Edexcel International A Level Maths · M1 · Mechanics 1
Moments
28 free, high-quality past-paper-style questions on Moments 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 · 239 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-M1-6-B-01M1 · ASEasy5 marks
A playground see-saw is modelled as a uniform plank AB of length 4.0 m and mass 20 kg, pivoted at its midpoint O. A child P of mass 32 kg sits at A. A second child Q, of mass 40 kg, sits on the plank between O and B, at a distance d metres from O. The children are modelled as particles and the plank rests in equilibrium in a horizontal position.
Figure: the see-saw AB pivoted at O, with the weights acting on it; not to scale.
(a)
Find the value of d.
(3)
(b)
Find the magnitude of the force exerted on the plank by the pivot.
(2)
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Answer
Mark
Notes
Part (a)
takes moments about O; the weight of the plank acts at O so has no moment
M1
32g × 2.0 = 40g × d
A1
d = 1.6
A1
Part (b)
resolving vertically: R = 20g + 32g + 40g
M1
R = 92g = 902 N (accept 900 N)
A1
Exam tip: Taking moments about the pivot removes both the pivot force and the plank's own weight from the equation, because both act at O.
A workbench top is modelled as a uniform plank AB of length 3.2 m and weight 150 N. It rests horizontally on two trestles, one at A and one at B. A toolbox of weight 90 N is placed on the plank at C, where AC = 0.8 m. The toolbox is modelled as a particle.
Figure: the plank on trestles at A and B, with the toolbox at C; not to scale.
(a)
Find the magnitude of the reaction on the plank at B.
(3)
(b)
Find the magnitude of the reaction on the plank at A.
(1)
(c)
A second toolbox, of weight 60 N, is now placed on the plank at E, so that the reactions at A and B are equal. Find the distance AE.
(2)
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Answer
Mark
Notes
Part (a)
takes moments about A, with all three relevant terms
M1
RB × 3.2 = 150 × 1.6 + 90 × 0.8
A1
RB = 97.5 N
A1
Part (b)
resolving vertically: RA = 240 − 97.5 = 142.5 N (ft from (a))
A1
Part (c)
each reaction = 300 ÷ 2 = 150 N, and moments about A: 150 × 3.2 = 150 × 1.6 + 90 × 0.8 + 60 × AE
M1
AE = 2.8 m
A1
Common error: Taking the 150 N weight at A or at B. A uniform plank has its weight at its midpoint, 1.6 m from each end.
A hanging decoration is made from a light rod PQ of length 60 cm. A bead of mass 0.30 kg is fixed at P and a bead of mass 0.50 kg is fixed at Q. The rod hangs in equilibrium in a horizontal position from a single light vertical thread attached to the rod at the point R. The beads are modelled as particles.
Figure: the light rod PQ hanging from a thread at R; the position of R is not to scale.
(a)
Find the tension in the thread.
(1)
(b)
Find the distance PR.
(3)
(c)
State how you have used the fact that the rod is light.
(1)
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Answer
Mark
Notes
Part (a)
T = 0.80g = 7.84 N (accept 7.8 N)
B1
Part (b)
takes moments about R (or about P with the tension)
M1
0.30g × PR = 0.50g × (0.60 − PR)
A1
PR = 0.375 m (37.5 cm)
A1
Part (c)
the weight of the rod is negligible, so it does not appear in the equations
B1
Exam tip: Work in metres throughout, even when the question gives centimetres — or in centimetres throughout. Mixing the two is the slip to avoid.
A uniform rod AB has length 1.8 m and mass 4 kg. It hangs in a horizontal position from two light vertical wires attached at C and D, where AC = 0.3 m and DB = 0.3 m. A lantern of mass 2 kg hangs from B. The lantern is modelled as a particle.
Figure: the rod AB hanging from wires at C and D, with the lantern's weight at B; not to scale.
(a)
Find the tension in the wire at D.
(3)
(b)
Find the tension in the wire at C.
(1)
(c)
The lantern is replaced by a heavier one, also hanging from B. Find the greatest mass this lantern can have if the wire at C is not to become slack.
Figure 1 shows a light rod AB of length 2.5 m. Four vertical forces act on the rod at A, C, D and B, with the magnitudes, directions and distances shown in the figure.
(a)
Find the magnitude of the total moment of the four forces about A, stating whether it is clockwise or anticlockwise.
(2)
(b)
Find the magnitude of the total moment of the four forces about B, stating whether it is clockwise or anticlockwise.
(1)
(c)
A single extra vertical force is now applied to the rod so that the rod is in equilibrium. Find the magnitude and direction of this force and its distance from A.
4 × 2.5 + 22 × 1.0 − 12 × 2.0 = 8, so 8 N m anticlockwise
B1
Part (c)
resultant of the four forces: 12 + 4 − 4 − 22 = −10, so the extra force is 10 N upwards
B1
its moment about A must cancel 17 N m clockwise: 10 × d = 17
M1
d = 1.7 m from A
A1
Exam tip: Choose one sense as positive, then write every moment with its sign in a single line. The sign of the total tells you the sense — no need to sort the forces into two lists first.
Figure 1 shows a wheelbarrow, modelled as a horizontal rod AB of length 1.5 m. The ground exerts a vertical force on the barrow at the wheel axle A. The barrow has mass 15 kg and its centre of mass is at G. It carries bricks of total mass 60 kg, modelled as a particle at L. A gardener holds the barrow at rest, with AB horizontal, by applying a vertical force of magnitude F newtons at B. The distances are shown in the figure.
(a)
Find the value of F.
(3)
(b)
Find the magnitude of the force exerted by the ground at A.
(1)
(c)
The gardener can apply a vertical force of at most 245 N. Find the greatest mass of bricks, placed at L, that the barrow can carry if he is still to hold it with AB horizontal.
(2)
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Answer
Mark
Notes
Part (a)
takes moments about A
M1
F × 1.5 = 15g × 0.5 + 60g × 0.3
A1
F = 17g = 167 (accept 166.6)
A1
Part (b)
resolving vertically: 75g − 17g = 58g = 568 N (ft from (a))
A1
Part (c)
245 × 1.5 = 15g × 0.5 + mg × 0.3
M1
m = 100 kg
A1
Exam tip: Take moments about the wheel axle. The ground's force acts there, so it has no moment and F is the only unknown in the equation.
Figure 1 shows a steelyard used for weighing sacks. It is modelled as a light rod AB of length 80 cm, hanging from a hook at O, where AO = 10 cm. A pan of mass 0.40 kg hangs from A. A counterweight of mass 1.2 kg can be hung from any point of OB, which is marked every 10 cm measured from O. A sack is placed in the pan, and the rod rests in equilibrium in a horizontal position with the counterweight at the mark shown. The pan, sack and counterweight are modelled as particles.
(a)
Find the mass of the sack.
(3)
(b)
Find the magnitude of the force exerted by the hook on the rod.
(1)
(c)
Find the greatest mass of sack that can be weighed with this steelyard.
(2)
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Answer
Mark
Notes
Part (a)
takes moments about O, including the pan
M1
(m + 0.40)g × 10 = 1.2g × 30
A1
m = 3.2 kg
A1
Part (b)
(3.2 + 0.40 + 1.2)g = 4.8g = 47.0 N (accept 47 N; ft from (a))
A1
Part (c)
counterweight at B, 70 cm from O: (m + 0.40)g × 10 = 1.2g × 70
M1
m = 8.0 kg
A1
Common error: Leaving out the pan. Its weight acts at A together with the sack's, so the moment of the load about O is (m + 0.40)g × 10, not mg × 10.
Figure 1 shows a car of mass 1300 kg at rest on horizontal ground. The car is modelled as a rigid body whose weight acts at G. The forces from the road on the car act vertically at the front axle F and at the rear axle R. The distances are shown in the figure.
(a)
Find the magnitudes of the forces on the car at F and at R.
(3)
(b)
Luggage of mass 80 kg, modelled as a particle, is placed in the boot at L. Find the new magnitude of the force on the car at F.
Exam tip: Luggage behind the rear axle turns the car the opposite way to its own weight about R, so it slightly lifts the front: the force at F goes down, not up.
A diving board is modelled as a uniform rod AB of length 4.0 m and mass 30 kg, held in a horizontal position. The end A is bolted to the ground, and the bolt can exert a vertical force on the board in either direction. The board also rests on a support at C, where AC = 1.2 m. A diver of mass 65 kg stands at rest at B. The diver is modelled as a particle.
Figure: the diving board, bolted at A and resting on the support at C; 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.
(3)
(c)
Explain, by taking moments about C, why the board could not stay horizontal if the bolt at A were removed.
(2)
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Answer
Mark
Notes
Part (a)
takes moments about A
M1
R × 1.2 = 30g × 2.0 + 65g × 4.0
A1
R = 266.7g = 2610 N (accept 2600 N)
A1
Part (b)
resolving vertically with the force at A unknown in sign
M1
force = 2613 − 95g = 1682 N, so 1680 N (accept 1700 N)
A1
acting vertically downwards
A1
Part (c)
about C, the diver (65g × 2.8 = 182g N m) and the board's weight, which acts 0.8 m beyond C (30g × 0.8 = 24g N m), both turn the board clockwise
M1
the support at C has no moment about C, so with no bolt nothing gives an anticlockwise moment and the board turns about C
A1
Common error: Assuming every support pushes upwards. A fixing bolt can pull down — if resolving gives a negative upward force, the force acts downwards.
Two gardeners carry a log, modelled as a non-uniform rod AB of length 2.8 m and mass 54 kg, in a horizontal position. One gardener supports the log at A and the other at B, each with a vertical force. The force at A is 1.25 times the force at B.
Figure: the log AB held at A and B; the position of G is not to scale.
(a)
Find the magnitude of the force at each end.
(2)
(b)
Find the distance from A of the centre of mass G of the log.
(3)
(c)
The gardener at A keeps holding A. The other gardener moves his hands to a point D, between G and B, so that the two forces are now equal. Find the distance BD.
(3)
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Answer
Mark
Notes
Part (a)
resolving: 2.25F = 54g
M1
force at B = 24g = 235 N, force at A = 30g = 294 N
A1
Part (b)
takes moments about A, with AG = x
M1
24g × 2.8 = 54g × x
A1
x = 1.244…, AG = 1.24 m
A1
Part (c)
each force = 27g; moments about A
M1
27g × AD = 54g × 1.244…, so AD = 2.488… m
A1
BD = 0.311 m
A1
Exam tip: Keep the unrounded value of AG (56/45 m) in your calculator for later parts; rounding it to 1.24 first would change the third significant figure of BD.
A gymnastics training beam is modelled as a uniform rod AB of length 3.6 m and mass 18 kg. It rests in a horizontal position on two blocks at C and D, where AC = 0.6 m and DB = 0.9 m. A gymnast of mass 72 kg stands on the beam. The gymnast is modelled as a particle.
Figure: the beam on blocks at C and D, with the gymnast at the midpoint of CD; not to scale.
(a)
The gymnast stands at the midpoint of CD. Find the magnitudes of the reactions at C and at D.
(3)
(b)
The gymnast walks along the beam towards B. Find her greatest distance from A for which the beam does not tilt.
(3)
(c)
State how you have used the fact that the beam is uniform.
(1)
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Answer
Mark
Notes
Part (a)
takes moments about C (or D) with the midpoint of CD 1.05 m from C
M1
RD × 2.1 = 18g × 1.2 + 72g × 1.05, RD = 454 N
A1
RC = 90g − RD = 428 N
A1
Part (b)
on the point of tilting about D, the reaction at C is zero
B1
moments about D: 18g × 0.9 = 72g × y, so y = 0.225 m beyond D
M1
distance from A = 2.7 + 0.225 = 2.925 m (accept 2.92 m, not 2.93 m)
A1
Part (c)
its weight acts at its midpoint, 1.8 m from A
B1
Exam tip: Write “about to tilt about D, so RC = 0” before any equation: it tells you exactly which point to take moments about.
In an experiment, a non-uniform rod AB of length 1.5 m and mass m kg balances in a horizontal position on a knife-edge placed at C, where AC = 0.9 m. A small weight of mass 0.4 kg is then fixed to the rod at A, and the rod now balances in a horizontal position when the knife-edge is at D, where AD = 0.7 m. The small weight is modelled as a particle.
Figure: the second balance, with the 0.4 kg weight at A and the knife-edge at D; the rod's own weight is not shown; not to scale.
(a)
Explain why the centre of mass of the rod is at C.
(1)
(b)
Find the value of m.
(3)
(c)
Find the magnitude of the force exerted by the knife-edge on the rod when it is at D.
(1)
(d)
The small weight is removed from A and fixed at B instead. Find the distance from A at which the knife-edge must now be placed for the rod to balance.
(3)
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Answer
Mark
Notes
Part (a)
with only the weight and the reaction acting, the weight must act through the knife-edge, otherwise it would have a moment about C
B1
Part (b)
takes moments about D
M1
0.4g × 0.7 = mg × 0.2
A1
m = 1.4
A1
Part (c)
1.8g = 17.6 N
B1
Part (d)
moments about the new balance point E, with AE = e
M1
1.4g(e − 0.9) = 0.4g(1.5 − e)
A1
e = 1.03 m (31/30)
A1
Exam tip: A rod balanced on a single knife-edge has its centre of mass vertically above the knife-edge — that one sentence locates G for you.
A short road bridge is modelled as a uniform beam AB of length 8 m and mass 600 kg, resting in a horizontal position on supports at A and at B. A vehicle of mass 1200 kg is on the bridge, at a distance x metres from A. The vehicle is modelled as a particle.
Figure: the bridge AB on supports at A and B, with the vehicle x m from A; not to scale.
(a)
Show that the reaction on the beam at B is (300 + 150x)g N.
(3)
(b)
Find, in terms of x and g, the reaction on the beam at A.
(2)
(c)
Find the value of x for which the reaction at A is twice the reaction at B.
(3)
(d)
State one way in which modelling the vehicle as a particle might make the answers inaccurate.
(1)
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Answer
Mark
Notes
Part (a)
takes moments about A
M1
RB × 8 = 600g × 4 + 1200g × x
A1
RB = (300 + 150x)g N, with no errors seen
A1
Part (b)
resolving vertically: RA = 1800g − RB
M1
RA = (1500 − 150x)g N
A1
Part (c)
sets up RA = 2RB using the expressions
M1
1500 − 150x = 600 + 300x
A1
x = 2
A1
Part (d)
a vehicle has its weight spread over two or more axles at different distances from A, not acting at one point
B1
Exam tip: With a “show that”, write the moments equation in full before simplifying, and finish with exactly the printed expression.
A car park barrier is modelled as a uniform rod AB of length 4.5 m and mass 12 kg, free to turn about a smooth horizontal axle at C, where AC = 0.5 m. A counterweight of mass 18 kg is fixed at A. When the barrier is closed, it rests in a horizontal position with the end B on a fixed post. The counterweight is modelled as a particle.
Figure: the barrier AB, free to turn about the axle at C, resting on the post at B; not to scale.
(a)
Find the magnitude of the force exerted on the barrier by the post.
(3)
(b)
Find the magnitude of the force exerted on the barrier by the axle.
(2)
(c)
The counterweight is replaced by a heavier one so that the barrier is on the point of lifting off the post. Find the mass of the new counterweight.
(3)
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Answer
Mark
Notes
Part (a)
takes moments about C; the barrier's weight acts 1.75 m from C
M1
R × 4.0 + 18g × 0.5 = 12g × 1.75
A1
R = 3g = 29.4 N
A1
Part (b)
resolving vertically: force = 30g − 3g
M1
27g = 265 N
A1
Part (c)
the force from the post is zero
B1
moments about C: Mg × 0.5 = 12g × 1.75
M1
M = 42 kg
A1
Common error: Measuring the barrier's weight from the end A. Its weight acts at the midpoint, 2.25 m from A, which is 1.75 m from the axle at C.
A metre rule is modelled as a uniform rod of length 1.00 m and mass m kg, with a scale marked in centimetres. Figure 1 shows the rule resting in equilibrium in a horizontal position on a knife-edge, with a particle of mass 100 g hanging from it by a light vertical string. The positions of the knife-edge and the string are read from the scale in the figure.
(a)
Find the value of m.
(3)
(b)
Find the magnitude of the force exerted by the knife-edge on the rule.
(1)
(c)
A second particle, also of mass 100 g, is hung from the 70 cm mark. The knife-edge is moved so that the rule again rests in a horizontal position. Find the mark at which the knife-edge is now placed.
(3)
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Answer
Mark
Notes
Part (a)
the weight of the rule acts at the 50 cm mark, 10 cm from the knife-edge
B1
moments about the knife-edge: 0.1g × 30 = mg × 10
M1
m = 0.30
A1
Part (b)
0.40g = 3.92 N (ft from (a))
A1
Part (c)
knife-edge at the k cm mark, with 10 < k < 50; moments about it
M1
0.1g(k − 10) = 0.30g(50 − k) + 0.1g(70 − k)
A1
k = 46, the 46 cm mark
A1
(or: centre of mass of the whole system at (0.1 × 10 + 0.3 × 50 + 0.1 × 70) ÷ 0.5)
—
Exam tip: Mark the rule's own weight at the 50 cm mark on your sketch before writing any equation — then every distance is just a difference of two scale readings.
A non-uniform plank AB has length 3.0 m. It rests in a horizontal position with the end A on one set of bathroom scales and the end B on another, as shown in Figure 1. Each set of scales displays the magnitude of the vertical force on it divided by g, in kilograms. The scales are modelled as exerting vertical forces at A and at B.
(a)
Write down the mass of the plank.
(1)
(b)
Find the distance from A of the centre of mass of the plank.
(3)
(c)
A child of mass 30 kg stands on the plank at the point P, and the two sets of scales now show the same reading. The child is modelled as a particle. Find the distance AP.
(3)
(d)
The child walks slowly from P towards B. State, with a reason, what happens to the reading on the scales at A.
(1)
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Answer
Mark
Notes
Part (a)
14.0 + 10.0 = 24 kg
B1
Part (b)
takes moments about A
M1
10g × 3.0 = 24g × d
A1
d = 1.25 m
A1
Part (c)
each reading is (24 + 30) ÷ 2 = 27 kg
B1
moments about A: 27g × 3.0 = 24g × 1.25 + 30g × AP
M1
AP = 1.7 m
A1
Part (d)
it decreases (to 14.0 kg when the child reaches B), because the child's moment about B gets smaller, so a smaller force at A balances it
B1
Exam tip: The two readings add up to the mass of everything on the scales, which gives you a free first equation; the moments equation then locates the centre of mass.
A shelf is modelled as a uniform rod AB of length 1.6 m and mass 4 kg. It rests in a horizontal position on two brackets at C and D, which exert vertical forces on it; the shelf is not fixed to the brackets. A box of books of mass 9 kg stands on the shelf with its centre of mass vertically above E, and a vase of mass 2 kg stands at B. The distances are shown in Figure 1. The box and the vase are modelled as particles.
(a)
Find the magnitudes of the forces exerted on the shelf by the brackets at C and at D.
(4)
(b)
The vase is replaced by a heavier vase, also at B. Find the greatest mass this vase can have if the shelf is not to tilt.
(3)
(c)
Explain why a heavy vase placed at A is much less likely to tilt the shelf than the same vase placed at B.
(1)
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Answer
Mark
Notes
Part (a)
takes moments about C (or D), with all four terms
M1
RD × 1.0 = 4g × 0.6 + 9g × 0.3 + 2g × 1.4
A1
RD = 7.9g = 77.4 N
A1
RC = 15g − 7.9g = 7.1g = 69.6 N
A1
Part (b)
on the point of tilting about D, so RC = 0
B1
moments about D: Mg × 0.4 = 4g × 0.4 + 9g × 0.7
M1
M = 19.75 kg (accept 19.7 or 19.8)
A1
Part (c)
A is only 0.2 m beyond C (compared with 0.4 m for B beyond D) and the shelf and the books are on the other side of C, so a vase at A has a small turning effect about C against a large opposing moment
B1
Common error: Measuring the distances of the loads from A when taking moments about C. Every arm must be measured from the point you take moments about: the books are 0.3 m from C, not 0.5 m.
Figure 1 shows a mobile hanging in equilibrium with both of its rods horizontal. The rods PQ and RS are light, and all the threads are light and vertical. PQ hangs from a thread attached at O. An ornament of mass M kg hangs from P, and a thread from Q is attached to RS at T. Ornaments of masses 0.05 kg and m kg hang from R and S. The ornaments are modelled as particles and the distances are shown in the figure.
(a)
Find the value of m.
(2)
(b)
Find the tension in the thread joining Q to T.
(1)
(c)
Find the value of M.
(2)
(d)
Find the tension in the thread at O.
(1)
(e)
The ornament at S is replaced by one of mass 0.15 kg, and the thread at T is moved along RS so that RS is again horizontal. Find the new distance RT.
(2)
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Answer
Mark
Notes
Part (a)
moments about T: 0.05g × 20 = mg × 10
M1
m = 0.1
A1
Part (b)
(0.05 + 0.1)g = 1.47 N (ft from (a))
B1
Part (c)
moments about O: Mg × 24 = 0.15g × 16
M1
M = 0.1
A1
Part (d)
(0.1 + 0.15)g = 0.25g = 2.45 N (ft)
A1
Part (e)
0.05 × RT = 0.15 × (30 − RT)
M1
RT = 22.5 cm
A1
Exam tip: Work from the bottom of a mobile upwards: the lower rod and its ornaments act on the upper rod as one single downward force — the tension in the connecting thread.
A non-uniform plank AB has length 4.0 m and weight 300 N. Its centre of mass is at G. The plank rests in a horizontal position on two supports at C and D, where AC = 0.5 m and AD = 3.0 m. A person of weight W newtons walks slowly along the plank, starting at A. The person is modelled as a particle at a distance x metres from A. The graph in Figure 1 shows how the magnitude RC newtons of the reaction at C varies with x; the line ends at x = 3.75.
(a)
By taking moments about D, show that the gradient of the line is −W ÷ 2.5, and use the graph to find the value of W.
(3)
(b)
Use the value of RC when the person is at D to find the distance AG.
(3)
(c)
Explain why the line ends at x = 3.75, and state what would happen if the person walked further towards B.
(2)
(d)
Find the magnitude of the reaction at D when the person stands at A.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
moments about D: 2.5RC = 300(3.0 − AG) + W(3.0 − x), so the gradient is −W ÷ 2.5
M1
gradient from the graph = −750 ÷ 3.75 = −200 (allow −195 to −205)
B1
W = 500 (allow 490–510)
A1
Part (b)
at x = 3.0, RC = 150 (from the graph)
B1
moments about D: 150 × 2.5 = 300 × (3.0 − AG)
M1
AG = 1.75 m
A1
Part (c)
at x = 3.75, RC = 0: the plank is on the point of tilting about D
B1
beyond this a negative reaction would be needed, which C cannot supply, so the plank would turn about D (the end A would lift off C)
B1
Part (d)
RD = 300 + 500 − 750 = 50 N (ft)
B1
Exam tip: A straight-line graph in a moments question comes from a moments equation that is linear in x. Write that equation first; then the gradient and the intercepts each mean something physical.
Figure 1 shows a uniform pole AB of length 2.0 m and mass 4 kg being carried in a horizontal position by two people. One person pushes vertically upwards on the pole at A with a force of magnitude FA newtons, and the other at B with a force of magnitude FB newtons. Buckets of mass 12 kg and 20 kg hang from the pole at C and D. The buckets are modelled as particles and the distances are shown in the figure.
(a)
Find the values of FA and FB.
(4)
(b)
The 20 kg bucket is moved along the pole, and the two people now push with forces of equal magnitude. Find the new distance of the 20 kg bucket from A.
(3)
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Answer
Mark
Notes
Part (a)
takes moments about A (or B), with all three weights
M1
FB × 2.0 = 4g × 1.0 + 12g × 0.4 + 20g × 1.5
A1
FB = 19.4g = 190
A1
FA = 36g − 19.4g = 16.6g = 163
A1
Part (b)
each force is 36g ÷ 2 = 18g
B1
moments about A: 18g × 2.0 = 4g × 1.0 + 12g × 0.4 + 20g × x
M1
x = 1.36 m
A1
Common error: Forgetting the pole itself. Its 4g weight acts at the midpoint of AB and has a moment about A of 4g × 1.0, which belongs in every equation.
A non-uniform plank AB has length 5.0 m and mass 30 kg. Its centre of mass is at G. The plank rests in a horizontal position on two supports at C and D, where AC = 1.0 m and DB = 1.0 m. When a boy of mass 36 kg stands at B, the plank is on the point of tilting about D. The boy, and later a girl, are modelled as particles.
Figure: the plank on supports at C and D with the boy at B; the position of G is not to scale.
(a)
Show that AG = 2.8 m.
(3)
(b)
The boy moves to a point E on the plank, and the reactions at C and D are now equal. Find the distance AE.
(4)
(c)
Show that the boy can stand at A without the plank tilting.
(3)
(d)
The boy stands at B and a girl of mass m kg stands at A. The reaction at C is now 400 N. Find m.
A window cleaner's platform is modelled as a uniform rod AB of length 6.0 m and mass 80 kg. It hangs in a horizontal position from two light vertical cables, one attached at A and the other at C, where CB = 1.5 m. Each cable is safe only while its tension is at most 1200 N. A window cleaner of mass 70 kg stands on the platform at a distance x metres from A. The window cleaner is modelled as a particle.
Figure: the platform hanging from cables at A and C, with the window cleaner x m from A; not to scale.
(a)
Show that the tension in the cable at C is (9.8/4.5)(240 + 70x) N.
(3)
(b)
Find, in terms of x, the tension in the cable at A.
(2)
(c)
Find the range of values of x for which both cables are safe.
(5)
(d)
Explain why the window cleaner must not stand at B.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
takes moments about A
M1
TC × 4.5 = 80g × 3.0 + 70g × x
A1
TC = (9.8/4.5)(240 + 70x) N, with no errors seen
A1
Part (b)
resolving vertically: TA = 150g − TC
M1
TA = (9.8/4.5)(435 − 70x) N
A1
Part (c)
sets TC ≤ 1200
M1
x ≤ 4.44
A1
considers the greatest value of TA, which occurs at x = 0
M1
TA ≤ 947 N < 1200 N, so the cable at A is always safe
A1
0 ≤ x ≤ 4.44
A1
Part (d)
at x = 6, TC = 1440 N (1437 N), which exceeds 1200 N, so the cable at C would not be safe
B1
Exam tip: For a range, apply every condition: each tension below its limit, each tension positive, and the person actually on the platform (0 ≤ x ≤ 6).
A non-uniform rod AB has length 2.0 m and mass M kg. Its centre of mass G is at a distance d metres from A. The rod is held in a horizontal position by two light vertical strings attached at C and D, where AC = 0.2 m and DB = 0.4 m. In a first experiment, a particle of mass 2 kg is attached to the rod at A, and the tensions in the two strings are equal. In a second experiment, the 2 kg particle is removed and a particle of mass 6 kg is attached at B; the rod is then on the point of turning about D.
Figure: the rod AB held by strings at C and D; the particles are not shown and G is not to scale.
(a)
Use the first experiment to show that Md = 0.9M + 1.8.
(4)
(b)
Use the second experiment to find a second equation connecting M and d.
(3)
(c)
Hence find the values of M and d.
(3)
(d)
Find the tension in the string at D in the second experiment.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
resolving vertically: each tension = ½(M + 2)g
M1
takes moments about C
M1
½(M + 2)g × 1.4 + 2g × 0.2 = Mg(d − 0.2)
A1
Md = 0.9M + 1.8, with no errors seen
A1
Part (b)
tension at C = 0
B1
moments about D
M1
Mg(1.6 − d) = 6g × 0.4, i.e. M(1.6 − d) = 2.4
A1
Part (c)
substitutes Md from (a): 1.6M − 0.9M − 1.8 = 2.4
M1
M = 6
A1
d = 1.2
A1
Part (d)
resolving vertically: T = (M + 6)g
M1
T = 12g = 118 N (accept 120 N)
A1
Exam tip: Two unknowns, two experiments: each situation gives one moments equation. Take moments about a point where an unknown tension acts so that each equation has only M and d in it.
A plank is laid across a ditch. It is modelled as a uniform rod AB of length 4.8 m and mass 25 kg, resting in a horizontal position on the two banks at C and D, where AC = 0.6 m and DB = 1.2 m. A bag of mass 15 kg is placed on the plank at A. A man of mass 80 kg steps onto the plank at C and walks slowly towards B. The bag and the man are modelled as particles.
Figure: the plank resting on the banks at C and D, with the man at the midpoint of CD; not to scale.
(a)
Find the magnitudes of the reactions at C and at D when the man is at the midpoint of CD.
(4)
(b)
Show that the man cannot reach B without the plank tilting, and find how far from A he is when the plank is about to tilt.
(4)
(c)
Find the smallest extra mass that must be added to the bag at A so that the man can reach B.
(3)
(d)
State two ways in which the model used here might be unrealistic.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
takes moments about C (or D), including the bag
M1
RD × 3.0 + 15g × 0.6 = 25g × 1.8 + 80g × 1.5
A1
RD = 52g = 510 N
A1
RC = 120g − 52g = 68g = 666 N
A1
Part (b)
about to tilt about D, so the reaction at C is zero
B1
moments about D
M1
25g × 1.2 + 15g × 3.6 = 80g × y, so y = 1.05 m beyond D
A1
distance from A = 4.65 m, which is less than 4.8 m, so he cannot reach B
A1
Part (c)
man at B, reaction at C = 0, moments about D with (15 + k)g at A
M1
25g × 1.2 + (15 + k)g × 3.6 = 80g × 1.2
A1
k = 10/3, so 3.34 kg (rounded up: 3.33 kg is not enough)
A1
Part (d)
a real plank bends (is not rigid) or is not uniform
B1
the man's weight is spread over his feet, not at one point / the banks touch the plank over a length, not at a single point
B1
Common error: Leaving out the bag at A when taking moments. Every force has a moment about the support unless it acts at that support — the bag works against the man.
A non-uniform beam AB has length 5.0 m. It rests in a horizontal position on two supports, one at A and one at B. A load of weight W newtons is hung from the beam at a fixed point E. For different values of W, the magnitudes of the reactions at A and at B are measured. The graph in Figure 1 shows the results, which lie on two straight lines. The load is modelled as a particle.
(a)
Use the graph to find the weight of the beam.
(2)
(b)
Find the distance from A of the centre of mass of the beam.
(3)
(c)
By considering the gradient of one of the lines, find the distance AE.
(3)
(d)
Explain why the gradients of the two lines add up to 1.
(2)
(e)
Find the value of W for which the reaction at A is twice the reaction at B.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
when W = 0 the reactions support the beam alone: 240 + 160
M1
weight = 400 N
A1
Part (b)
with W = 0, takes moments about A
M1
160 × 5.0 = 400 × AG
A1
AG = 2.0 m
A1
Part (c)
moments about A: 5.0RB = 400 × 2.0 + W × AE, so the gradient of the line for B is AE ÷ 5.0
M1
gradient from the graph = (280 − 160) ÷ 400 = 0.3 (allow 0.29–0.31)
B1
AE = 1.5 m
A1
(or: gradient of the line for A = 0.7 = (5.0 − AE) ÷ 5.0)
—
Part (d)
resolving vertically: RA + RB = 400 + W
B1
so each extra newton of load adds exactly 1 N to the total of the two reactions, and the sum of the gradients is 1 (0.7 + 0.3)
B1
Part (e)
240 + 0.7W = 2(160 + 0.3W)
M1
W = 800
A1
Exam tip: Read the intercepts first: at W = 0 the graph is describing the beam on its own, which gives its weight and its centre of mass before the load enters the problem.
Figure 1 shows a uniform plank AB, of length 4.0 m and mass 35 kg, resting in a horizontal position on two supports at C and D. A second uniform plank PQ, of length 2.0 m and mass 10 kg, lies in line with AB: its end Q rests on top of AB at B, and its end P rests on a ledge, so that PQ is horizontal. A person of mass 50 kg stands on PQ at a distance y metres from P. The person is modelled as a particle, and all the contact forces are vertical.
(a)
Given that y = 0.8, find the magnitude of the force exerted on PQ by AB at Q, and the magnitude of the force exerted on PQ by the ledge.
(4)
(b)
With y = 0.8, find the magnitudes of the forces exerted on AB by the supports at C and at D.
(4)
(c)
The person now walks slowly along PQ towards Q. Find the greatest value of y for which AB does not tilt.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
takes moments about P for PQ alone
M1
N × 2.0 = 10g × 1.0 + 50g × 0.8
A1
N = 25g = 245 N
A1
ledge: 60g − 25g = 35g = 343 N
A1
Part (b)
PQ pushes down on AB at B with 25g (equal and opposite to N)
B1
moments about C: RD × 2.5 = 35g × 1.5 + 25g × 3.5
M1
RD = 56g = 549 N
A1
RC = 60g − 56g = 4g = 39.2 N
A1
Part (c)
AB on the point of tilting about D, so RC = 0; moments about D: N × 1.0 = 35g × 1.0
M1
N = 35g
A1
moments about P for PQ: 35g × 2.0 = 10g × 1.0 + 50g × y
M1
y = 1.2
A1
Exam tip: With two bodies, draw a separate force diagram for each. The force at Q appears in both — upwards on PQ and downwards on AB — and it is the link between the two moments equations.
A scaffold board is modelled as a uniform rod AB of length 6.0 m and mass 30 kg. It rests in a horizontal position on two trestles at C and D, which exert vertical forces on it. A tin of paint of mass 10 kg stands at A, and a painter of mass 75 kg stands on the board at a distance x metres from A, as shown in Figure 1 (the painter's position in the figure is only one possibility). The tin and the painter are modelled as particles.
(a)
The painter stands at the midpoint of AB. Find the magnitudes of the forces exerted on the board by the trestles at C and at D.
(4)
(b)
Find the range of values of x for which the board does not tilt.
(6)
(c)
The tin is moved from A to B. Show that the painter can now stand at A without the board tilting, and find the greatest distance from A at which the painter can stand.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
takes moments about C, with the tin's moment in the opposite sense
M1
RD × 3.3 + 10g × 1.2 = 30g × 1.8 + 75g × 1.8
A1
RD = 177g ÷ 3.3 = 526 N
A1
RC = 115g − 526 = 601 N
A1
Part (b)
on the point of tilting about C, RD = 0; moments about C
M1
75g(1.2 − x) + 10g × 1.2 = 30g × 1.8
A1
x = 0.64
A1
on the point of tilting about D, RC = 0: 75g(x − 4.5) = 30g × 1.5 + 10g × 4.5
M1
x = 5.7
A1
0.64 ≤ x ≤ 5.7
A1
Part (c)
painter at A, tin at B, moments about C: RD × 3.3 = 30g × 1.8 + 10g × 4.8 − 75g × 1.2
M1
RD × 3.3 = 12g > 0, so the board stays in contact with D and does not tilt
A1
on the point of tilting about D: 75g(x − 4.5) + 10g × 1.5 = 30g × 1.5
M1
x = 4.9, so the greatest distance is 4.9 m
A1
Exam tip: For a range, find both tipping positions separately: tilting about C (reaction at D zero) and tilting about D (reaction at C zero). Then check that each one is actually on the board.
A workshop girder is modelled as a uniform rod AB of length 12 m and weight 24 kN, resting in a horizontal position on two columns, one at A and one at B. A hoist trolley runs along the top of the girder. Its rear wheel is at a distance x metres from A and its front wheel is 2 m further towards B. With its load, the trolley presses vertically down on the girder with the forces shown in Figure 1, which act at the wheels.
(a)
Show that, while both wheels are on the girder, the magnitude of the force on the girder from the column at B is (14 + 2.5x) kN.
(3)
(b)
Find, in terms of x, the magnitude of the force on the girder from the column at A.
(2)
(c)
Find the value of x for which the two columns carry forces of equal magnitude.
(2)
(d)
The column at B can safely carry a force of at most 36 kN. Find the range of values of x for which the trolley can be used safely with both wheels on the girder.
(3)
(e)
Explain why the expression in part (a) does not apply when x > 10.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
takes moments about A, with the front wheel at (x + 2)
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