Edexcel A level Maths (9MA0) · Mechanics › Moments
Practise Ladders, tilting and limiting equilibrium. 3 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy8 marks
A ladder \(AB\), of length 5 m and mass 25 kg, rests in equilibrium with its end \(A\) on rough horizontal ground and its end \(B\) against a smooth vertical wall. The ladder makes an angle \(\theta\) with the ground, where \(\tan \theta = \frac{12}{5}\), and lies in a vertical plane perpendicular to the wall. The ladder is modelled as a uniform rod. Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.
[object Object]
(a) Find the magnitude of the reaction of the wall on the ladder.[3]
(b) Find the magnitude of the frictional force exerted by the ground on the ladder.[2]
(c) Find the magnitude of the normal reaction of the ground on the ladder.[1]
(d) Show that the coefficient of friction between the ladder and the ground is at least \(\frac{5}{24}\).[2]
Show the answer and mark scheme
(a)Answer: 51.0 N (accept 51 N) N
M1 for taking moments about \(A\) with the correct terms
A1 for \(S \times 5\sin\theta = 25g \times 2.5\cos\theta\)
A1 for 51.0 or 51 (N) or \(\frac{125}{24}g\)
Worked solution: Moments about \(A\): \(S \times 5\sin\theta = 25g \times 2.5\cos\theta\), so \(S = \frac{25g}{2\tan\theta} = \frac{125}{24}g = 51.0\) N.
(b)Answer: 51.0 N (accept 51 N) N
M1 for resolving horizontally: \(F = S\)
A1ft for 51.0 or 51 (N)
Worked solution: Resolving horizontally: \(F = S = 51.0\) N.
(c)Answer: 245 N (accept 250 N) N
B1 for \(R = 25g\) = 245 or 250 (N)
Worked solution: Resolving vertically: \(R = 25g = 245\) N.
A ladder of length 5 m and mass 12 kg rests with its foot on rough horizontal ground and its top against a smooth vertical wall. The ladder makes an angle of \(65^\circ\) with the ground and is on the point of slipping. The ladder is modelled as a uniform rod, and it lies in a vertical plane perpendicular to the wall. Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.
[object Object]
(a) Find the magnitude of the reaction of the wall on the ladder.[3]
(b) Find the coefficient of friction between the ladder and the ground, giving your answer to 3 significant figures.[3]
(c) Explain how you have used the modelling assumption that the ladder is uniform.[1]
(d) In reality the ladder is not uniform: it is heavier near its foot. State, with a reason, whether the true value of the coefficient of friction needed would be greater or smaller than your answer to part (b).[2]
(c)Answer: The weight of the ladder acts at its midpoint, 2.5 m from each end.
B1 for the weight acts at the midpoint of the ladder
Worked solution: A uniform ladder has its centre of mass at its midpoint, so its weight acts 2.5 m from the foot.
(d)Answer: Smaller: the centre of mass is nearer the foot, so the moment of the weight about the foot is smaller, the wall reaction is smaller and less friction is needed.
B1 for smaller
B1 for the weight acts closer to the foot, reducing its moment about the foot and hence the reaction at the wall and the friction required
Worked solution: If the centre of mass is lower, \(S \times 5\sin 65^\circ = 12g \times d\cos 65^\circ\) with \(d \lt 2.5\), so \(S\), and hence the friction required, is smaller.
Question 3Hard9 marks
A uniform ladder \(AB\), of length 6 m and mass 15 kg, rests with its end \(A\) on rough horizontal ground and its end \(B\) against a smooth vertical wall. The ladder makes an angle \(\theta\) with the ground, where \(\tan\theta = 3\), and lies in a vertical plane perpendicular to the wall. A decorator needs to put a pot of paint of mass 5 kg on the ladder, either at \(A\) or at \(B\). The pot of paint is modelled as a particle. Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.
[object Object]
(a) Find the least coefficient of friction between the ladder and the ground needed for equilibrium when the pot is at \(A\).[4]
(b) Find the least coefficient of friction needed when the pot is at \(B\).[3]
(c) The coefficient of friction between the ladder and the ground is 0.15. Advise the decorator where to place the pot of paint, giving a reason.[2]
Show the answer and mark scheme
(a)Answer: \(\mu = \frac{1}{8} = 0.125\)
M1 for moments about \(A\) (the pot at \(A\) has no moment): \(S \times 6\sin\theta = 15g \times 3\cos\theta\)
M1 for \(\mu \geqslant \frac{F}{R} = \frac{S}{20g}\)
A1 for \(\frac{1}{8}\)
Worked solution: Moments about \(A\): \(6S\sin\theta = 45g\cos\theta\), so \(S = \frac{45g}{6\tan\theta} = 2.5g\). With the pot at \(A\), \(R = 20g\) and \(F = S\), so \(\mu \geqslant \frac{2.5g}{20g} = \frac{1}{8}\).
(b)Answer: \(\mu = \frac{5}{24} \approx 0.208\)
M1 for moments about \(A\) including the pot: \(6S\sin\theta = 15g \times 3\cos\theta + 5g \times 6\cos\theta\)
A1 for \(S = \frac{75g}{18} = \frac{25g}{6}\)
A1 for \(\mu \geqslant \frac{25g/6}{20g} = \frac{5}{24}\)
Worked solution: \(6S \times 3 = 45g + 30g\) (dividing by \(\cos\theta\)), so \(S = \frac{75g}{18} = \frac{25g}{6}\). \(R = 20g\), so \(\mu \geqslant \frac{25}{120} = \frac{5}{24}\).
(c)Answer: At \(A\) (the foot): 0.15 exceeds the \(\frac{1}{8}\) needed, but is less than the \(\frac{5}{24}\) needed with the pot at \(B\), so with the pot at \(B\) the ladder would slip.
M1 for comparing 0.15 with both of their values
A1 for place it at \(A\), since \(0.125 \lt 0.15 \lt 0.208\) (the ladder would slip if the pot were at \(B\))
Worked solution: Only the arrangement with the pot at the foot needs \(\mu \leqslant 0.15\): \(\frac{1}{8} = 0.125 \lt 0.15\), whereas \(\frac{5}{24} \approx 0.208 \gt 0.15\). So the pot should be placed at \(A\).