Edexcel International A Level Maths · M2 · Mechanics 2
Statics of rigid bodies
28 free, high-quality past-paper-style questions on Statics of rigid bodies for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 244 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-M2-6-B-01M2 · A2Easy7 marks
A mechanic loosens a wheel nut with a wheel brace. The nut is at O and she pushes at the point A on the handle, where OA = 0.40 m. All forces act in the plane in which the brace turns. Model the brace as a light rod.
Figure: the wheel brace OA with the 150 N force at 65° to OA, as in part (b); not to scale.
(a)
She pushes at A with a force of 150 N perpendicular to OA. Find the moment of this force about O.
(1)
(b)
She now pushes at A with a force of 150 N at an angle of 65° to OA. Find the moment of this force about O.
(2)
(c)
The nut starts to turn when the moment about O reaches 75 N m. Find the least force that must be applied at A to turn the nut, and state its direction.
(2)
(d)
She pushes at A with a force of 220 N. Find the smallest angle between this force and OA for which the nut turns.
(2)
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Answer
Mark
Notes
Part (a)
150 × 0.40 = 60 N m
B1
Part (b)
moment = 150 × 0.40 sin 65° (or perpendicular component 150 sin 65° × 0.40)
M1
= 54.4 N m
A1
Part (c)
least force when the force is perpendicular to OA: F × 0.40 = 75
M1
F = 187.5 N (accept 188 N), perpendicular to OA
A1
Part (d)
220 × 0.40 sin θ = 75, so sin θ = 0.852…
M1
θ = 58.5°
A1
Common error: Using force × length of the handle whatever the angle. Only the component of the force perpendicular to OA turns the nut.
A light rectangular frame ABCD has AB = 0.6 m and AD = 0.4 m. It lies in a vertical plane with AB horizontal and D vertically above A, and it is free to turn about a fixed smooth axis through A, perpendicular to the frame. Three forces act on the frame: a force of 20 N at B acting vertically upwards, a force of 15 N at D acting horizontally in the direction from D to C, and a force of 25 N at C acting at 30° above the horizontal, directed away from the side AD.
Figure: the light frame ABCD, free to turn about A, with the three forces; not to scale.
(a)
Find the moment about A of the 20 N force, stating its sense.
(1)
(b)
Find the moment about A of the 15 N force, stating its sense.
(1)
(c)
By resolving the 25 N force into horizontal and vertical components, find its moment about A.
(3)
(d)
Find the total moment of the three forces about A, stating its sense.
(1)
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Answer
Mark
Notes
Part (a)
20 × 0.6 = 12 N m anticlockwise
B1
Part (b)
15 × 0.4 = 6 N m clockwise
B1
Part (c)
components: 25 cos 30° = 21.65 N horizontal and 25 sin 30° = 12.5 N vertical
M1
moment = 12.5 × 0.6 − 21.65 × 0.4
M1
= −1.16, so 1.16 N m clockwise
A1
Part (d)
12 − 6 − 1.16 = 4.84 N m anticlockwise (ft their (a), (b), (c))
A1
Exam tip: With a force at a corner, split it into components: each component has an easy perpendicular distance. Decide the sense of each before adding.
A diving board AB is modelled as a uniform rod of length 4.0 m and mass 30 kg. It is held horizontal by a bolt at A, which exerts a vertical force on the board, and by a support at C, where AC = 1.2 m, which pushes vertically upwards. A diver of mass 60 kg stands on the board at B. Model the diver as a particle. g = 9.8 m s−2.
Figure: the diving board, bolted at A and resting on the support at C, with the diver at B; not to scale.
(a)
Find the magnitude of the force exerted on the board by the support at C.
(3)
(b)
Find the magnitude and direction of the force exerted on the board by the bolt at A.
(2)
(c)
Explain, using moments about C, why the force at A must act downwards.
(1)
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Answer
Mark
Notes
Part (a)
moments about A: R × 1.2 = 30g × 2 + 60g × 4
M1
R × 1.2 = 300g
A1
R = 250g = 2450 N
A1
Part (b)
resolving vertically: 2450 = 90g + P, where P acts downwards
M1
P = 160g = 1570 N downwards (1568 N)
A1
Part (c)
the weight of the board (0.8 m beyond C) and the diver (2.8 m beyond C) both turn the board the same way about C, so the force at A must turn it the other way, which needs a downward force at A
B1
Exam tip: Take moments about the point where the unknown force you do not want acts. Here, moments about A remove the bolt force completely.
A uniform ladder of length 4 m and mass 18 kg rests in equilibrium with its foot on rough horizontal ground and its top against the smooth vertical wall of a shed. The ladder makes an angle of 70° with the ground and lies in a vertical plane perpendicular to the wall. g = 9.8 m s−2.
Figure: the uniform ladder AB against the smooth wall, at 70° to the rough ground; not to scale.
(a)
Find the magnitude of the force exerted on the ladder by the wall.
(3)
(b)
Find the friction force and the normal reaction at the ground.
(1)
(c)
Find the least possible value of the coefficient of friction between the ladder and the ground.
(2)
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Answer
Mark
Notes
Part (a)
moments about the foot: S × 4 sin 70° = 18g × 2 cos 70°
M1
correct equation, e.g. S = 9g ÷ tan 70°
A1
S = 32.1 N
A1
Part (b)
F = S = 32.1 N and R = 18g = 176.4 N
B1
Part (c)
F ≤ μR, so μ ≥ 32.1 ÷ 176.4
M1
least μ = 0.182
A1
Common error: Putting friction at the smooth wall. A smooth contact gives a normal reaction only; the friction is at the ground, towards the wall.
Figure 1 shows a light lamina PABC on a coordinate grid, with distances in metres. The lamina lies in a vertical plane and is free to turn about a fixed smooth axis through P, perpendicular to the lamina. Forces act on the lamina at A, B and C as shown. The force at C has magnitude Q newtons and acts in the positive x-direction. i and j are unit vectors in the positive x- and y-directions.
(a)
Find the moment about P of the force at A, stating its sense.
(2)
(b)
Hence find the perpendicular distance from P to the line of action of the force at A.
(2)
(c)
Show that the moment about P of the force at B is 16 N m anticlockwise.
(1)
(d)
The lamina is in equilibrium. Find the value of Q.
(2)
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Answer
Mark
Notes
Part (a)
the position of A relative to P is 3i, so moment = 3 × 3 − 0 × 4 (or: only the 3 N component has a moment, 3 N × 3 m)
M1
= 9 N m anticlockwise
A1
Part (b)
magnitude of the force = √(42 + 32) = 5 N, and moment = force × perpendicular distance
M1
d = 9 ÷ 5 = 1.8 m
A1
Part (c)
B relative to P is 2i + 3j: moment = 2 × 5 − 3 × (−2) = 10 + 6 = 16, anticlockwise
B1
Part (d)
C relative to P is −i + 2j, so the force at C has moment 2Q clockwise; 9 + 16 − 2Q = 0
M1
Q = 12.5
A1
Exam tip: For a force (Xi + Yj) acting at a point (x, y) relative to the pivot, the anticlockwise moment is xY − yX. Each component times its own perpendicular distance, with signs, gives the same result.
A trapdoor is modelled as a uniform rod AB of length 1.2 m and mass 15 kg, freely hinged at A to a horizontal floor. The door is held open at an angle of 50° to the floor by a force of magnitude P newtons applied at B perpendicular to AB, as shown in Figure 1. The force and the door lie in a vertical plane. g = 9.8 m s−2.
(a)
Find the value of P.
(2)
(b)
Find the magnitude and direction of the force exerted on the door by the hinge.
(3)
(c)
The force at B is replaced by a horizontal force that holds the door in the same position. Find its magnitude, and explain why it is greater than P.
(2)
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Answer
Mark
Notes
Part (a)
moments about A: P × 1.2 = 15g × 0.6 cos 50°
M1
P = 47.2
A1
Part (b)
resolving: X = P sin 50° = 36.2 N, horizontally towards the side on which B lies, and Y + P cos 50° = 15g, so Y = 116.6 N upwards
M1
magnitude √(36.22 + 116.62) = 122 N
A1
at tan−1(116.6/36.2) = 72.8° above the horizontal
A1
Part (c)
moments about A: H × 1.2 sin 50° = 15g × 0.6 cos 50°
M1
H = 61.7 N; its perpendicular distance from A (1.2 sin 50° = 0.919 m) is less than 1.2 m, so a larger force is needed for the same moment
A1
Exam tip: A force perpendicular to the rod has the whole rod length as its perpendicular distance, so it is the smallest force that can hold the rod. Any other direction needs a bigger force.
A uniform ladder AB has length 5 m and mass 20 kg. Its foot A rests on rough horizontal ground and its top B rests against a smooth vertical wall. A bag of mass 12 kg hangs from the top of the ladder, so that its weight acts at B. The ladder lies in a vertical plane perpendicular to the wall. The lengths are shown in Figure 1. Model the bag as a particle. g = 9.8 m s−2.
(a)
Find the magnitude of the force exerted on the ladder by the wall.
(3)
(b)
Find the magnitude of the total force exerted on the ladder by the ground, and the angle it makes with the ground.
(2)
(c)
The coefficient of friction between the ladder and the ground is 0.18. Show that the ladder cannot rest in this position.
(2)
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Answer
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Notes
Part (a)
cos θ = 1.4/5 = 0.28 and sin θ = 4.8/5 = 0.96, or use the horizontal and vertical distances directly
B1
moments about A: S × 4.8 = 20g × 0.7 + 12g × 1.4
M1
S = 30.8g/4.8 = 62.9 N
A1
Part (b)
R = 32g = 313.6 N and F = S = 62.9 N; total = √(313.62 + 62.92) and angle tan−1(313.6/62.9)
M1
320 N at 78.7° to the ground, sloping towards the wall
A1
Part (c)
maximum friction = 0.18 × 313.6 = 56.4 N
M1
56.4 N < 62.9 N, the friction needed, so the ladder slips
A1
Common error: Putting the bag's weight at the middle of the ladder. A load hung from the top acts at B, so its horizontal distance from A is the full 1.4 m.
A shelf AB is modelled as a uniform rod of length 0.9 m and mass 4 kg, freely hinged at A to a vertical wall. The shelf is held horizontal, perpendicular to the wall, by a light rigid strut CD, freely jointed to the shelf at C and to the wall at D, as shown in Figure 1. A box of mass 6 kg, modelled as a particle, rests on the shelf at B. The shelf and the strut lie in a vertical plane. g = 9.8 m s−2.
(a)
Find the thrust in the strut.
(3)
(b)
Find the magnitude of the force exerted on the shelf by the hinge at A.
(3)
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Answer
Mark
Notes
Part (a)
CD = 0.75 m, so the strut makes an angle with sine 0.45/0.75 = 0.6 with the shelf
B1
moments about A: T × 0.6 × 0.6 = 4g × 0.45 + 6g × 0.9
M1
T = 20g = 196 N
A1
Part (b)
horizontally: X = 0.8T = 156.8 N, towards the wall
M1
vertically: Y = 10g − 0.6T = −2g, so 19.6 N downwards
A1
magnitude √(156.82 + 19.62) = 158 N
A1
Exam tip: A light strut jointed at both ends acts like a string that can push: its force lies along the strut. Here it pushes up and away from the wall, so the hinge must pull the shelf towards the wall.
A non-uniform pole AB has length 3.0 m and mass 12 kg. It rests horizontally on two trestles, one at A and one at B. The trestle at A exerts a vertical force of 70 N on the pole. Model the pole as a non-uniform rod. g = 9.8 m s−2.
Figure: the pole on trestles at A and B; the position of G is not to scale.
(a)
Find the force exerted on the pole by the trestle at B.
(2)
(b)
Find the distance of the centre of mass of the pole from A.
(2)
(c)
The trestles are moved to C and D, where AC = 0.5 m and DB = 0.5 m, and the pole again rests horizontally on them. A bag of mass m kg is hung from B. Find the greatest value of m for which the pole remains in equilibrium.
(4)
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Answer
Mark
Notes
Part (a)
resolving vertically: 70 + RB = 12g
M1
RB = 47.6 N
A1
Part (b)
moments about A: 47.6 × 3.0 = 12g × x
M1
x = 1.21 m (17/14 m)
A1
Part (c)
on the point of tilting about D, the force at C is zero
B1
moments about D: 12g × (2.5 − 17/14) = mg × 0.5
M1
correct equation, e.g. 12 × 1.286 = 0.5m
A1
m = 30.86, so 30.8 (rounded down: 30.9 would tilt the pole)
A1
Common error: Taking the weight of a non-uniform rod at its midpoint. Find the centre of mass first, from the given forces, and use that position in every later moment.
A shop sign hangs from a bracket. The bracket is a uniform rod AB of length 1.5 m and mass 6 kg, freely hinged at A to a vertical wall. The rod is held horizontal, perpendicular to the wall, by a light inextensible string attached to the rod at C, where AC = 1.0 m, and to a point on the wall vertically above A. The string makes an angle of 40° with the rod. The sign has weight 40 N and hangs from B. The rod and the string lie in a vertical plane. g = 9.8 m s−2.
Figure: the bracket AB hinged at A, held by the string from C, with the sign at B; not to scale.
(a)
Find the tension in the string.
(3)
(b)
Find the horizontal and vertical components of the force exerted on the rod at A.
(4)
(c)
Hence find the magnitude and direction of the force exerted on the rod at A.
(2)
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Answer
Mark
Notes
Part (a)
moments about A: T sin 40° × 1.0 = 6g × 0.75 + 40 × 1.5
M1
T sin 40° = 104.1
A1
T = 162 N
A1
Part (b)
horizontally: X = T cos 40°
M1
X = 124 N, away from the wall
A1
vertically: Y + T sin 40° = 6g + 40
M1
Y = 98.8 − 104.1 = −5.3, so 5.3 N downwards
A1
Part (c)
magnitude √(124.12 + 5.32) = 124 N
M1
at tan−1(5.3/124.1) = 2.4° below the horizontal
A1
Exam tip: If a hinge component comes out negative, keep the working and say which way it really acts. A string attached part-way along the rod can lift more than the total weight.
A car-park barrier arm AB is modelled as a uniform rod of length 3.0 m and mass 40 kg. The end A is freely hinged to a fixed point on horizontal ground. The arm points up and to the right of A, making an angle of 35° with the horizontal. It is held in equilibrium by a light rope attached at B. The rope is horizontal, lies in the vertical plane containing the arm, and pulls B to the left. g = 9.8 m s−2.
Figure: the barrier arm AB hinged at A and held by the horizontal rope at B; not to scale.
(a)
Find the tension in the rope.
(3)
(b)
Find the magnitude of the force exerted on the arm at A, and the angle it makes with the horizontal.
(3)
(c)
Only three forces act on the arm. Show that the line of action of the force at A passes through the point where the lines of action of the weight and the rope tension meet.
(2)
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Answer
Mark
Notes
Part (a)
moments about A: T × 3 sin 35° = 40g × 1.5 cos 35°
M1
correct equation, e.g. T = 20g ÷ tan 35°
A1
T = 280 N
A1
Part (b)
horizontal component = T = 279.9 N to the right; vertical component = 40g = 392 N upwards
M1
magnitude √(279.92 + 3922) = 482 N
A1
angle tan−1(392/279.9) = 54.5° above the horizontal
A1
Part (c)
the weight acts on the vertical line 1.5 cos 35° = 1.229 m from A; the rope acts on the horizontal line 3 sin 35° = 1.721 m above A, so they meet at (1.229, 1.721)
M1
tan−1(1.721/1.229) = 54.5°, the same direction as the force at A, so the three forces are concurrent
A1
Exam tip: When exactly three non-parallel forces keep a body in equilibrium, their lines of action meet at one point. It is a quick check on a hinge direction.
A uniform ladder AB has length 5 m and mass 20 kg. Its foot A is on rough horizontal ground and its top B rests against a smooth vertical wall. The ladder is in a vertical plane perpendicular to the wall and makes an angle θ with the ground, where tan θ = 3. A painter of mass 70 kg climbs the ladder. When she is 4 m from A, the ladder is on the point of slipping. Model the painter as a particle. g = 9.8 m s−2.
Figure: the ladder AB (mass 20 kg) against the smooth wall, tan θ = 3, with the painter 4 m from A; not to scale.
(a)
Find the coefficient of friction between the ladder and the ground.
(6)
(b)
So that the painter can reach the top of the ladder safely, her assistant stands on the bottom of the ladder at A. Find the least mass of the assistant, modelled as a particle at A.
(3)
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Answer
Mark
Notes
Part (a)
resolving vertically: R = 90g
B1
moments about A: S × 5 sin θ = 20g × 2.5 cos θ + 70g × 4 cos θ
M1
5S tan θ = 330g
A1
S = 22g
A1
limiting: F = S = μR, so 22g = 90μg
M1
μ = 11/45 = 0.244
A1
Part (b)
painter at B: 5S tan θ = (50 + 70 × 5)g, so S = 80g/3
M1
need S ≤ μR with R = (90 + m)g: 80/3 ≤ (11/45)(90 + m)
M1
m ≥ 19.1, so least mass 19.1 kg
A1
Exam tip: A load at the foot of the ladder has no moment about A, but it increases R and so increases the friction available.
A uniform ladder of length 6 m and mass 25 kg stands on a smooth horizontal floor with its top against a smooth vertical wall. It makes an angle of 65° with the floor and lies in a vertical plane perpendicular to the wall. The foot of the ladder is tied to the bottom of the wall by a light inextensible rope, which is horizontal. A librarian of mass 70 kg stands on the ladder 4.5 m from its foot. Model the librarian as a particle. g = 9.8 m s−2.
Figure: the ladder AB (mass 25 kg) on the smooth floor, its foot tied to the wall by the horizontal rope; not to scale.
(a)
Find the normal reaction between the ladder and the floor.
(1)
(b)
Find the tension in the rope.
(4)
(c)
The rope breaks if its tension exceeds 350 N. Find how far up the ladder the librarian can stand before the rope breaks.
(3)
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Answer
Mark
Notes
Part (a)
R = 95g = 931 N
B1
Part (b)
moments about the foot: S × 6 sin 65° = 25g × 3 cos 65° + 70g × 4.5 cos 65°
M1
6S tan 65° = 390g
A1
resolving horizontally: T = S
B1
T = 297 N
A1
Part (c)
350 × 6 sin 65° = (75 + 70d)g cos 65°
M1
75 + 70d = 459.5
A1
d = 5.49 m from the foot
A1
Exam tip: With smooth floor and smooth wall, the only horizontal force at the foot is the rope. It plays exactly the part that friction plays on rough ground.
A uniform pole AB of length 3 m and mass 8 kg rests with its end A on rough horizontal ground and with the pole resting on a smooth horizontal rail at C. The rail is perpendicular to the vertical plane containing the pole, and is 1.5 m above the ground. AC = 2.5 m. The pole is in limiting equilibrium. g = 9.8 m s−2.
Figure: the pole AB (3 m, 8 kg) with A on rough ground, resting on the smooth rail at C; not to scale.
(a)
Find the magnitude of the force exerted on the pole by the rail.
(3)
(b)
Find the normal reaction and the friction force at A.
(3)
(c)
Find the coefficient of friction between the pole and the ground.
(2)
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Answer
Mark
Notes
Part (a)
sin θ = 1.5/2.5 = 0.6, cos θ = 0.8, where θ is the angle between the pole and the ground
B1
moments about A: N × 2.5 = 8g × 1.5 × 0.8
M1
N = 3.84g = 37.6 N
A1
Part (b)
resolving vertically: R + N cos θ = 8g
M1
R = 4.928g = 48.3 N
A1
resolving horizontally: F = N sin θ = 2.304g = 22.6 N
A1
Part (c)
F = μR, so μ = 2.304 ÷ 4.928
M1
μ = 0.468
A1
Common error: Drawing the rail's force vertically. A smooth rail can only push perpendicular to the pole, so its force has a horizontal component that friction must balance.
A non-uniform beam AB of length 3 m and mass 24 kg hangs in equilibrium with AB horizontal, supported by two light inextensible strings attached to the beam at A and B. The strings make the angles with the vertical shown in Figure 1, and lie in the same vertical plane as the beam. g = 9.8 m s−2.
(a)
Find the tension in each string.
(4)
(b)
Find the distance of the centre of mass of the beam from A.
(3)
(c)
The lines of the two strings, extended below the beam, meet at the point X. Explain why X must lie vertically below the centre of mass of the beam.
(1)
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Answer
Mark
Notes
Part (a)
resolving horizontally: TA sin 40° = TB sin 25°
M1
resolving vertically: TA cos 40° + TB cos 25° = 24g
M1
TA = 110 N (109.7)
A1
TB = 167 N (166.8)
A1
Part (b)
moments about A: TB cos 25° × 3 = 24g × x
M1
166.8 × 0.9063 × 3 = 235.2x
A1
x = 1.93 m
A1
Part (c)
only three forces act on the beam and they are not parallel, so their lines of action meet at one point; the weight acts vertically through the centre of mass, so it must pass through X
B1
Exam tip: When the tensions are unknown and the centre of mass is unknown, resolve first: two resolving equations give both tensions, and a single moments equation then locates the centre of mass.
A jib AB is modelled as a uniform rod of length 4 m and mass 30 kg. The end A is freely hinged to a fixed point on horizontal ground. The jib is held at 60° to the ground by a light inextensible cable joining B to a point C on the ground, where CA = 3 m and C is on the opposite side of A from B, as shown in Figure 1. A load of mass 50 kg hangs from B. The jib, the cable and the load lie in one vertical plane. Model the load as a particle. g = 9.8 m s−2.
(a)
Show that the cable makes an angle of 34.7° with the ground, to 3 significant figures.
(2)
(b)
Find the tension in the cable.
(3)
(c)
Find the magnitude of the force exerted on the jib by the hinge, and the angle it makes with the horizontal.
(4)
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Answer
Mark
Notes
Part (a)
B is 4 cos 60° = 2 m horizontally from A and 4 sin 60° = 3.464 m above the ground, so 5 m horizontally from C
M1
tan−1(3.464/5) = 34.7°
A1
Part (b)
moments about A, using the perpendicular distance from A to BC, 3 sin 34.7° = 1.71 m, or components of T at B
M1
T × 1.708 = 30g × 1 + 50g × 2
A1
T = 746 N
A1
Part (c)
horizontally: X = T cos 34.7° = 613 N, directed away from C
M1
vertically: Y = 80g + T sin 34.7° = 1209 N upwards
A1
magnitude √(6132 + 12092) = 1360 N (1355)
A1
angle tan−1(1209/613) = 63.1° above the horizontal
A1
Common error: Forgetting that the cable pulls B downwards as well as sideways. Its vertical component adds to the weights, so the hinge has to push up with more than the total weight.
A garden gate ABCD is modelled as a uniform rectangular lamina of weight 200 N with the dimensions shown in Figure 1. It hangs in a vertical plane on a vertical post, from two hinges at A and D, with AB horizontal. Model the hinge at D as exerting a horizontal force only, so that the hinge at A supports all the vertical load. g = 9.8 m s−2.
(a)
Find the magnitude of the force exerted on the gate by the hinge at D, and state its direction.
(2)
(b)
Find the magnitude and direction of the force exerted on the gate by the hinge at A.
(3)
(c)
A child of mass 30 kg stands on the bottom of the gate at B. Model the child as a particle. Find the new magnitudes of the forces exerted by the hinges at D and at A.
(3)
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Answer
Mark
Notes
Part (a)
moments about A: H × 1.0 = 200 × 0.75
M1
H = 150 N, horizontally towards the post
A1
Part (b)
components: 150 N horizontally away from the post and 200 N vertically upwards
M1
magnitude √(1502 + 2002) = 250 N
A1
at tan−1(200/150) = 53.1° above the horizontal, away from the post
A1
Part (c)
moments about A: H × 1.0 = 200 × 0.75 + 30g × 1.5
M1
H = 591 N at D
A1
at A: components 591 N and 200 + 294 = 494 N, so √(5912 + 4942) = 770 N
A1
Exam tip: Take moments about the hinge that carries the unknown vertical force. Only the horizontal force at the other hinge then appears, with the full distance between the hinges as its arm.
A uniform rod AB of length 2.4 m and mass 6 kg is held at an angle of 30° to the horizontal by two fixed rough pegs, P and Q, which are perpendicular to the vertical plane containing the rod. The end A presses against the underside of P, and the rod rests on top of Q, where AQ = 0.8 m, as shown in Figure 1. The coefficient of friction between the rod and each peg is μ. The rod is about to slide down. g = 9.8 m s−2.
(a)
Find the normal reaction between the rod and the peg P.
(3)
(b)
Find the normal reaction between the rod and the peg Q.
(2)
(c)
Find the value of μ.
(3)
(d)
Explain why the rod could not be held in this way if the peg Q were further from A than the midpoint of the rod.
(1)
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Answer
Mark
Notes
Part (a)
the normal reactions are perpendicular to the rod, and friction acts along it, so take moments about Q
M1
NP × 0.8 = 6g cos 30° × 0.4
A1
NP = 25.5 N
A1
Part (b)
resolving perpendicular to the rod: NQ = NP + 6g cos 30°
M1
NQ = 76.4 N
A1
Part (c)
resolving along the rod: FP + FQ = 6g sin 30° = 29.4 N
M1
limiting at both pegs: μ(25.5 + 76.4) = 29.4
M1
μ = 0.289 (√3/6)
A1
Part (d)
the weight would then have a moment about Q turning A downwards, away from P, so NP would have to be negative: a peg cannot pull the rod
B1
Exam tip: With a rod on pegs, resolve along and perpendicular to the rod, not horizontally and vertically. Every normal reaction is then in one equation and every friction force in the other.
A uniform ladder AB has mass 20 kg. Its foot A rests on rough horizontal ground and its top B rests against a smooth vertical wall, in the position shown in Figure 1. The ladder lies in a vertical plane perpendicular to the wall. The coefficient of friction between the ladder and the ground is 0.2. A horizontal force of magnitude P newtons, acting towards the wall in the same vertical plane, is applied to the ladder at A. g = 9.8 m s−2.
(a)
Find the magnitude of the force exerted on the ladder by the wall.
(3)
(b)
Show that the ladder would slip if P = 0.
(2)
(c)
Find the least value of P for which the ladder stays in equilibrium.
(2)
(d)
Find the greatest value of P for which the ladder stays in equilibrium.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
B is 3.2 m above the ground (Pythagoras)
B1
moments about A: S × 3.2 = 20g × 1.2
M1
S = 7.5g = 73.5 N
A1
Part (b)
R = 20g = 196 N, so the maximum friction is 0.2 × 196 = 39.2 N
M1
the friction needed is 73.5 N > 39.2 N, so it slips
A1
Part (c)
the foot is about to slide away from the wall, so friction acts towards the wall: P + 39.2 = 73.5
M1
P = 34.3
A1
Part (d)
the foot is about to slide towards the wall, so friction acts away from it: P = 73.5 + 39.2
M1
P = 112.7 (exact; do not round up to 113)
A1
Common error: Keeping the friction in the same direction for both limits. At the least P the foot tends to slide away from the wall; at the greatest P it tends to slide towards it, so the friction reverses.
A uniform log AB of length 3 m and mass 40 kg has its end A on rough horizontal ground. It is held at 20° to the ground by a light rope attached at B, which pulls at 75° to the horizontal, as shown in Figure 1. The log and the rope lie in a vertical plane. Model the log as a rod. The log is on the point of slipping. g = 9.8 m s−2.
(a)
Find the tension in the rope.
(3)
(b)
Find the normal reaction and the friction force on the log at A.
(3)
(c)
Find the coefficient of friction between the log and the ground.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
the angle between the rope and the log is 75° − 20° = 55°, so the moment of T about A is T × 3 sin 55°
M1
3T sin 55° = 40g × 1.5 cos 20°
A1
T = 225 N (224.8)
A1
Part (b)
resolving vertically: R + T sin 75° = 40g
M1
R = 175 N (174.8)
A1
resolving horizontally: F = T cos 75° = 58.2 N, horizontally in the direction from B towards A
A1
Part (c)
limiting: F = μR, so μ = 58.2 ÷ 174.8
M1
μ = 0.333
A1
Exam tip: The moment of a force at the end of a rod is force × length × sin(angle between the force and the rod). Find that angle from the figure first: here 75° − 20°.
A hanging basket is supported by a bracket. The bracket is a uniform rod AB of length 2 m and mass 5 kg, freely hinged at A to a vertical wall. The rod points away from the wall and slopes upwards at 20° above the horizontal. A light inextensible string joins B to a point C on the wall vertically above A, so that angle ABC = 70°. The basket, of mass 8 kg, hangs from B. The rod, the string and the basket lie in a vertical plane perpendicular to the wall. Model the basket as a particle. g = 9.8 m s−2.
Figure: the bracket AB hinged at A, the string BC and the basket hanging from B; not to scale.
(a)
Show that the tension in the string is 10.5g N, and find its value.
(3)
(b)
Show that the string makes an angle of 50° with the horizontal.
(1)
(c)
Find the magnitude and direction of the force exerted on the rod at A.
(6)
(d)
The string will break if its tension exceeds 150 N. Find the greatest mass of basket that the bracket can support.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
moments about A: T × 2 sin 70° = 5g × 1 cos 20° + 8g × 2 cos 20°
M1
sin 70° = cos 20°, so 2T = 21g, T = 10.5g
A1
T = 103 N (102.9 N)
A1
Part (b)
BA is 20° below the horizontal (towards the wall); BC is 70° from BA, so it is 70° − 20° = 50° above the horizontal
B1
Part (c)
horizontally: X = T cos 50°
M1
X = 66.1 N, away from the wall
A1
vertically: Y + T sin 50° = 13g
M1
Y = 48.6 N upwards
A1
magnitude √(66.12 + 48.62) = 82.1 N
A1
direction tan−1(48.6/66.1) = 36.3° above the horizontal
A1
Part (d)
T = (5 + 2m)g/2 ≤ 150
M1
m ≤ 12.8, so 12.8 kg
A1
Exam tip: The moment of the tension about A is T × AB × sin(angle between the string and the rod). Look for angles that make sines and cosines cancel.
A uniform ladder AB has length 5 m and mass 20 kg. The foot A is on rough horizontal ground and the top B rests against a rough vertical wall. The coefficient of friction between the ladder and the wall is 0.25. The ladder lies in a vertical plane perpendicular to the wall and makes an angle α with the ground, where tan α = 4/3. A window cleaner of mass 80 kg stands on the ladder at a point 3.5 m from A, and the ladder is about to slip. Model the window cleaner as a particle. g = 9.8 m s−2.
Figure: the ladder AB (mass 20 kg) against the rough wall, tan α = 4/3, with the window cleaner 3.5 m from A; not to scale.
(a)
Explain why the friction force exerted by the wall on the ladder acts upwards.
(1)
(b)
Taking S as the normal reaction at the wall, R as the normal reaction at the ground and μ as the coefficient of friction between the ladder and the ground, write down equations from resolving horizontally and vertically, using the fact that friction is limiting at both ends.
(4)
(c)
By taking moments about A, show that S = 41.7g, to 3 significant figures.
(3)
(d)
Find R.
(1)
(e)
Find the coefficient of friction between the ladder and the ground.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
if the ladder slips, B moves down the wall, and friction opposes this motion, so it acts upwards on the ladder
B1
Part (b)
horizontally: F = S, where F is the friction at the ground
M1
limiting at the ground: F = μR, so S = μR
B1
vertically: R + (wall friction) = 100g
M1
limiting at the wall: wall friction = 0.25S, so R + 0.25S = 100g
A1
Part (c)
S × 5 sin α + 0.25S × 5 cos α = 20g × 2.5 cos α + 80g × 3.5 cos α
M1
with sin α = 0.8, cos α = 0.6: 4S + 0.75S = 30g + 168g
A1
S = 198g/4.75 = 41.7g
A1
Part (d)
R = 100g − 0.25 × 41.68g = 89.6g = 878 N
A1
Part (e)
μ = S/R = 41.68g ÷ 89.58g
M1
μ = 0.465
A1
Common error: Writing R = total weight when the wall is rough. The upward friction from the wall carries part of the weight, so R is smaller.
A uniform rod AB has length 2 m and mass 4 kg. The end A rests against a rough vertical wall, and the rod is held horizontal and perpendicular to the wall by a light inextensible string joining B to the point D on the wall, 1.5 m vertically above A. The coefficient of friction between the rod and the wall is 0.6. A particle of mass 3 kg is attached to the rod at P, where AP = x m. The rod and the string lie in a vertical plane. g = 9.8 m s−2.
Figure: the rod AB (2 m, 4 kg) held horizontal by the string BD, with the particle at P; not to scale.
(a)
Show that the tension in the string is 5(4 + 3x)g/6 N.
(3)
(b)
Find, in terms of x and g, the normal reaction at A.
(2)
(c)
Find, in terms of x and g, the friction force at A, and show that it acts upwards for every position of P on the rod.
(2)
(d)
Find the least value of x for which the rod can rest in equilibrium.
(3)
(e)
State what happens if the particle is placed closer to A than this, and why.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
sin(angle DBA) = 1.5/2.5 = 0.6 and cos = 0.8
B1
moments about A: T × 0.6 × 2 = 4g × 1 + 3g × x
M1
T = (4 + 3x)g/1.2 = 5(4 + 3x)g/6
A1
Part (b)
resolving horizontally: R = T × 0.8
M1
R = 2(4 + 3x)g/3
A1
Part (c)
resolving vertically: F + T × 0.6 = 7g, so F = (10 − 3x)g/2
M1
0 ≤ x ≤ 2, so 10 − 3x > 0 and F is positive, i.e. upwards
A1
Part (d)
F ≤ 0.6R: (10 − 3x)/2 ≤ 0.4(4 + 3x)
M1
10 − 3x ≤ 3.2 + 2.4x, so 5.4x ≥ 6.8
A1
x ≥ 34/27, least x = 1.26
A1
Part (e)
the rod slips down the wall: the friction needed is greater than the maximum available 0.6R
B1
Exam tip: When a position is unknown, keep it as a letter all the way through and use F ≤ μR to turn the equilibrium equations into an inequality.
A uniform plank AB of length 4 m and weight 60 N is placed with its end A on rough horizontal ground and the plank resting on a smooth peg C fixed to a garden fence, where AC = 2.5 m. The plank lies in a vertical plane perpendicular to the peg and makes an angle of 25° with the ground. The coefficient of friction between the plank and the ground is 0.3. Model the plank as a rod.
Figure: the uniform plank AB (4 m) with A on the rough ground, resting on the smooth peg at C; not to scale.
(a)
Find the magnitude of the force exerted on the plank by the peg.
(3)
(b)
Show that the plank cannot rest in this position on its own.
(4)
(c)
A heavy stone of weight W newtons is placed on the plank at A, so that its weight acts at A. Find the least value of W for which the plank rests in this position.
(3)
(d)
Explain why placing the stone at any other point of AC would require a heavier stone.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the force is perpendicular to the plank; moments about A: N × 2.5 = 60 × 2 cos 25°
M1
N = 48 cos 25°
A1
N = 43.5 N
A1
Part (b)
resolving vertically: R = 60 − N cos 25°
M1
R = 20.6 N
A1
resolving horizontally: friction needed F = N sin 25° = 18.4 N
B1
maximum friction 0.3 × 20.6 = 6.17 N < 18.4 N, so the plank slips
A1
Part (c)
W acts at A, so N (and so F = 18.4 N) is unchanged
B1
F ≤ 0.3R: 18.39 ≤ 0.3(20.57 + W)
M1
W ≥ 40.71, least W = 40.8 (rounded up: 40.7 is not enough)
A1
Part (d)
a stone away from A has a moment about A, which increases N and so increases the friction needed, N sin 25°, while adding less to R
B1
Common error: Forgetting that the peg's force has a horizontal component. That component is what friction at the foot has to balance.
A uniform ladder AB has mass 20 kg. Its foot A rests on rough horizontal ground and its top B rests against a rough vertical wall, in the position shown in Figure 1. The ladder lies in a vertical plane perpendicular to the wall. The coefficient of friction is 0.4 between the ladder and the ground and 0.3 between the ladder and the wall. A person of mass 70 kg climbs the ladder. When she reaches the point C, a distance d metres from A, the ladder is about to slip. Model the person as a particle. g = 9.8 m s−2.
(a)
State the directions of the friction forces on the ladder at A and at B, giving a reason.
(1)
(b)
Using limiting friction at both ends, find the normal reactions at the ground and at the wall.
(4)
(c)
Find the value of d.
(3)
(d)
The wall is unchanged. Find the least coefficient of friction at the ground that would allow the person to reach the top of the ladder.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
the foot tends to slide away from the wall and the top to slide down, so friction at A acts towards the wall and friction at B acts vertically upwards
B1
Part (b)
horizontally: S = 0.4R
M1
vertically: R + 0.3S = 90g
M1
1.12R = 90g, R = 80.4g = 787.5 N (accept 788 N)
A1
S = 315 N
A1
Part (c)
moments about A: S × 4.8 + 0.3S × 3.6 = 20g × 1.8 + 70g × 0.6d
M1
1512 + 340.2 = 352.8 + 411.6d (or in terms of g: 189 = 36 + 42d)
A1
d = 3.64
A1
Part (d)
with the person at B and both ends limiting: S(4.8 + 0.3 × 3.6) = (36 + 252)g
M1
S = 49.0g
A1
R = 90g − 0.3S = 75.3g
A1
μ = S/R = 0.651 (80/123 = 0.6504…, rounded up so that it is large enough)
A1
Common error: Taking R as the total weight. With a rough wall, the upward friction at B carries part of the load, so R = 90g − 0.3S, and that friction also has a moment about A.
A uniform rod AB of length 2.5 m and mass 4 kg is freely hinged at A to a vertical wall. The rod is held in equilibrium, sloping downwards away from the wall, by a light inextensible string joining B to the point D on the wall, vertically above A. The distances are shown in Figure 1. A particle of mass 6 kg is attached to the rod at P, where AP = x metres. The rod and the string lie in a vertical plane perpendicular to the wall. g = 9.8 m s−2.
(a)
Show that the tension in the string is (3.2 + 3.84x)g N.
(3)
(b)
Find, in terms of x and g, the horizontal and vertical components of the force exerted on the rod by the hinge.
(3)
(c)
Find the value of x for which the force exerted by the hinge is horizontal.
(2)
(d)
The string breaks if its tension exceeds 120 N. Find the greatest value of x, and state whether the hinge force can be horizontal without the string breaking.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
BD = √(2.42 + 3.22) = 4 m, so the string pulls along (−0.6, 0.8)
horizontally: X = 0.6T = (1.92 + 2.304x)g, away from the wall
M1
vertically: Y = 10g − 0.8T
M1
Y = (7.44 − 3.072x)g, upwards
A1
Part (c)
Y = 0: 7.44 = 3.072x
M1
x = 2.42 (155/64)
A1
Part (d)
(3.2 + 3.84x) × 9.8 ≤ 120
M1
x ≤ 2.355…, so the greatest value is 2.35 (rounded down, since 2.36 would break the string)
A1
2.355 < 2.42, so the string breaks before the hinge force becomes horizontal: it cannot
A1
Exam tip: For a sloping rod, write the string's direction as a unit vector from the figure's lengths. Its moment about the hinge is then xFy − yFx with the coordinates of B, with no angles needed.
A uniform rod AB of length 2 m and mass 5 kg has its end A against a rough vertical wall. The rod rests on a smooth horizontal peg C, where AC = 0.8 m, and rises away from the wall at an angle θ to the horizontal, where tan θ = 3/4, as shown in Figure 1. The peg is parallel to the wall and the rod lies in a vertical plane perpendicular to the wall. At first no particle is attached; the particle shown hanging from B is used from part (d) onwards. g = 9.8 m s−2.
(a)
Find the magnitude of the force exerted on the rod by the peg.
(2)
(b)
Find the normal reaction and the friction force on the rod at A.
(2)
(c)
Show that the rod can rest in this position only if the coefficient of friction at the wall is at least 1/3.
(1)
(d)
The coefficient of friction at the wall is in fact 0.25. A particle of mass m kg is hung from B. Find the least value of m for which the rod rests in the same position.
(3)
(e)
Find the greatest value of m for which the rod rests in the same position.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
moments about A, with the peg force perpendicular to the rod: N × 0.8 = 5g × 1 × cos θ = 5g × 0.8
M1
N = 5g = 49 N
A1
Part (b)
horizontally: R = N sin θ = 29.4 N
B1
vertically: F = 5g − N cos θ = 9.8 N upwards
B1
Part (c)
F ≤ μR: 9.8 ≤ 29.4μ, so μ ≥ 1/3
B1
Part (d)
moments about A: 0.8N = (4 + 1.6m)g, so N = (5 + 2m)g; R = 0.6N and F = (1 − 0.6m)g upwards
M1
about to slip down: 1 − 0.6m = 0.25 × 0.6(5 + 2m)
M1
m = 5/18 = 0.278
A1
Part (e)
for larger m the friction needed acts downwards: 0.6m − 1 ≤ 0.15(5 + 2m)
M1
0.3m ≤ 1.75
A1
m ≤ 35/6, so the greatest value is 5.83
A1
Common error: Assuming the friction at the wall always acts upwards. A heavy load beyond the peg tips the rod so that A tends to slide up the wall, and the friction then acts downwards.
A crate is modelled as a uniform rectangular lamina ABCD of mass 50 kg, with the dimensions shown in Figure 1. It stands on rough horizontal ground with AB on the ground. A rope attached at C pulls with a force of magnitude P newtons at 30° above the horizontal, in the plane ABCD, as shown. The coefficient of friction between the crate and the ground is 0.4. P is increased gradually from zero. g = 9.8 m s−2.
(a)
Assuming the crate does not tip, find the value of P at which it is about to slide.
(4)
(b)
Assuming the crate does not slide, find the value of P at which it is about to tip about B.
(3)
(c)
State what happens to the crate as P is increased, giving a reason.
(1)
(d)
Find the least coefficient of friction for which the crate would tip about B before it slides.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
resolving vertically: R = 50g − P sin 30°
M1
resolving horizontally, limiting: P cos 30° = 0.4R
M1
P(cos 30° + 0.4 sin 30°) = 196
A1
P = 184
A1
Part (b)
about to tip: the whole reaction from the ground acts at B
B1
moments about B: P cos 30° × 1.6 = 50g × 0.6
M1
P = 212
A1
Part (c)
184 < 212, so the crate slides before it tips
B1
Part (d)
tipping value is still 0.6 × 50g/(1.6 cos 30°), which does not depend on μ
B1
need μ(50g − P sin 30°) ≥ P cos 30° at that value of P
M1
μ(1.6 cos 30° − 0.6 sin 30°) ≥ 0.6 cos 30°
A1
μ ≥ 0.479
A1
Exam tip: For slide-or-tip, work out the two values of P separately: one from limiting friction, one from the moment about the tipping corner with the whole reaction at that corner. The smaller value happens first.
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