication between scalars. To take the dot product of two vectors, you just multiply their magnitudes to get out a scalar answer. The mathematical definition of the dot product is:

a b = | a | · | b | cos θ a b = | a | · | b | cos θ
(15)

Take two vectors aa and bb:

Figure 12
Figure 12 (PG11C1_069.png)

You can draw in the component of bb that is parallel to aa:

Figure 13
Figure 13 (PG11C1_070.png)

In this way we can arrive at the definition of the dot product. You find how much of bb is lined up with aa by finding the component of bb parallel to aa. Then multiply the magnitude of that component, |b||b|cosθcosθ, with the magnitude of aa to get a scalar.

The second type of multiplication, the cross product, is more subtle and uses the directions of the vectors in a more complicated way. The cross product of two vectors, aa and bb, is written a×ba×b and the result of this operation on aa and bb is another vector. The magnitude of the cross product of these two vectors is:

| a × b | = | a | | b | sin θ | a × b | = | a | | b | sin θ
(16)

We still need to find the direction of a×ba×b. We do this by applying the right hand rule.

Method: Right Hand Rule

  1. Using your right hand:
  2. Point your index finger in the direction of aa.
  3. Point the middle finger in the direction of bb.
  4. Your thumb will show the direction of a×ba×b.
Figure 14
Figure 14 (PG11C1_071.png)

Figure 15

Summary

  1. A scalar is a physical quantity with magnitude only.
  2. A vector is a physical quantity with magnitude and direction.
  3. Vectors may be represented as arrows where the length of the arrow indicates the magnitude and the arrowhead indicates the direction of the vector.
  4. The direction of a vector can be indicated by referring to another vector or a fixed point (eg. 30 from the river bank); using a compass (eg. N 30 W); or bearing (eg. 053).
  5. Vectors can be added using the head-to-tail method, the parallelogram method or the component method.
  6. The resultant of a number of vectors is the single vector whose effect is the same as the individual vectors acting together.

End of chapter exercises: Vectors

  1. An object is suspended by means of a light string. The sketch shows a horizontal force FF which pulls the object from the vertical position until it reaches an equilibrium position as shown. Which one of the following vector diagrams best represents all the forces acting on the object?
    Figure 16
    Figure 16 (PG11C1_072.png)
    Table 1
    ABCD
    Figure 17
    Figure 17 (PG11C1_073.png)
    Figure 18
    Figure 18 (PG11C1_074.png)
    Figure 19
    Figure 19 (PG11C1_075.png)
    Figure 20
    Figure 20 (PG11C1_076.png)
  2. A load of weight WW is suspended from two strings. F1F1 and F2F2 are the forces exerted by the strings on the load in the directions show in the figure above. Which one of the following equations is valid for this situation?
    1. A: W=F12+F22W=F12+F22
    2. B: F1sin50=F2sin30F1sin50=F2sin30
    3. C: F1cos50=F2cos30F1cos50=F2cos30
    4. D: W=F1+F2W=F1+F2
    Figure 21
    Figure 21 (PG11C1_077.png)
  3. Two spring balances PP and QQ are connected by means of a piece of string to a wall as shown. A horizontal force of 100 N is exerted on spring balance Q. What will be the readings on spring balances PP and QQ?
    Figure 22
    Figure 22 (PG11C1_078.png)
    Table 2
     PQ
    A100 N0 N
    B25 N75 N
    C50 N50 N
    D100 N100 N
  4. 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.
  5. 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?
    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
    Figure 23
    Figure 23 (PG11C1_079.png)
  6. Which of the following contains two vectors and a scalar?
    1. A: distance, acceleration, speed
    2. B: displacement, velocity, acceleration
    3. C: distance, mass, speed
    4. D: displacement, speed, velocity
  7. Two vectors act on the same point. What should the angle between them be so that a maximum resultant is obtained?
    1. A: 0
    2. B: 90
    3. C: 180
    4. D: cannot tell
  8. Two forces, 4 N and 11 N, act on a point. Which one of the following cannot be the magnitude of a resultant?
    1. A: 4 N
    2. B: 7 N
    3. C: 11 N
    4. D: 15 N

End of chapter exercises: Vectors - Long questions

  1. A helicopter flies due east with an air speed of 150 km.h-1-1. It flies through an air current which moves at 200 km.h-1-1 north. Given this information, answer the following questions:
    1. In which direction does the helicopter fly?
    2. What is the ground speed of the helicopter?
    3. Calculate the ground distance covered in 40 minutes by the helicopter.
  2. A plane must fly 70 km due north. A cross wind is blowing to the west at 30 km.h-1-1. In which direction must the pilot steer if the plane flies at a speed of 200 km.h-1-1 in windless conditions?
  3. A stream that is 280 m wide flows along its banks with a velocity of 1.80m.s-1-1. A raft can travel at a speed of 2.50 m.s-1-1 across the stream. Answer the following questions:
    1. What is the shortest time in which the raft can cross the stream?
    2. How far does the raft drift downstream in that time?
    3. In what direction must the raft be steered against the current so that it crosses the stream perpendicular to its banks?
    4. How long does it take to cross the stream in part c?
  4. A helicopter is flying from place XX to place YY. YY is 1000 km away in a direction 5050 east of north and the pilot wishes to reach it in two hours. There is a wind of speed 150 km.h-1-1 blowing from the northwest. Find, by accurate construction and measurement (with a scale of 1 cm =50 km .h-11 cm =50 km .h-1), the
    1. the direction in which the helicopter must fly, and
    2. the magnitude of the velocity required for it to reach its destination on time.
  5. An aeroplane is flying towards a destination 300 km due south from its present position. There is a wind blowing from the north east at 120 km.h-1-1. The aeroplane needs to reach its destination in 30 minutes. Find, by accurate construction and measurement (with a scale of 1 cm =30 km .s-11 cm =30 km .s-1), or otherwise,
    1. the direction in which the aeroplane must fly and
    2. the speed which the aeroplane must maintain in order to reach the destination on time.
    3. Confirm your answers in the previous 2 subquestions with calculations.
  6. 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 by calculation.
    Figure 24
    Figure 24 (PG11C1_080.png)
  7. In a map-work exercise, hikers are required to walk from a tree marked A on the map to another tree marked B which lies 2,0 km due East of A. The hikers then walk in a straight line to a waterfall in position C which has components measured from B of 1,0 km E and 4,0 km N.
    1. Distinguish between quantities that are described as being vector and scalar.
    2. Draw a labelled displacement-vector diagram (not necessarily to scale) of the hikers' complete journey.
    3. What is the total distance walked by the hikers from their starting point at A to the waterfall C?
    4. What are the magnitude and bearing, to the nearest degree, of the displacement of the hikers from their starting point to the waterfall?
  8. An object XX is supported by two strings, AA and BB, attached to the ceiling as shown in the sketch. Each of these strings can withstand a maximum force of 700 N. The weight of XX is increased gradually.
    1. Draw a rough sketch of the triangle of forces, and use it to explain which string will break first.
    2. Determine the maximum weight of XX which can be supported.
    Figure 25
    Figure 25 (PG11C1_081.png)
  9. 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 in any part is 1000 N. Find the largest mass (mm), in kg, that can be carried at the midpoint of the rope, without breaking the rope. Include a vector diagram in your answer.
    Figure 26
    Figure 26 (PG11C1_082.png)

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