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Adding and subtracting vectors

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

Techniques of Vector Addition

Now that you have learned about the mathematical properties of vectors, we return to vector addition in more detail. There are a number of techniques of vector addition. These techniques fall into two main categories - graphical and algebraic techniques.

Graphical Techniques

Graphical techniques involve drawing accurate scale diagrams to denote individual vectors and their resultants. We next discuss the two primary graphical techniques, the head-to-tail technique and the parallelogram method.

The Head-to-Tail Method

In describing the mathematical properties of vectors we used displacements and the head-to-tail graphical method of vector addition as an illustration. The head-to-tail method of graphically adding vectors is a standard method that must be understood.

Method: Head-to-Tail Method of Vector Addition

  1. Draw a rough sketch of the situation.
  2. Choose a scale and include a reference direction.
  3. Choose any of the vectors and draw it as an arrow in the correct direction and of the correct length – remember to put an arrowhead on the end to denote its direction.
  4. Take the next vector and draw it as an arrow starting from the arrowhead of the first vector in the correct direction and of the correct length.
  5. Continue until you have drawn each vector – each time starting from the head of the previous vector. In this way, the vectors to be added are drawn one after the other head-to-tail.
  6. The resultant is then the vector drawn from the tail of the first vector to the head of the last. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram.
Exercise 1: Head-to-Tail Addition I

A ship leaves harbour H and sails 6 km north to port A. From here the ship travels 12 km east to port B, before sailing 5,5 km south-west to port C. Determine the ship's resultant displacement using the head-to-tail technique of vector addition.

Solution
  1. Step 1. Draw a rough sketch of the situation :

    Its easy to understand the problem if we first draw a quick sketch. The rough sketch should include all of the information given in the problem. All of the magnitudes of the displacements are shown and a compass has been included as a reference direction. In a rough sketch one is interested in the approximate shape of the vector diagram.

    Figure 1
    Figure 1 (PG11C1_040.png)

  2. Step 2. Choose a scale and include a reference direction. :

    The choice of scale depends on the actual question – you should choose a scale such that your vector diagram fits the page.

    It is clear from the rough sketch that choosing a scale where 1 cm represents 2 km (scale: 1 cm = 2 km) would be a good choice in this problem. The diagram will then take up a good fraction of an A4 page. We now start the accurate construction.

  3. Step 3. Choose any of the vectors to be summed and draw it as an arrow in the correct direction and of the correct length – remember to put an arrowhead on the end to denote its direction. :

    Starting at the harbour H we draw the first vector 3 cm long in the direction north.

    Figure 2
    Figure 2 (PG11C1_041.png)

  4. Step 4. Take the next vector and draw it as an arrow starting from the head of the first vector in the correct direction and of the correct length. :

    Since the ship is now at port A we draw the second vector 6 cm long starting from point A in the direction east.

    Figure 3
    Figure 3 (PG11C1_042.png)

  5. Step 5. Take the next vector and draw it as an arrow starting from the head of the second vector in the correct direction and of the correct length. :

    Since the ship is now at port B we draw the third vector 2,25 cm long starting from this point in the direction south-west. A protractor is required to measure the angle of 45∘∘.

    Figure 4
    Figure 4 (PG11C1_043.png)

  6. Step 6. The resultant is then the vector drawn from the tail of the first vector to the head of the last. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram. :

    As a final step we draw the resultant displacement from the starting point (the harbour H) to the end point (port C). We use a ruler to measure the length of this arrow and a protractor to determine its direction.

    Figure 5
    Figure 5 (PG11C1_044.png)

  7. Step 7. Apply the scale conversion :

    We now use the scale to convert the length of the resultant in the scale diagram to the actual displacement in the problem. Since we have chosen a scale of 1 cm = 2 km in this problem the resultant has a magnitude of 9,2 km. The direction can be specified in terms of the angle measured either as 072,3∘∘ east of north or on a bearing of 072,3∘∘.

  8. Step 8. Quote the final answer :

    The resultant displacement of the ship is 9,2 km on a bearing of 072,3∘∘.

Exercise 2: Head-to-Tail Graphical Addition II

A man walks 40 m East, then 30 m North.

  1. What was the total distance he walked?
  2. What is his resultant displacement?
Solution
  1. Step 1. Draw a rough sketch :

    Figure 6
    Figure 6 (PG11C1_045.png)

  2. Step 2. Determine the distance that the man traveled :

    In the first part of his journey he traveled 40 m and in the second part he traveled 30 m. This gives us a total distance traveled of 40 m + 30 m = 70 m.

  3. Step 3. Determine his resultant displacement :

    The man's resultant displacement is the vector from where he started to where he ended. It is the vector sum of his two separate displacements. We will use the head-to-tail method of accurate construction to find this vector.

  4. Step 4. Choose a suitable scale :

    A scale of 1 cm represents 10 m (1 cm = 10 m) is a good choice here. Now we can begin the process of construction.

  5. Step 5. Draw the first vector to scale :

    We draw the first displacement as an arrow 4 cm long in an eastwards direction.

    Figure 7
    Figure 7 (PG11C1_046.png)

  6. Step 6. Draw the second vector to scale :

    Starting from the head of the first vector we draw the second vector as an arrow 3 cm long in a northerly direction.

    Figure 8
    Figure 8 (PG11C1_047.png)

  7. Step 7. Determine the resultant vector :

    Now we connect the starting point to the end point and measure the length and direction of this arrow (the resultant).

    Figure 9
    Figure 9 (PG11C1_048.png)

  8. Step 8. Find the direction :

    To find the direction you measure the angle between the resultant and the 40 m vector. You should get about 37∘∘.

  9. Step 9. Apply the scale conversion :

    Finally we use the scale to convert the length of the resultant in the scale diagram to the actual magnitude of the resultant displacement. According to the chosen scale 1 cm = 10 m. Therefore 5 cm represents 50 m. The resultant displacement is then 50 m 37∘∘ north of east.

Figure 10
Phet simulation for Vectors

The Parallelogram Method

The parallelogram method is another graphical technique of finding the resultant of two vectors.

Method: The Parallelogram Method

  1. Make a rough sketch of the vector diagram.
  2. Choose a scale and a reference direction.
  3. Choose either of the vectors to be added and draw it as an arrow of the correct length in the correct direction.
  4. Draw the second vector as an arrow of the correct length in the correct direction from the tail of the first vector.
  5. Complete the parallelogram formed by these two vectors.
  6. The resultant is then the diagonal of the parallelogram. The magnitude can be determined from the length of its arrow using the scale. The direction too can be determined from the scale diagram.
Exercise 3: Parallelogram Method of Vector Addition I

A force of F1=5NF1=5N is applied to a block in a horizontal direction. A second force F2=4NF2=4N is applied to the object at an angle of 30∘∘ above the horizontal.

Figure 11
Figure 11 (PG11C1_049.png)

Determine the resultant force acting on the block using the parallelogram method of accurate construction.

Solution
  1. Step 1. Firstly make a rough sketch of the vector diagram :

    Figure 12
    Figure 12 (PG11C1_050.png)

  2. Step 2. Choose a suitable scale :

    In this problem a scale of 1 cm = 1 N would be appropriate, since then the vector diagram would take up a reasonable fraction of the page. We can now begin the accurate scale diagram.

  3. Step 3. Draw the first scaled vector :

    Let us draw F1F1 first. According to the scale it has length 5 cm.

    Figure 13
    Figure 13 (PG11C1_051.png)

  4. Step 4. Draw the second scaled vector :

    Next we draw F2F2. According to the scale it has length 4 cm. We make use of a protractor to draw this vector at 30∘∘ to the horizontal.

    Figure 14
    Figure 14 (PG11C1_052.png)

  5. Step 5. Determine the resultant vector :

    Next we complete the parallelogram and draw the diagonal.

    Figure 15
    Figure 15 (PG11C1_053.png)

    The resultant has a measured length of 8,7 cm.

  6. Step 6. Find the direction :

    We use a protractor to measure the angle between the horizontal and the resultant. We get 13,3∘∘.

  7. Step 7. Apply the scale conversion :

    Finally we use the scale to convert the measured length into the actual magnitude. Since 1 cm = 1 N, 8,7 cm represents 8,7 N. Therefore the resultant force is 8,7 N at 13,3∘∘ above the horizontal.

The parallelogram method is restricted to the addition of just two vectors. However, it is arguably the most intuitive way of adding two forces acting on a point.

Algebraic Addition and Subtraction of Vectors

Vectors in a Straight Line

Whenever you are faced with adding vectors acting in a straight line (i.e. some directed left and some right, or some acting up and others down) you can use a very simple algebraic technique:

Method: Addition/Subtraction of Vectors in a Straight Line

  1. Choose a positive direction. As an example, for situations involving displacements in the directions west and east, you might choose west as your positive direction. In that case, displacements east are negative.
  2. Next simply add (or subtract) the magnitude of the vectors using the appropriate signs.
  3. As a final step the direction of the resultant should be included in words (positive answers are in the positive direction, while negative resultants are in the negative direction).

Let us consider a few examples.

Exercise 4: Adding vectors algebraically I

A tennis ball is rolled towards a wall which is 10 m away from the ball. If after striking the wall the ball rolls a further 2,5 m along the ground away from the wall, calculate algebraically the ball's resultant displacement.

Solution
  1. Step 1. Draw a rough sketch of the situation :

    Figure 16
    Figure 16 (PG11C1_054.png)

  2. Step 2. Decide which method to use to calculate the resultant :

    We know that the resultant displacement of the ball (x→Rx→R) is equal to the sum of the ball's separate displacements (x→1x→1 and x→2x→2):

    x → R = x → 1 + x → 2 x → R = x → 1 + x → 2
    (1)

    Since the motion of the ball is in a straight line (i.e. the ball moves towards and away from the wall), we can use the method of algebraic addition just explained.

  3. Step 3. Choose a positive direction :

    Let's choose the positive direction to be towards the wall. This means that the negative direction is away from the wall.

  4. Step 4. Now define our vectors algebraically :

    With right positive:

    x → 1 = + 10 , 0 m · s - 1 x → 2 = - 2 , 5 m · s - 1 x → 1 = + 10 , 0 m · s - 1 x → 2 = - 2 , 5 m · s - 1
    (2)
  5. Step 5. Add the vectors :

    Next we simply add the two displacements to give the resultant:

    x → R = ( + 10 m · s - 1 ) + ( - 2 , 5 m · s - 1 ) = ( + 7 , 5 ) m · s - 1 x → R = ( + 10 m · s - 1 ) + ( - 2 , 5 m · s - 1 ) = ( + 7 , 5 ) m · s - 1
    (3)
  6. Step 6. Quote the resultant :

    Finally, in this case towards the wall is the positive direction, so: x→Rx→R = 7,5 m towards the wall.

Exercise 5: Subtracting vectors algebraically I

Suppose that a tennis ball is thrown horizontally towards a wall at an initial velocity of 3 m··s-1-1to the right. After striking the wall, the ball returns to the thrower at 2 m··s-1-1. Determine the change in velocity of the ball.

Solution
  1. Step 1. Draw a sketch :

    A quick sketch will help us understand the problem.

    Figure 17
    Figure 17 (PG11C1_055.png)

  2. Step 2. Decide which method to use to calculate the resultant :

    Remember that velocity is a vector. The change in the velocity of the ball is equal to the difference between the ball's initial and final velocities:

    Δ v → = v → f - v → i Δ v → = v → f - v → i
    (4)

    Since the ball moves along a straight line (i.e. left and right), we can use the algebraic technique of vector subtraction just discussed.

  3. Step 3. Choose a positive direction :

    Choose the positive direction to be towards the wall. This means that the negative direction is away from the wall.

  4. Step 4. Now define our vectors algebraically :
    v → i = + 3 m · s - 1 v → f = - 2 m · s - 1 v → i = + 3 m · s - 1 v → f = - 2 m · s - 1
    (5)
  5. Step 5. Subtract the vectors :

    Thus, the change in velocity of the ball is:

    Δ v → = ( - 2 m · s - 1 ) - ( + 3 m · s - 1 ) = ( - 5 ) m · s - 1 Δ v → = ( - 2 m · s - 1 ) - ( + 3 m · s - 1 ) = ( - 5 ) m · s - 1
    (6)
  6. Step 6. Quote the resultant :

    Remember that in this case towards the wall means a positive velocity, so away from the wall means a negative velocity: Δv→=5m·s-1Δv→=5m·s-1 away from the wall.

Resultant Vectors
  1. Harold walks to school by walking 600 m Northeast and then 500 m N 40∘∘ W. Determine his resultant displacement by using accurate scale drawings.
  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 135∘∘ and then at a velocity of 1,2 m··s-1-1 on a bearing of 230∘∘. Calculate her resultant velocity by using accurate scale drawings.
  3. A squash ball is dropped to the floor with an initial velocity of 2,5 m··s-1-1. It rebounds (comes back up) with a velocity of 0,5 m··s-1-1.
    1. What is the change in velocity of the squash ball?
    2. What is the resultant velocity of the squash ball?

Remember that the technique of addition and subtraction just discussed can only be applied to vectors acting along a straight line. When vectors are not in a straight line, i.e. at an angle to each other, the following method can be used:

A More General Algebraic technique

Simple geometric and trigonometric techniques can be used to find resultant vectors.

Exercise 6: An Algebraic Solution I

A man walks 40 m East, then 30 m North. Calculate the man's resultant displacement.

Solution
  1. Step 1. Draw a rough sketch :

    As before, the rough sketch looks as follows:

    Figure 18
    Figure 18 (PG11C1_056.png)

  2. Step 2. Determine the length of the resultant :

    Note that the triangle formed by his separate displacement vectors and his resultant displacement vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let xRxR represent the length of the resultant vector. Then:

    x R 2 = ( 40 m · s - 1 ) 2 + ( 30 m · s - 1 ) 2 x R 2 = 2 500 m · s - 1 2 x R = 50 m · s - 1 x R 2 = ( 40 m · s - 1 ) 2 + ( 30 m · s - 1 ) 2 x R 2 = 2 500 m · s - 1 2 x R = 50 m · s - 1
    (7)
  3. Step 3. Determine the direction of the resultant :

    Now we have the length of the resultant displacement vector but not yet its direction. To determine its direction we calculate the angle αα between the resultant displacement vector and East, by using simple trigonometry:

    tan α = opposite side adjacent side tan α = 30 40 α = tan - 1 ( 0 , 75 ) α = 36 , 9 ∘ tan α = opposite side adjacent side tan α = 30 40 α = tan - 1 ( 0 , 75 ) α = 36 , 9 ∘
    (8)
  4. Step 4. Quote the resultant :

    The resultant displacement is then 50 m at 36,9∘∘ North of East.

    This is exactly the same answer we arrived at after drawing a scale diagram!

In the previous example we were able to use simple trigonometry to calculate the resultant displacement. This was possible since the directions of motion were perpendicular (north and east). Algebraic techniques, however, are not limited to cases where the vectors to be combined are along the same straight line or at right angles to one another. The following example illustrates this.

Exercise 7: An Algebraic Solution II

A man walks from point A to point B which is 12 km away on a bearing of 45∘∘. From point B the man walks a further 8 km east to point C. Calculate the resultant displacement.

Solution
  1. Step 1. Draw a rough sketch of the situation :

    Figure 19
    Figure 19 (PG11C1_057.png)

    BA^F=45∘BA^F=45∘ since the man walks initially on a bearing of 45∘∘. Then, AB^G=BA^F=45∘AB^G=BA^F=45∘ (parallel lines, alternate angles). Both of these angles are included in the rough sketch.

  2. Step 2. Calculate the length of the resultant :

    The resultant is the vector AC. Since we know both the lengths of AB and BC and the included angle AB^CAB^C, we can use the cosine rule:

    A C 2 = A B 2 + B C 2 - 2 · A B · B C cos ( A B ^ C ) = ( 12 ) 2 + ( 8 ) 2 - 2 · ( 12 ) ( 8 ) cos ( 135 ∘ ) = 343 , 8 A C = 18 , 5 km A C 2 = A B 2 + B C 2 - 2 · A B · B C cos ( A B ^ C ) = ( 12 ) 2 + ( 8 ) 2 - 2 · ( 12 ) ( 8 ) cos ( 135 ∘ ) = 343 , 8 A C = 18 , 5 km
    (9)
  3. Step 3. Determine the direction of the resultant :

    Next we use the sine rule to determine the angle θθ:

    sin θ 8 = sin 135 ∘ 18 , 5 sin θ = 8 × sin 135 ∘ 18 , 5 θ = sin - 1 ( 0 , 3058 ) θ = 17 , 8 ∘ sin θ 8 = sin 135 ∘ 18 , 5 sin θ = 8 × sin 135 ∘ 18 , 5 θ = sin - 1 ( 0 , 3058 ) θ = 17 , 8 ∘
    (10)

    To find FA^CFA^C, we add 45∘∘. Thus, FA^C=62,8∘FA^C=62,8∘.

  4. Step 4. Quote the resultant :

    The resultant displacement is therefore 18,5 km on a bearing of 062,8∘∘.

More Resultant Vectors
  1. A frog is trying to cross a river. It swims at 3 m··s-1-1in a northerly direction towards the opposite bank. The water is flowing in a westerly direction at 5 m··s-1-1. Find the frog's resultant velocity by using appropriate calculations. Include a rough sketch of the situation in your answer.
  2. Sandra walks to the shop by walking 500 m Northwest and then 400 m N 30∘∘ E. Determine her resultant displacement by doing appropriate calculations.

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