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GCSE Physics notes: Drawing scale diagrams and calculating a resultant force

Forces 3. Calculating resultant forces using graphs and vector diagrams

Doc Brown's Physics Revision Notes

Suitable for GCSE/IGCSE Physics/Science courses or their equivalent

What is a resultant force? Why is it a vector?

How do we draw scale diagrams to deduce a resultant force?

What do we mean by balanced and unbalanced forces?

When dealing with acting forces, what is an equilibrium?

Introduction to resultant force problem solving calculations

Forces were introduced in "What is a force?" including contact and non-contact forces AND, importantly for this page, free body diagrams showing multiple forces on an object.

Force data is useless without its direction of action.

You not only need to know the value of a force in newtons (N) but the precise direction or angle of the line of action of one force with respect to at least one other force.

That is why force is always a vector - it has magnitude and direction!

When a body is subjected to multiple known forces (usually >= vectors in newtons) how can we deduce and calculate the net resultant force and its direction?

It is possible to replace multiple forces acting at a single point with a single force known as the resultant force.

Some resultant forces are easily deduced with a simple addition or subtraction.

For example image an object of weight 12 N falling vertically in air. If the air resistance opposing the fall was 5 N, what would be the resultant force? Quite simply it would be 12 - 5 = 7 N.

So the object would continue to fall and accelerate due to the 7 N resultant force.

Other more complicated situations require a scale drawing showing all the forces involved and the direction (e.g. angle) in which each individual force acts.

From such graph work you can measure and calculate the resultant force and its direction of action.

If the resultant force is zero, this is described as an equilibrium situation - a position of balance.

Examples of using graphical scale drawings to determine a resultant force and its direction and to test for an equilibrium situation are explained below.

Example 1. Two forces at 90o to each other (left scale diagram)

Suppose two forces act on an object, a 25 N force in the north direction, and 20 N force at 90o to the east.

In this case, using a scale of 2 mm = 1 N, you draw vertical line 50 mm long and connect to it a horizontal line of 40 mm.

The two lines form half a rectangle, so, to get the resultant line, imagine the other half of the rectangle (or draw it in faintly) and draw the resultant line diagonally across the rectangle.

In this case I made it 71 mm long, giving a resultant force of 35.5 N.

With a protractor you can then measure the angle of the resultant force, which is 43o from the north.

Example 2. Three forces acting on an object (right scale diagram)

This is a bit more tricky.

Here an object is subjected to a 200 N force to the north, a 300 N force to the east, and a 100 N force to the south.

This is illustrated by the 'not to scale' sketch on the side.

You might need to draw a starting sketch like this if all the information comes as text only!

Using a scale of 1.0 cm = 100 N.

You draw a vertical line 4.0 cm long in the north direction.

Then at top of this line draw a horizontal line 6.0 cm long for the east force of 300 N.

OK, so far, as in example 1., but, now draw the line of the 3rd (south force) at the end of the east force line.

You now draw the resultant diagonal line from the bottom of the north line to the bottom of the south line.

(You should note the reduction of the north force is 100 N because of the opposing 100 N south force).

The resultant diagonal line is 6.7 cm long, which equals 335 N.

With a protractor you should get a measured angle of 70-71o from the north

(I calculated it to be 70.5o, using trigonometry - see APPENDIX 1, which is another way of solving these problems if it can be reduced to a right angled triangle situation).

Example 3. An equilibrium or 'state of balance' situation

In example 3 I want you to imagine the 300 N force acting in the opposite direction to two other forces, 225 N and 140 N at an angle to each other.

In this case there is NO net resultant force (other than zero!). The object is then described as being in a state of equilibrium (or balanced) and would remain stationary.

This illustrates a method of deducing a force needed to produce an equilibrium situation involving three forces e.g. you might be given two and have to work out the 3rd force needed to balance the other two forces.

Example 4. Pulling an object along (but watch out for forces that don't count!)

Imagine a car of mass 800 kg being towed by a breakdown truck.

The tension in the towing cable is 500 N at the angle indicated in the diagram.

The weight of the car will be 800 x 10 = 8000 N (gravity is ~10 N/kg).

Resolve the towing force into its vertical and horizontal components.

The graphical drawing uses a scale of 54 mm = 500 N.

This comes out of the way I've done the diagram i.e. I didn't choose a more convenient scale!

Scaling down for the vertical force component of length 20 mm: 500 x 20 / 54 = 185 N

Scaling down for the horizontal force component of length 50 mm: 500 x 50 / 54 = 463 N (in the towing direction)

Note that the weight of the car acting in a vertical direction does NOT affect these force calculations.

Forces that act at 90o to the direction of motion, do not contribute to the horizontal force you are calculating here in the direction of motion.

Unless the front of the car is physically lifted off the ground, it cannot affect the vertical component force.

Example 5. Multiple forces acting on an object - is the resultant force zero (equilibrium) or otherwise

5a. An equilibrium situation, a state of balance with no resultant force other than zero from multiple forces

Consider four forces of 31, 36, 22 and 28 N, all acting on the same object and at four different directions.

This is shown as the left vector diagram.

To check if there is a non-zero resultant force, you draw all the forces tip-to-tail to create a loop - note all the forces are drawn clock-wise to fit in with the force arrow direction.

This is the right vector diagram.

Here there is a complete loop i.e. you end up at the starting point, this means there is no effective resultant force and the object is in a state of equilibrium.

5b. A non-zero resultant force from multiple forces acting on an object, not an equilibrium

Consider four forces of 32, 42, 45 and 36 N, all acting on the same object and at four different directions.

This is shown as the left vector diagram.

Again, to check if there is a non-zero resultant force, you draw all the forces tip-to-tail to create a loop - note all the forces are drawn clock-wise to fit in with the force arrow direction.

This is the right vector diagram.

Here there is NO complete loop i.e. you do NOT end up at the starting point, this means there is an effective resultant force and the object is NOT in a state of equilibrium.

Example 6.

Examples of problem solving to calculate a resultant force using graphs or other vector diagrams

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APPENDIX 1: right angle triangle rules

Important formulae in the trigonometry of a right angled triangle:

for angle θ shown on the right diagram

tangent rule: tan θ = opposite / adjacent

sine rule: sin θ = opposite / hypotenuse

cosine rule: cos θ = adjacent / hypotenuse

APPENDIX 2 Calculating work done from a resultant force

If a source of energy is available, you can calculate the work done from the acting force and the distance the force acts through.

work done (joules) = force (newtons) x distance along the line of action of the force (metres)

W (J) = F (N) x d (m),   F = W/d,   d = W/F

Q1 If you drag a heavy box with a force of 200 N across a floor for 3 m, what work is done?

work done = 200 x 3 = 600 J

Q2 If a machine part does 500 J of work moving linearly 2.5 m, what force was applied by the machine?

work done = force x distance, rearranging,  force (N) = work done (J) ÷ distance (m)

force = 500 ÷ 2.6 = 200 N

Q3 Part of a machine requires a continuous resultant force of 500 N from a motor to move it in a linear direction.

(a) How much work is done in moving it a distance of 50 m?

work done = force x distance = 500 x 50 = 25000 J (25 kJ)

Q4 A toy model car has a clockwork motor, whose spring can store 8.75 J of elastic potential energy.

On release the clockwork motor can deliver a continuous force of 2.5 N.

How far will the car travel in one go?

energy store = total work done = force x distance

distance = energy store / force = 8.75 / 2.5 = 3.5 m

These examples were 'borrowed' from  Types of energy stores, mechanical work done and power calculations

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Forces revision notes index

FORCES 1. What are contact forces & non-contact forces?, scalar & vector quantities, free body force diagrams

FORCES 4. Elasticity and energy stored in a spring

FORCES 5. Turning forces and moments - from spanners to wheelbarrows and equilibrium situations

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