Moments: rods and beams - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 16 exam-style questions on Moments: rods and beams 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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3–9 marks each
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16 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

4 marks

A non-uniform metre rule AB has length 100 cm. When the rule is placed on a knife-edge at the point 42 cm from A, it rests horizontally in equilibrium.

A small object of mass 50 g is then hung from the rule at the point 10 cm from A. The knife-edge is moved so that the rule again rests horizontally in equilibrium, and the knife-edge is now 34 cm from A.

The object is modelled as a particle.

Find the mass of the rule.

Model answer

The rule balances on its own at 42 cm, so its centre of mass G is 42 cm from A.
With the object, moments about the knife-edge (34 cm from A):
\(Mg \times (42 - 34) = 0.05g \times (34 - 10)\)
\(8M = 1.2\)
\(M = 0.15\) kg, so the mass of the rule is 150 g.

Explanation

When the rule balances on its own, its weight must act through the knife-edge, which locates the centre of mass at 42 cm. With the object added, the moments of the rule's weight and the object's weight about the new knife-edge balance, and the distances are measured from the knife-edge, not from A. A common error is to use distances from A in the moments equation.

Mark scheme

1 mark
B1 The centre of mass of the rule is 42 cm from A (the weight acts through the knife-edge when the rule balances on its own); or a correct final answer of 150 g (0.15 kg) with no working
An answer of 150 g (or 0.15 kg) with no working scores 4 marks.
1 mark
M1 Takes moments about the knife-edge in its new position (34 cm from A): \(Mg \times (42 - 34) = 0.05g \times (34 - 10)\)
Two terms, each a weight times a distance from the knife-edge; masses in grams with the object as 50 is fine. g may be omitted throughout (it cancels). An omitted distance is a method error: M0.
1 mark
A1 \(8M = 0.05 \times 24\) oe
1 mark
A1 Mass of the rule 150 g (or 0.15 kg)
Depends on the M mark in point 2 (not on the A mark just before). Units needed here, since the answer could be in grams or kilograms: 150 with no unit is A1 only if the working is clearly in grams.

Question 2

5 marks

A uniform scaffold board AB, of length 6 m and mass 40 kg, hangs horizontally from two vertical ropes, one attached at C and the other at D, where AC = 1.5 m and DB = 0.5 m. A bag of tools of mass 20 kg is placed on the board at the point E.

The tension in the rope at C is 98 N greater than the tension in the rope at D.

The board is modelled as a uniform rod, the bag as a particle and the ropes as light.

Find the distance AE.

Use \(g = 9.8\ \text{m s}^{-2}\).

Model answer

Resolving vertically: \(T_C + T_D = 40g + 20g = 60g\)
\(T_C - T_D = 98 = 10g\)
Adding: \(2T_C = 70g\), so \(T_C = 35g\) and \(T_D = 25g\)
Moments about C (G is 1.5 m from C and D is 4 m from C):
\(T_D \times 4 = 40g \times 1.5 + 20g \times (AE - 1.5)\)
\(100g = 60g + 20g(AE - 1.5)\)
\(AE - 1.5 = 2\), so \(AE = 3.5\) m

Explanation

Resolving vertically and using the given difference gives both tensions, since 98 N is 10g. Then a moments equation gives the position of the bag; taking moments about C keeps the distances simple. A common error is to treat 98 N as a mass or to measure every distance from A while taking moments about C.

Mark scheme

1 mark
M1 Resolves vertically and uses the difference of the tensions: \(T_C + T_D = 60g\) and \(T_C - T_D = 98\) (= 10g); or a correct final answer of 3.5 m with no working
An answer of AE = 3.5 m with no working scores 1 mark (the final A1 only): the equations must be seen for the method marks.
1 mark
A1 \(T_C = 35g\) (343 N) and \(T_D = 25g\) (245 N)
Both needed.
1 mark
M1 Takes moments about C (or A, or any point) with their tensions: \(T_D \times 4 = 40g \times 1.5 + 20g \times (AE - 1.5)\)
All forces with a moment, each multiplied by its distance; an omitted distance is a method error: M0. Moments about A: \(T_C \times 1.5 + T_D \times 5.5 = 40g \times 3 + 20g \times AE\). An omitted or extra g is an accuracy error.
1 mark
A1 \(100g = 60g + 20g(AE - 1.5)\) oe
1 mark
A1 \(AE = 3.5\) m
Depends on the M mark in point 3 (not on the A mark just before). Depends on both M marks. Units not needed. Working with g = 9.8 numerically (343 N and 245 N) also gives exactly 3.5 m. Use of g = 9.81 (which gives 3.50 m) is penalised once per question.

Question 3

7 marks

A non-uniform pole AB, of length 2.5 m, hangs in a horizontal position from two vertical light strings attached at A and at B. Each string is attached to a spring balance. The balance attached at A reads 30 N and the balance attached at B reads 45 N.

The pole is modelled as a non-uniform rod.

Find the mass of the pole and the distance of its centre of mass from A.

A lantern of mass 2 kg is then hung from the pole at the midpoint of AB, and the pole remains horizontal. The lantern is modelled as a particle. Find the new reading on each balance.

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

Model answer

Resolving vertically: weight \(= 30 + 45 = 75\) N, so the mass is \(\dfrac{75}{9.8} = 7.7\) kg (2 s.f.)
Moments about A, with G a distance x m from A: \(45 \times 2.5 = 75x\), so \(x = 1.5\) m
With the lantern (weight \(2g = 19.6\) N) at the midpoint, moments about A:
\(T_B \times 2.5 = 75 \times 1.5 + 19.6 \times 1.25 = 112.5 + 24.5 = 137\)
\(T_B = 54.8\) N
Resolving vertically: \(T_A = 75 + 19.6 - 54.8 = 39.8\) N

Explanation

The two readings together balance the weight, and moments about one end locate the centre of mass. Adding the lantern changes both readings; moments about A give the new reading at B and resolving gives the reading at A. A common error is to give 75 as the mass in kilograms, or to split the lantern's weight equally between the balances without taking moments (which happens to be wrong here because the pole's weight is not at the midpoint).

Mark scheme

1 mark
M1 Resolves vertically: weight \(= 30 + 45 = 75\) N, and divides by g; or correct final answers of 7.7 kg (7.65), 1.5 m, 54.8 N at B and 39.8 N at A with no working
Correct answers of 7.7 kg (or 7.65 kg), 1.5 m, 54.8 N at B and 39.8 N at A with no working score 4 marks (the four A1 marks). An answer of 75 kg for the mass is A0. Accept any equivalent form.
1 mark
A1 Mass \(= \dfrac{75}{9.8} = 7.7\) kg or 7.65 kg
Answers from g = 9.8 should be given to 2 or 3 s.f.: 7.653 loses this mark (once per question). Use of g = 9.81 is penalised once per question. 75 kg is A0. Accept awrt 7.65, i.e. [7.645, 7.655), or 7.7.
1 mark
M1 Takes moments about A: \(45 \times 2.5 = 75x\)
Or about B: \(30 \times 2.5 = 75(2.5 - x)\). An omitted distance is a method error: M0. Accept any equivalent form.
1 mark
A1 Centre of mass 1.5 m from A
Units not needed. cao (1.5 m exactly).
1 mark
M1 With the lantern, takes moments about A (or B): \(T_B \times 2.5 = 75 \times 1.5 + 2g \times 1.25\)
Their centre of mass. The lantern's weight is 2g = 19.6 N. Accept any equivalent form. An omitted or extra g on the lantern's weight is an accuracy error.
1 mark
A1 Reading at B: \(T_B = 54.8\) N
Accept 55 (2 s.f.).
1 mark
A1 Reading at A: \(T_A = 39.8\) N
Depends on the third M mark. From resolving, \(T_A = 75 + 19.6 - T_B\), or from moments about B. Accept 40 (2 s.f.).

Question 4

9 marks

Figure 1 shows a model of a tower crane. The jib AB is horizontal, has length 40 m and mass 6000 kg, and rests on two supports at C and D at the top of the tower, where AC = 10 m and CD = 4 m. A counterweight of mass 9000 kg is fixed to the jib at A. A load of mass L kg hangs from the jib at B on a vertical cable.

In the model, the jib is a uniform rod, the counterweight and the load are particles, and the cable is light.

Find the range of values of L for which the jib does not tilt.

Explain why, according to the model, the crane must not be left with no load at B.

A horizontal jib AB resting on two supports C and D on top of a tower. A counterweight block is fixed at A and a load hangs from B on a vertical cable. AC is 10 m, CD is 4 m and AB is 40 m.

Model answer

Positions from A: C 10 m, D 14 m, G 20 m, B 40 m.
Tilting about C (A end down) happens when the force at D is zero. No tilting needs
\(9000g \times 10 \leqslant 6000g \times 10 + Lg \times 30\)
\(90\,000 \leqslant 60\,000 + 30L\), so \(L \geqslant 1000\)
Tilting about D (B end down) happens when the force at C is zero. No tilting needs
\(Lg \times 26 + 6000g \times 6 \leqslant 9000g \times 14\)
\(26L + 36\,000 \leqslant 126\,000\), so \(L \leqslant 3461.5\)
So \(1000 \leqslant L \leqslant 3460\) (3 s.f.)
With no load, the moment of the counterweight about C (90 000g N m) is greater than the moment of the jib's weight about C (60 000g N m), so the jib would tilt about C, with the counterweight end A going down.

Explanation

The jib can tilt either way: about C when the load is too small and about D when it is too large, and in each case the force from the other support is zero at the point of tilting. Taking moments about the support it would tilt about gives each bound, and g cancels throughout. A common error is to look at only one way of tilting, which misses the lower bound that explains why the crane needs a load.

Mark scheme

1 mark
B1 States the two tilting conditions: about to tilt about C (end A going down) when the force at D is zero, and about to tilt about D (end B going down) when the force at C is zero; or a correct final answer of \(1000 \leqslant L \leqslant 3460\) with no working
An answer of \(1000 \leqslant L \leqslant 3460\) with no working scores 3 marks (the A1 marks for the two bounds and the final range); the B1 for the explanation needs its reason.
1 mark
M1 Takes moments about C with the force at D zero: \(9000g \times 10 = 6000g \times 10 + Lg \times 30\) (or the inequality \(9000 \times 10 \leqslant 6000 \times 10 + 30L\))
G is 20 m from A, so 10 m from C; B is 30 m from C. g may be omitted throughout (it cancels). An omitted distance is a method error: M0.
1 mark
A1 \(L \geqslant 1000\)
1 mark
M1 Takes moments about D with the force at C zero: \(Lg \times 26 + 6000g \times 6 = 9000g \times 14\) (or the corresponding inequality)
D is 14 m from A: G is 6 m from D and B is 26 m from D.
1 mark
A1 \(26L + 36\,000 = 126\,000\) oe
1 mark
A1 \(L \leqslant 3460\) (3461.5, or \(\dfrac{45\,000}{13}\))
Depends on the M mark in point 4 (not on the A mark just before). Here g cancels, so accept 3460, 3461 or 3462, or \(\dfrac{45\,000}{13}\); accept 3500 (2 s.f.).
1 mark
A1 \(1000 \leqslant L \leqslant 3460\)
Depends on both earlier A1 bounds. The inequalities must allow equality (on the point of tilting is still not tilting): strict inequalities score A0 for this mark only.
1 mark
B1 With no load (L = 0 < 1000), the moment of the counterweight about C (90 000g) is greater than the moment of the jib's weight about C (60 000g)
Must give a reason in terms of moments about C (or refer to their lower bound for L). 'It would be unbalanced' alone is B0.
1 mark
B1 so the jib would tilt about C, with the end A (the counterweight end) going down
Needs the conclusion that it tilts (or falls) towards A / about C.

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