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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

 This page will answer questions such as ...

 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?

See also What are contact forces & non-contact forces?, scalar & vector quantities, free body force diagrams



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)

scale drawing graphs to deduce resultant forces gcse physics igcse science

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)

scale drawing graphs to deduce resultant forces gcse physics igcse science

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

See also work done calculations in Appendix 2.

Q1. Two forces acting in the same direction (parallel) on an object

What is the resultant force on the object?

In this case you simply add up the two forces acting from left to right:

resultant force = force 1 + force 2 = 70 N + 45 N = 115 N

So the object experiences an acting force of 115 N to the right (vector!).

Q2. Two forces acting in parallel but opposite directions

What is the resultant force on the object?

Above you added the forces, but here, you subtract one from the other.

resultant force = force 1 - force 2 = 780 - 330 = 450 N to the right

Since the force to the right is greater than the opposing force from the left, the net force (resultant force of 330 N) must act from left to right.

Q3. Multiple force acting on an object

Given the free body diagram, what is the magnitude and direction of the resultant force.

This situation is not quite as complicated as it seems because acting forces 2 and 3 are equal and cancel each other out and are at 90o to the line of action of the resultant force on which they have no effect (*). Whatever the motion of the object, it will not rise or fall.

Therefore the situation is actually the same as example Q2. above.

resultant force = force 1 - force 4 = 120 - 55 = 65 N horizontally to the right

 This free body diagram could represent a moving vehicle.

(*) Forces that act at 90o to the direction of motion, do not contribute to the resultant force you are calculating here.

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.

Using graph paper and a suitable scale, you draw the two forces out at 90o to each other in the manner shown in the left diagram ('tip to tip') - which produces a triangle when you join up the start of the north force to the right hand tip of the east force.

You then measure the length of the hypotenuse and from the chosen scale you get 7.2 cm which translates to 7.2 x 15 = 108 N

Using a protractor you can then measure the angle which I found to be 33o (033o) with respect to north. I made both measurements as a student would in class using 2 mm graph paper.

Solving Q4.  using trigonometric calculations (see APPENDIX 1):

In GCSE exams you will have to solve it by the graphical method described above, but you can solve it just from a simple non-scale sketch by trigonometry using the known direction of the two forces and the angle between them. This is really advanced A level maths using a scientific calculator, but I decided to check my own 'honest' graph work just using the forces and directions given and ignoring the graph completely.

tan θ = O / A = 60/90 = 0.667, tan-1(0.667) = 33.7o (so I made a small manual graph error of 0.7o)

You can work out the magnitude of the resultant force using the accurate angle above and either of the sine or cosine rule equations e.g. if R = resultant force (= H)

sin θ = O / H = sin (33.7o) = 60/R = 0.555,   R = 60/0.555 = 108 N

cos θ = A / H = cos (33.7o) = 90/R = 0.832,   R = 90/0.832 = 108 N

So my resultant force measurement was spot on!

 

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'.

Q Deduce the component vectors for due north and due east.

You draw the resultant force line to scale at 39o (90o - 51o) from the horizontal.

Use a convenient triangle e.g. in this case the hypotenuse of the triangle is 7.8 cm. and the other two sides are 6.0 to the east and 5.0 cm to the north.

I've then chosen a scale of 20.0 N/cm on 2 mm graph paper.

( Note from the scale chose 7.8 x 20.0 = 156 N)

You can then split this 'resultant' force into two components by draw lines for the northerly contribution (5.0 cm) and the easterly contribution (6.0 cm). The convenient numbers of 5.0 and 6.0 was pure coincidence!

From the scale this gives the two contributing forces of:

5.0 x 20.0 = 100 N to the north and 6.0 x 20.0 = 120 N to the east.


Again you can work out the answers from trigonometry just using the angle of 39o and the force of 156 N (the original information): Let N be the north force and E be the east force

sin θ = O / H = sin (39o) = 0.629 = N/156,   N = 0.629 x 156 = 98.1 N to the north

cos θ = A / H = cos (39o) = 0.777 = E/156,  E = 0.777 x 156 = 121.2 N to the east

So I'd made a 1-2% error on my graph work, and note the trigonometrical calculations are absolutely correct based on the original information. This level of calculation is covered by GCSE maths courses.

Q6. Variations on two forces.

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).

With graph paper, protractor and ruler, draw a scale diagram to deduce the magnitude and direction of the resultant force. The graph is drawn to a scale of 1 mm = 5 N.

This can be solved using the principle of 'parallelogram of forces'. Its quite simple to do. The dotted line are drawn parallel to the two known vectors and where they intersect gives you the length of the resultant force of 60 mm.

Therefore the resultant force is 60 x 5 = 300 N horizontally to the right.

 

However, you can avoid the parallelogram of forces diagram by drawing the force diagram in a simpler way as shown in diagram B.

You draw the first force of 225 N at 27o to the horizontal (45 mm) and from its top tip draw the 2nd force downwards at 45o to the horizontal line (28 mm).

The horizontal line of 60 mm gives you the same resultant force of 300 N.

 

In situation C I want you to imagine the 300 N force acting in the opposite direction.

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. more examples needed?

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.

Using a graph calculate the tension force in (a) the wire (T1) and (b) in the string (T2).

The triangle of forces is drawn on 1 mm graph paper. Tension T3 is the weight of the object = 196 N (20 x 9.8, g = 9.8 N/kg). This is the vertical force.

The diagonal side of the triangle is tension T1 in the wire holding up the weight at an angle of 32o.

The horizontal side of the triangle is the tension T2 in the string pulling from the left.

(a) The graph is drawn to a scale of 2 N per mm. The diagonal (hypotenuse) was found to be 115 mm long.

From the scale tension T1 = 115 x 2 = 230 N

(b) The opposite side of the triangle = 60 mm, this equates to 60 x 2 = 120 N for tension T2 in the string

Trigonometric calculation check (perfect answers!):

Tension T1: cos (32o) = adjacent / hypotenuse = 0.848 = T3/T1 = 196/T1, so T1 = 196/0.848 = 231.1 N

Tension T2: tan (32o) = opposite / adjacent = 0.625 = T2/T3 = T2/196, so T2 = 0.625 x 196 = 122.5 N

So, I think my graph work was pretty good and an 'ok' with only a 0.5% to 2% error!

Q8. Resolving two converging forces

Two 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.

Using the principle of the parallelogram of forces, and focus on the left of the diagram: If you draw the lines a-b (6.0 cm) parallel to action line o-c and line b-c (5.0 cm) parallel to action line a-o, then the diagonal of 9.5 cm gives you the resultant distance which equates to a resultant force of 9.5 N (1 cm = 1 N).

However, drawing true parallel lines is awkward, so you can deduce the answer without graph paper and on plain white paper with just a mm ruler and protractor. If you extend the line a-o a distance of 5 cm giving line o-d, you get exactly the same resultant distance of 9.5 cm = resultant force of 9.5 N by joining up the relevant tip to tip points c-d.

Its simple pure geometric logic, no problem! Just compare situations A and B in Q6.

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.

Using the principle of the parallelogram of forces, and focus on the left of the diagram: If you draw the lines a-b (6.0 cm) parallel to action line o-c and line b-c (5.0 cm) parallel to action line a-o, then the diagonal of 9.0 cm gives you the resultant distance. BUT, the direction of action is opposite to that in Q8, so this equates to a resultant force of 36.0 N (1 cm = 14N) and the arrow points towards the right balancing and opposing the forces acting to the left.

As in Q8, drawing true parallel lines is awkward, so you can deduce the answer without graph paper and on plain white paper with just a mm ruler and protractor. If you extend the line a-o a distance of 5 cm giving line o-d, you get exactly the same resultant distance of 9.0 cm = resultant force of 9.0 x 4 =36 N by joining up the relevant tip to tip points c-d.

Note that although the total force to the left is 44 N, the opposing force is less (36 N), because the forces are not acting parallel to each other.

Q10. A push and pull grass roller!

A roller of mass 80 kg 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.

The scale drawing on the right shows you how to obtain the horizontal pulling force.

At 45o on the graph paper, the diagonal line for 300 N equals 56.5 mm.

The horizontal line has a length of 40 mm, so scaling down, gives ..

300 x 40 / 56.5 = 212 N

Note that, neglecting friction effects, the downward force of 800 N of the weight of the roller has no effect on the calculation.

This is because it acts at 90o to the direction of the motion and doesn't contribute to the resultant force you are calculating here.

(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).

You can do this simply with a graph as illustrated above, or a trigonometric calculation shown below.

sin (45o) = opposite/hypotenuse = 0.707 = Fp/300, so Fp = 0.707 x 300 = 212 N in a vertical direction.

You also need the weight of the roller = m x g = 80 x 10 = 800 N (taking gravity as 10 N/kg)

When pushing the roller you are increasing the overall downward force of the roller, therefore

total vertical force exerted by the roller = weight of roller + vertical pushing force

= 800 + 212 = 1012 N, the vertical normal contact force due to the roller.

(c) Calculate the vertical force of the roller on the grass when it is pushed.

When pulling the roller you are decreasing the overall downward force of the roller, therefore

total vertical force exerted by the roller = weight of roller - vertical pulling force

= 800 - 212 = 588 N, the vertical normal contact force due to the roller.

AND, now you can see why its easier to pull a roller than push it and how accurate was your graph work!

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).

(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?

You can do this simply with a graph as in Q7, but I'll leave you to practice that and I'll go straight for trigonometric calculations, also at the end of Q7 and as in Q11. above.

750 g = 0.75 g, so m = m x g = 0.75 x 10 = 7.5 N for the weight of the microphone.

The diagram shows how the forces will operate.

(a) The tension in the wire T2

cos (25o) = adjacent/hypotenuse = 0.9063 = 7.5/T2, so T2 = 7.5/0.9063 = wire tension = 8.27 N (3 sf, 2 dp)

Note that the tension on the wire is greater than the weight of the microphone because it is being pulled both downwards and to one side.

(b) Force T1 needed to stabilise microphone

tan 25o) = opposite/adjacent = 0.466 = T1/7.5, so T1 = 0.466 x 7.5 = cord tension required = 3.50 N (3 sf)

How accurate was your graph work!

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.

By drawing a graph diagram or from a trigonometric calculation you should get a resultant force of 269 N.

Ignoring the rudder! the boat will travel away from the bank at an angle of 68o to it (or 22o from the perpendicular line from the bank).

(b) With reference to your answer to (b), how should the ferry be steered to minimise the distance travelled to cross the river?

To counteract the flow of the river, the boat should be steered at an angle opposite to the direction indicated above in (a). This means working against the flow of the river to counteract the enforced sideways movement of the boat.

You don't save energy but you can reduce the distance travelled.

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 forces involved and explain their origin.

F1 = the contact force of the road material pushing up

F2 = contact force from the weight of the bus

F3 = forces of resistance due to friction - air resistance, wheels on road, moving parts of bus etc.

F4 = driving force generated by the engine of the bus

(b) Calculate the resultant force

Unless the bus is moving up and down F1 will equal F2, so no resultant in the vertical direction.

However, there is a resultant force to the right of F4 - F3 = 2000 - 1500 = 500 N

(c) Calculate the acceleration of the bus (this requires much more advanced knowledge

Assume gravity force is 10 N/kg.

W = m x g, m = W / g,  and from Newton's 2nd Law: F = ma,  a = F/m

F = resultant force = 500 N

weight = mass x g, mass = weight /g = 20 000/10 = 2000 kg

a = F/m = 500 / 2000 = 0.25 m/s2

 

 

See 4. Newton's First, Second and Third Laws of Motion, inertia and F = ma calculations gcse physics revision notes

 
 

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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 2. Mass and the effect of gravity force on it - weight, (mention of work done and GPE)

FORCES 3. Calculating resultant forces using vector diagrams and work done

FORCES 4. Elasticity and energy stored in a spring

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

FORCES 6. Pressure in liquid fluids and hydraulic systems

FORCES 7. Pressure & upthrust in liquids, why do objects float or sink in a fluid?, variation of atmospheric pressure with height


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