Resultant forces - AQA GCSE Physics Exam Questions

Practise 36 exam-style questions on Resultant forces for AQA GCSE Physics. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

36 questions
1–6 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

State what is meant by the resultant force on an object.

Model answer

The resultant force is the single force that has the same effect as all the forces acting on the object together.

Explanation

When several forces act on an object, you can replace them all with one force - the resultant - without changing what happens to the object. If the resultant is zero, the forces are balanced.

Mark scheme

1 mark
a single force
1 mark
that has the same effect as all the (original) forces acting together
allow replaces all the forces acting on the object

Question 2

2 marks

The diagram shows a force of 15 N drawn to scale on squared paper. It has been resolved into a horizontal component X and a vertical component Y.

Determine the sizes of components X and Y.

Model answer

The scale is 1 square = 1 N.
The horizontal component X is 9 squares long, so X = 9 N.
The vertical component Y is 12 squares long, so Y = 12 N.

Explanation

Count the squares along each dashed component and multiply by the scale. Notice that 9 N + 12 N is more than 15 N: components at right angles do not simply add up to the original force, because they act in different directions.

Mark scheme

1 mark
X = 9 N
allow 8.8-9.2 N; correct answer with no working scores full marks
1 mark
Y = 12 N
allow 11.8-12.2 N

Question 3

3 marks

A picture with a weight of 20 N hangs at rest from a nail. It is held by two equal strings, each at an angle of 30° to the vertical.

Explain why the tension in each string must be greater than 10 N.

Model answer

The picture is at rest, so it is in equilibrium and the resultant force on it is zero. The tension in each string can be resolved into a vertical component and a horizontal component. The horizontal components cancel, and the two vertical components together must balance the 20 N weight, so each vertical component must be 10 N. Because each string is at an angle, only part of its tension acts upwards, so the tension itself must be greater than 10 N.

Explanation

When a force acts at an angle, only its component in a particular direction has an effect in that direction. The more the strings slope away from the vertical, the smaller the upward part of each tension, so the tension has to be larger to hold up the same weight.

Mark scheme

1 mark
the picture is in equilibrium / the resultant force on it is zero
max 3
1 mark
the tension in each string can be resolved into a vertical component and a horizontal component
1 mark
the two vertical components together must balance the 20 N weight, so each vertical component is 10 N
1 mark
the string is at an angle so the tension is bigger than its vertical component
allow only part of the tension acts upwards

Question 4

4 marks

A child pulls a sledge across flat snow using a rope at an angle to the ground. The pull in the rope has a horizontal component of 80 N and a vertical component of 60 N.

The friction force on the sledge is 50 N. The weight of the sledge is 400 N.

Calculate the resultant horizontal force on the sledge.
Calculate the normal contact force from the ground on the sledge.

Model answer

Horizontal: resultant = 80 N − 50 N = 30 N forwards.
Vertical: the sledge does not move up or down, so the vertical forces are balanced.
Upward forces = downward forces
normal contact force + 60 N = 400 N
normal contact force = 400 N − 60 N = 340 N

Explanation

Treat the horizontal and vertical directions separately, using the components of the angled pull. Because the rope pulls partly upwards, the ground does not have to support the whole weight, so the normal contact force is less than 400 N.

Mark scheme

1 mark
80 − 50
a horizontal resultant of 30 N forwards scores 2 marks
1 mark
30 N forwards
allow 30 N in the direction of the pull
1 mark
normal contact force + 60 = 400 / normal contact force = 400 − 60
the sledge does not move vertically so the vertical forces balance; a normal contact force of 340 N scores 2 marks
1 mark
340 N
do not accept 400 N or 460 N

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