In adding vectors we have to consider their directions too.  

If they act in the same direction or in exactly opposite they can be simply added considering their plus or minus signs.  

If they are acting at different angles we can use ‘vector diagrams’.

The figure shows a vector diagram to get the direction of a boat. It is sailing to East but a cross current is acting to North.  The velocity of the boat is 4 meters per second while the current is flowing at 3 meters per second. The red arrow gives the velocity.  The magnitude is the length which can be obtained by measurement or by using the Pythagoras theorem. This also gives the direction of the resultant force.
 

Using Pythagoras theorem we get:-    
      R ² =   3² + 4²   ........>   R     =  √  ( 9 +  16).......>     R      =  5 ms^-1
      
      

    Using trigonometry to get the angle giving the direction:-
There are 3 basic ratios that would be better to get familiar with.
In a right angle triangle as shown

1.     The Sine  ϴ          =    Opposite side / Hypotenuse.                       N/R 2.    The Cosine   ϴ       =   Adjacent  / Hypotenuse.                    S/R 3.    The Tangent   ϴ     =   Opposite side / adjacent side             N/S

 

 

Taking the Tangent according to fig. 1 we get:-  

  Tan  ϴ   =    N/S   =  ¾  =0.75

For this value of Tan, we can get the angle using  a Calculator.
     Tangent 0.75 =  37 °   
So the boat will travel at an angle of 37 ° North of East at a velocity of 10m S^-1

• Parallelogram method of Adding Vectors.

 

A Force has a direction and as such it is a Vector.

A force has to be represented by a line in length and direction. Consider the two forces in the given example. The wind F-1, is acting towards East by a value of 800 N. The water is pushing it to North by a force of 600N.  We cannot draw lines of 800 or 600 cm.
We now decide on a suitable scale. Let us take 1cm for every 100 N. Then it works out as follows;

                      F-1…………..800N     ………..8cm.  to East
                      F-2…………..600N     ………..6cm   to North.

These two forces can be represented as follows;

Fig.3 Adding two forces F1 and F2.

The blue line represents the wind while the black line shows the current. The dotted lines are drawn parallel to these. Once the parallelogram, which is a rectangle in this case, is completed the diagonal will indicate the resulting force in direction and magnitude.
Now it is only a case of measuring the Resultant (Resulting force) and converting it back to newtons.

In this case the diagonal, showed by the red line, is only 10cm long. Using the scale we get it as 1000 N.

 Subtracting  vectors

In subtracting a vector the same method can be used after taking the quantity to be subtracted as  a negative quantity.  When the changing angle is a right angle the magnitude of the difference will be the same as the sum, but the direction will change.

Example 1

 

The flight path of a bird is shown in red. It flies at a constant speed and takes a right angle turn. Find the change in velocity.

Answer.

The change in velocity is = Final velocity - Initial velocity.

The diagram on the right shows the vector diagram obtained by changing the direction of the flight path.

                     Δ,V =  Vf - Vi

                            =   √  62 - (-62) ms-1

                            =  √   36 + 36

                                    =  √    72      =    8.5 ms-1

Example 2

 Fig.4 Parallelogram method to subtract.

A fish is swimming to east with a speed of 30 kms^-1 while the river is flowing in a north westerly direction at 20s^-1. Find the change in these two vectors.

 Calculation

  

Fig.5.

h/20= sine 45°  = 0.71

h =  0.71x 20    = 14.2

y is also = 14.2  ( two sides of an isosceles triangle)

As x + y = 30 ........       x   =  (30 – 14.2)  =  15.8    ......                                                                                                                                                                       

Using Pythagoras theorem we get

R² = x² + x^{2...............}R² =  14.2 ²   +15.8 ²

Therefore R  =  √(249 + 201.64)  =√  450.64.............R =  21.2

                            Resolving forces

When a force acts in a certain direction, we very often have to resolve the vertical and horizontal effects of this.

Man  is puling the weight as shown but it moves horizontally. How to determine the effective force parallel to the ground?

 Horizontal force =  F cos 30°   and the Vertical force = F sin 30°.

 If the pulling force is 100 N   the horizontal force will be = 100 cos 30°.

                                                                                            = 100 x0.87  =  87 N.

An  example.

A man is pulling a log exerting a force of 40 N towards East. His son is pulling at an angle of 30 °  to his line.  Find the resulting force.

   Answer

  

  When you complete the parallelogram , the diagonal becomes the resultant.

  Or you can use trigonometry by extending the section shown in red.

      Sin 30 = h/20...........  h/20 = 0.5     Therefor h= 10

      Tan 30 = h/x  .............h/x = 0.58     Therefore x=  10/x = 0.58   = 17.2

      Now consider  the big triangle with the base 40 + 17.2

      Tan  ϴ   =  10/ 57.2...........0.17

     Therefre  ϴ   = 

 

Question   1. 0

 

Fig.  Airplane in flight. 

1. to  4.  Name the forces shown by the coloured arrows. Select the answers from the given words.
a.  Weight    b. Lift     c.  Friction (Drag)   d. Thrust   e.  Reaction.

1.1. Red arrow -   .......    1.2.  Blue arrow- ....... 1.3.   Black arrow - ..........              1.4.Green arrow-.............                                                    (20 marks)

2.0 Which one out of the following cannot be considered as a force?

a. Pull   b. Kick   c. Magnet  d. Repulsion   e. Attraction             (10 marks)

3.0 What two qualities out of the following should a vector must have?
a. Magnitude  b. Direction   c. Speed    d. Velocity  e .Colour           (10 marks)

 4.0

John and Jerry are pulling a boat.

4.1 Find the total force on the boat.                                        (6 marks)

4.2  using the same forces suggest a way to increase the total pull on the boat    ( 6 marks).

5.0  

     Fig.4  Ali and Nelly pulling a cow.

Look at this picture. Ali is pulling the cow by a force of 600N. Nelly is using a force of 400N. The angle between the two forces is 60º.

5.1 Draw a free body force diagram for the forces, using a suitable scale.         (12 marks)

5.2 Complete the parallelogram and get the diagonal.           (12 marks)

5.3 Obtain Resultant force.                                                   (12 marks)

6.0

If the cow does not wish to be dragged, what force should it use?
    Give the direction and the magnitude.                              (12)

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