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School Physics Notes: Drawing scale diagrams and calculating a resultant force

Forces 3. Problem solving - calculating resultant forces using graphs, vector diagrams and work done

Doc Brown's GCSE level Physics exam study revision notes

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 scaled 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 scaled 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 due to gravity 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 (scale 1 mm = 1 N, 2 mm squares, 2 mm = 2 N).

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. There is actually a resultant force of 10 N (10 mm) to the horizontal right.

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.

Set A QUESTIONS

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

 Q1. Two forces acting on an object. What is the resultant force and direction of movement of the object? Q2. Two forces acting in parallel but opposite directions What is the resultant force and direction on the object? Q3. Multiple forces acting on an objectGiven the free body diagram, what is the magnitude and direction of the resultant force. Q4. Resolving two forces that are not parallel Imagine an object is subjected to a northward force of 90 N and a due east force of 60 N. Deduce the magnitude of the resultant force and its angle with respect to due north. There are two approaches you can follow. (i) Using graph paper and a suitable scale or (ii) using the trigonometric calculation method (see APPENDIX 1): Q5. Resolving a forces into two forces at right angles to each other Imagine a force of 156 N acting at an angle of 51o from the vertical 'north'. Deduce the component vectors for due north and due east quoted on the diagram. Q6. Variations on two forces. (a) Situation A: A pulling force of 225 N is acting on an object at an angle of 27o to the horizontal. At the same time it is also subjected to another pulling force at 45o to the horizontal (as shown in the diagram). Deduce the resultant force and direction quoted on the diagram. (b) Situation B: However, you can avoid the parallelogram of forces diagram by drawing the force diagram in a simpler way - can you arrive at the same answer as (a)?. (c) Situation C:  I want you to imagine the 300 N force acting in the opposite direction.Deduce the resultant force and direction? Q7. A hanging weight, tensioned horizontally A mass of 20 kg is suspended from a beam by a wire at angle of 32o away from the vertical (see diagram). The tension in this wire is T2. The same 20 kg weight is pulled to one side by a horizontal string with a tension T2. (g = 9.8 N/kg) Using a graph and suitable scale calculate the tension force in (a) the wire (T1) (b) in the string (T2). Q8. Resolving two converging forcesTwo forces, 5.0 N and 6.0 N, act on an object at an angle of 60o between the lines of action (as in the diagram). Calculate the resultant force on the object O at point o. Q9. Calculating a 3rd force required to establish an equilibrium Two wires with pulling force tensions of 20 N and 24 N are pulling on a metal ring. If the angle between the lines of action is 70o calculate the force needed on the 3rd wire on the right, to stabilise the ring to give an equilibrium situation. Q10. A push and pull grass roller!A roller of mass 80 kg (~800 N) is pushed or pulled with a force of 300 N acting at an angle of 45o to the horizontal grass surface. Ignore friction effects in this question.   (a) Calculate the pulling force in the horizontal direction. (b) Calculate the vertical force of the roller on the grass when it is pushed. You first need to calculate the vertical force involved due to the person pushing or pulling the roller (call it Fp). (c) Calculate the vertical force of the roller on the grass when it is pushed. Q11. A suspended microphone!A microphone of mass 750 g is suspended by a wire at 25o to the vertical (tension T2, diagram) and pulled to the right by a horizontal cord (tension T1, diagram). (Take g = 10N/kg) (a) Calculate the tension in the wire holding up the microphone. (b) What tension must be applied to the horizontal cord to maintain the microphone in a stable position? Q12 The motor of a ferry boat provides driving force of 250 N perpendicular to the bank of a river.If the flow of the river acts in the same direction as the river bank with a force of 100 N ... (a) Calculate the magnitude and direction of the resultant force. (b) With reference to your answer to (b), how should the ferry be steered to minimise the distance travelled to cross the river? Use graph or pure calculation method. Q13 The engine of a double-decker bus weighing 20 000 N generates a driving force of 2000 N against opposing friction forces of 1500 N. (a) Draw a free body diagram of the four forces involved and explain their origin. (b) Calculate the resultant force (c) Calculate the acceleration of the bus (this requires much more advanced knowledge

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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

SET B QUESTIONS

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

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

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

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.

These examples were 'borrowed' from

Types of energy stores, mechanical work done and power calculations

Keywords, phrases and learning objectives for forces

Problem solving calculating resultant forces using graphs and vector diagrams

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What next? Forces revision notes index

FORCES 4. Elasticity and energy stored in a spring

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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.

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

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