Edexcel A level Maths (9MA0) · Mechanics › Moments
Practise Moments, beams and rods in equilibrium. 2 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 1Easy7 marks
A uniform plank \(AB\) has length 4 m and mass 30 kg. It rests horizontally on two supports, at \(A\) and at \(C\), where \(AC = 3\) m. A child of mass 25 kg sits on the plank at \(B\). The plank is modelled as a uniform rod and the child 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 magnitude of the reaction at \(C\).[3]
(b) Find the magnitude of the reaction at \(A\).[2]
(c) State how you have used the modelling assumption that the child is a particle.[1]
(d) State how you have used the modelling assumption that the plank is a rod.[1]
Show the answer and mark scheme
(a)Answer: 523 N (accept 520 N or \(\frac{160g}{3}\)) N
M1 for taking moments about \(A\)
A1 for \(3R_C = 30g \times 2 + 25g \times 4\)
A1 for 523 or 520 (N)
Worked solution: Moments about \(A\): \(3R_C = 30g \times 2 + 25g \times 4 = 160g\), so \(R_C = \frac{160g}{3} = 523\) N (3 s.f.).
(b)Answer: 16.3 N (accept 16 N or \(\frac{5g}{3}\)) N
M1 for resolving vertically: \(R_A + R_C = 55g\)
A1ft for 16.3 or 16 (N)
Worked solution: \(R_A = 55g - \frac{160g}{3} = \frac{5g}{3} = 16.3\) N (3 s.f.). (The reaction at \(A\) is small: the plank is close to tilting about \(C\).)
(c)Answer: The weight of the child acts at the single point \(B\).
B1 for the child's weight acts at a single point (at \(B\))
Worked solution: Modelling the child as a particle means the child's weight acts at one point, \(B\), at a known distance from \(A\).
(d)Answer: The plank is treated as rigid and one-dimensional (it does not bend and its thickness is ignored), so all the forces act along one straight line.
B1 for the plank does not bend (rigid) / has no thickness, so the distances along it are fixed and the forces act on a line
Worked solution: A rod is rigid and has negligible thickness: it stays straight, so the distances used in the moments equation are fixed.
Question 2Medium8 marks
A uniform plank \(AB\) has length 6 m and mass 40 kg. It rests horizontally on two supports, at \(C\) and \(D\), where \(AC = 1\) m and \(AD = 4\) m. A man of mass 60 kg walks along the plank from \(D\) towards \(B\). The plank is modelled as a uniform rod and the man as a particle.
(a) Find the greatest distance from \(A\) that the man can reach before the plank starts to tilt.[3]
(b) A bag of mass 20 kg is placed on the plank at \(A\). Show that the man can now just reach \(B\).[3]
(c) Explain why modelling the man as a particle may not be appropriate when he is close to \(B\).[1]
(d) State how the assumption that the plank is uniform has been used.[1]
Show the answer and mark scheme
(a)Answer: \(\frac{14}{3}\) m (4.67 m) m
M1 for the reaction at \(C\) being zero when the plank is about to tilt about \(D\)
M1 for moments about \(D\): \(40g \times 1 = 60g(x - 4)\)
A1 for \(x = \frac{14}{3}\) (m)
Worked solution: When the plank is about to tilt about \(D\), the reaction at \(C\) is zero. The plank's weight acts at its midpoint, 3 m from \(A\), i.e. 1 m on the \(A\) side of \(D\). Moments about \(D\): \(40g \times 1 = 60g \times (x - 4)\), so \(x - 4 = \frac{2}{3}\) and \(x = \frac{14}{3}\) m (about 4.67 m from \(A\)).
(b)Answer: With the man at \(B\) and the reaction at \(C\) zero, moments about \(D\): \(40g \times 1 + 20g \times 4 = 120g\) and \(60g \times 2 = 120g\), so the plank is on the point of tilting but does not tilt.
M1 for taking moments about \(D\) with the man at \(B\) (2 m from \(D\)) and the bag at \(A\) (4 m from \(D\))
A1 for anticlockwise \(40g + 80g = 120g\) and clockwise \(120g\)
A1* for the moments balance with the reaction at \(C\) equal to zero, so the plank is only just on the point of tilting when the man reaches \(B\)
Worked solution: Take moments about \(D\) with the man at \(B\) and \(R_C = 0\). Anticlockwise (on the \(A\) side): plank \(40g \times 1\) + bag \(20g \times 4 = 120g\). Clockwise: man \(60g \times 2 = 120g\). These are equal, so with the man at \(B\) the plank is just on the point of tilting (\(R_C = 0\)): he can just reach \(B\).
(c)Answer: His weight is spread over the length of his feet, so it does not act at a single point; close to the point of tilting, the exact position of his weight matters.
B1 for the man's weight is not concentrated at a point (it is spread over his feet), and near the tilting position small changes in where it acts matter
Worked solution: A real person's weight is spread over the area of their feet, so it does not act at a single point. When the plank is on the point of tilting, a small shift in where his weight acts decides whether it tilts, so the particle model is least reliable there.
(d)Answer: The weight of the plank acts at its midpoint, 3 m from \(A\).
B1 for the weight acts at the midpoint of the plank (3 m from \(A\))
Worked solution: A uniform plank has its centre of mass at its midpoint, so its weight was taken to act 3 m from \(A\).
Question 3Hard9 marks
A non-uniform rod \(AB\) has length 5 m and mass 10 kg. The rod rests in a horizontal position on two supports at \(C\) and \(D\), where \(AC = 1\) m and \(DB = 1\) m. The reaction at \(C\) is twice the reaction at \(D\). The centre of mass of the rod is at the point \(G\). 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 distance \(AG\).[4]
(b) A particle of mass \(m\) kg is now placed on the rod at \(B\), and the rod is on the point of tilting about \(D\). Find the value of \(m\).[4]
(c) Explain how the modelling assumption that the rod is non-uniform affects your working.[1]
Show the answer and mark scheme
(a)Answer: 2 m
M1 for resolving vertically with the given ratio: \(R_C + R_D = 10g\) and \(R_C = 2R_D\)
A1 for \(R_D = \frac{10}{3}g\)
M1 for taking moments about a point (e.g. \(C\)) with \(AG\) unknown
A1 for \(AG = 2\) m
Worked solution: \(R_C + R_D = 10g\) with \(R_C = 2R_D\) gives \(R_D = \frac{10}{3}g\). Moments about \(C\): \(\frac{10}{3}g \times 3 = 10g(AG - 1)\), so \(AG = 2\) m.
(b)Answer: \(m = 20\)
B1 for the reaction at \(C\) is zero
M1 for moments about \(D\) with their \(AG\)
A1ft for \(10g \times 2 = mg \times 1\)
A1 for \(m = 20\)
Worked solution: About to tilt about \(D\), so \(R_C = 0\). Moments about \(D\): \(10g \times (4 - 2) = mg \times 1\), so \(m = 20\).
(c)Answer: The weight of the rod does not act at its midpoint, so its position (\(G\)) has to be found and used when taking moments.
B1 for the centre of mass (point of action of the weight) is not at the midpoint of the rod
Worked solution: For a non-uniform rod the weight acts at \(G\), which is not the midpoint; this is why \(AG\) had to be found.