Forces, resultants and equilibrium - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 19 exam-style questions on Forces, resultants and equilibrium for Edexcel A-Level Maths (Paper 3). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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3–10 marks each
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19 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

3 marks

A particle moves with constant velocity \((2\mathbf{i} - 5\mathbf{j})\ \text{m s}^{-1}\) under the action of three forces only: \(\mathbf{F}_1 = (3\mathbf{i} + 4\mathbf{j})\) N, \(\mathbf{F}_2 = (-7\mathbf{i} + \mathbf{j})\) N and \(\mathbf{F}_3\).

Find the exact magnitude of \(\mathbf{F}_3\).

Model answer

The velocity is constant, so the acceleration is zero and by Newton's first law the resultant force is zero.
\(\mathbf{F}_3 = -(\mathbf{F}_1 + \mathbf{F}_2) = -(-4\mathbf{i} + 5\mathbf{j}) = 4\mathbf{i} - 5\mathbf{j}\)
\(|\mathbf{F}_3| = \sqrt{4^2 + (-5)^2} = \sqrt{41}\) N

Explanation

A particle moving with constant velocity is in equilibrium, exactly as if it were at rest, so the three forces add to zero. The velocity itself is not used. A common error is to bring the velocity into the force equation, or to give 6.40 when the exact value is asked for.

Mark scheme

1 mark
M1 Uses Newton's first law: constant velocity, so \(\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0}\); or a correct final answer of \(\sqrt{41}\) N with no working
An answer of \(\sqrt{41}\) N with no working scores 3 marks. M0 for using the velocity in a force equation, e.g. \(\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = 2\mathbf{i} - 5\mathbf{j}\).
1 mark
A1 \(\mathbf{F}_3 = (4\mathbf{i} - 5\mathbf{j})\) N
Accept a column vector.
1 mark
A1 \(|\mathbf{F}_3| = \sqrt{41}\) N
Depends on the M mark. Exact answer required: 6.40 alone is A0 (isw if \(\sqrt{41}\) is seen first).

Question 2

6 marks

A particle is in equilibrium under the action of three forces \(\mathbf{P}\), \(\mathbf{Q}\) and \(\mathbf{R}\).
The force \(\mathbf{P}\) has magnitude 39 N and acts in the direction of the vector \((5\mathbf{i} - 12\mathbf{j})\).
The force \(\mathbf{Q} = (-3\mathbf{i} + 20\mathbf{j})\) N.

Find \(\mathbf{R}\), giving your answer in the form \((a\mathbf{i} + b\mathbf{j})\) N, and find the magnitude of \(\mathbf{R}\).

Model answer

\(|5\mathbf{i} - 12\mathbf{j}| = \sqrt{25 + 144} = 13\)
\(\mathbf{P} = 39 \times \dfrac{5\mathbf{i} - 12\mathbf{j}}{13} = 15\mathbf{i} - 36\mathbf{j}\)
Equilibrium: \(\mathbf{P} + \mathbf{Q} + \mathbf{R} = \mathbf{0}\)
\(\mathbf{R} = -(15\mathbf{i} - 36\mathbf{j}) - (-3\mathbf{i} + 20\mathbf{j}) = (-12\mathbf{i} + 16\mathbf{j})\) N
\(|\mathbf{R}| = \sqrt{(-12)^2 + 16^2} = 20\) N

Explanation

A force of given magnitude in a given direction is the magnitude times the unit vector in that direction, so \(\mathbf{P}\) is \(\dfrac{39}{13}\) times \(5\mathbf{i} - 12\mathbf{j}\). In equilibrium the three forces add to zero, which gives \(\mathbf{R}\), and Pythagoras gives its magnitude. Common errors are multiplying the direction vector by 39 and giving the resultant of \(\mathbf{P}\) and \(\mathbf{Q}\) instead of the force that balances it.

Mark scheme

1 mark
M1 Writes \(\mathbf{P}\) as 39 times the unit vector in the direction of \(5\mathbf{i} - 12\mathbf{j}\): \(\mathbf{P} = \dfrac{39}{13}(5\mathbf{i} - 12\mathbf{j})\); or correct answers of \(\mathbf{R} = (-12\mathbf{i} + 16\mathbf{j})\) N and 20 N with no working
Correct answers of \(\mathbf{R} = (-12\mathbf{i} + 16\mathbf{j})\) N and \(|\mathbf{R}| = 20\) N with no working score 2 marks (the A marks for \(\mathbf{R}\) and its magnitude). A common error is to take \(\mathbf{P} = 39(5\mathbf{i} - 12\mathbf{j})\) or \(\mathbf{P} = 5\mathbf{i} - 12\mathbf{j}\): M0.
1 mark
A1 \(\mathbf{P} = (15\mathbf{i} - 36\mathbf{j})\) N
Accept a column vector.
1 mark
M1 Uses equilibrium: \(\mathbf{P} + \mathbf{Q} + \mathbf{R} = \mathbf{0}\) with their \(\mathbf{P}\)
Sign slips condoned for the M mark.
1 mark
A1 \(\mathbf{R} = (-12\mathbf{i} + 16\mathbf{j})\) N
The answer must be in the form asked for: a column vector as the final answer is A0. \((12\mathbf{i} - 16\mathbf{j})\) (the resultant of \(\mathbf{P}\) and \(\mathbf{Q}\), not the balancing force) is A0.
1 mark
M1 Uses Pythagoras, with the square root, on their \(\mathbf{R}\)
1 mark
A1 \(|\mathbf{R}| = 20\) N
cao

Question 3

6 marks

Figure 1 shows a hanging basket of mass 4.5 kg suspended from the end \(B\) of a light rod \(AB\). The rod is horizontal and its end \(A\) is freely hinged to a vertical wall. A light chain \(BC\) is attached to the wall at the point \(C\), vertically above \(A\), and the chain makes an angle of \(50^\circ\) with the wall. The basket hangs from \(B\) on a short light chain and the system is in equilibrium. The basket is modelled as a particle, and the force exerted by the light rod on \(B\) acts along the rod.

Find the tension in the chain \(BC\) and the magnitude of the force in the rod \(AB\), stating whether the rod is in tension or in thrust.

Use \(g = 9.8\ \text{m s}^{-2}\) and give your answers to 2 or 3 significant figures.

Figure 1: a vertical wall on the left. A horizontal rod AB sticks out from the wall at A. A straight chain runs from B up to a point C on the wall above A, and the angle between the chain and the wall at C is marked 50 degrees. A particle hangs from B on a short vertical chain.

Model answer

Forces on \(B\): the tension \(T\) along \(BC\), the force \(S\) from the rod along \(AB\), and the weight \(4.5g\) of the basket.
Resolving vertically: \(T\cos 50^\circ = 4.5g = 44.1\)
\(T = \dfrac{44.1}{\cos 50^\circ} = 68.61\)
Tension in the chain = 68.6 N (3 s.f.)
Resolving horizontally: \(S = T\sin 50^\circ = 68.61 \times \sin 50^\circ = 52.56\)
Force in the rod = 52.6 N (3 s.f.)
The chain pulls \(B\) towards the wall, so the rod must push \(B\) away from the wall: the rod is in thrust.

Explanation

Treat \(B\) as a particle in equilibrium under three forces: the vertical resolution gives the chain's tension, and the horizontal resolution gives the force in the rod. The horizontal component of the chain's tension pulls \(B\) towards the wall, so the rod must push outwards, which means it is in thrust. Common errors are resolving the \(50^\circ\) angle the wrong way (it is measured from the vertical wall) and calling the rod's force a tension.

Mark scheme

1 mark
M1 Resolves vertically at \(B\): \(T\cos 50^\circ = 4.5g\) (tension resolved; condone sin/cos confusion); or correct answers of 68.6 N (or 69 N) and 52.6 N (or 53 N) with no working
Correct answers of 68.6 N (or 69 N) for the chain and 52.6 N (or 53 N) for the rod, with thrust stated, with no working score 3 marks (the A marks for the two values and the B1). An omitted \(g\) is an accuracy error (M1 A0). Accept any equivalent form.
1 mark
A1 \(T\cos 50^\circ = 4.5g = 44.1\)
Accept any equivalent form.
1 mark
A1 \(T = 68.6\) N or 69 N
Depends on the first M mark. Answers from g = 9.8 to 2 or 3 s.f. (68.61 is over-accurate; penalise over-accuracy once in the question). Use of g = 9.81 is penalised once in this question.
1 mark
M1 Resolves horizontally at \(B\): \(S = T\sin 50^\circ\) with their \(T\) (or \(S = 4.5g\tan 50^\circ\))
M0 if a reaction from the wall is wrongly included at \(B\). Accept any equivalent form.
1 mark
A1 Force in the rod \(= 52.6\) N or 53 N
52.56 is over-accurate (only penalised if not already penalised in this question). Accept awrt 52.6, i.e. [52.55, 52.65), or 53.
1 mark
B1 The rod is in thrust (compression)
Accept 'thrust', 'compression', 'the rod pushes \(B\) away from the wall' (the chain pulls \(B\) towards the wall, so the rod must push outwards). 'Tension' is B0. A contradictory reason is B0.

Question 4

8 marks

At a school sports day, four teams take part in a four-way tug of war. Each team pulls on a horizontal rope attached to a small metal ring, which is modelled as a particle. At one instant the ring is at rest and the forces exerted on it by the ropes are
400 N due north,
250 N due east,
350 N on a bearing of \(240^\circ\),
\(P\) N on a bearing of \(\theta^\circ\).

Find the value of \(P\) and the value of \(\theta\). Give \(P\) to 3 significant figures and \(\theta\) to one decimal place.

Model answer

Let \(X\) and \(Y\) be the east and north components of the fourth force.
East: \(250 + 350\sin 240^\circ + X = 0\)
\(250 - 303.11 + X = 0\), so \(X = 53.11\)
North: \(400 + 350\cos 240^\circ + Y = 0\)
\(400 - 175 + Y = 0\), so \(Y = -225\)
\(P = \sqrt{53.11^2 + 225^2} = 231.18\), so \(P = 231\) (3 s.f.)
The fourth force points east of south: \(\tan\phi = \dfrac{53.11}{225}\), \(\phi = 13.28^\circ\)
\(\theta = 180 - 13.28 = 166.7\)

Explanation

The ring is in equilibrium, so the components of all four forces add to zero in each of two perpendicular directions; east and north suit bearings. The fourth force has the components that cancel the other three, so Pythagoras gives \(P\) and the signs of the components show it points south-east. Common errors are resolving \(240^\circ\) as if it were measured from east and giving the acute angle \(13.3^\circ\) instead of the bearing.

Mark scheme

1 mark
M1 Resolves east-west, with all four forces and the 350 N force resolved: \(250 + 350\sin 240^\circ + X = 0\) (or \(250 - 350\cos 30^\circ + X = 0\)), where \(X\) is the east component of the fourth force; or correct answers of \(P = 231\) and \(\theta = 166.7\) with no working
Correct answers of \(P = 231\) and \(\theta = 166.7\) with no working score 2 marks (the final A marks for P and for theta). Condone sin/cos confusion and sign errors for the M marks. Alternative: write each force as a vector (\(\mathbf{i}\) east, \(\mathbf{j}\) north) and set the sum to zero; the M and A marks are the same for the two components.
1 mark
A1 East component of the fourth force \(= 350\cos 30^\circ - 250 = 53.1\) N
Accept \(P\sin\theta = 53.1\).
1 mark
M1 Resolves north-south, with the 350 N force resolved: \(400 + 350\cos 240^\circ + Y = 0\) (or \(400 - 350\sin 30^\circ + Y = 0\))
1 mark
A1 North component of the fourth force \(= -225\) N (225 N to the south)
Accept \(P\cos\theta = -225\).
1 mark
M1 (dM1) Uses Pythagoras on their two components: \(P = \sqrt{53.1^2 + 225^2}\)
Depends on both resolving M marks. Accept their components to 3 s.f. or better.
1 mark
A1 \(P = 231\) (awrt 231, i.e. [230.5, 231.5))
No \(g\) is used: 3 s.f. as asked.
1 mark
M1 Uses trigonometry for a relevant angle, e.g. \(\tan\phi = \dfrac{53.1}{225}\) (\(\phi = 13.3^\circ\), the angle east of south)
With their components; accept 3 s.f. or better.
1 mark
A1 \(\theta = 166.7\) (awrt 166.7, i.e. [166.65, 166.75))
The bearing must be in the south-east quadrant. \(13.3\) or \(076.7\) as the final answer is A0.

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