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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-01 M1 · 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.

AB32g N20g N40g NOQ2.0 md m
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)
Show mark scheme
AnswerMarkNotes
Part (a)
takes moments about O; the weight of the plank acts at O so has no momentM1
32g × 2.0 = 40g × dA1
d = 1.6A1
Part (b)
resolving vertically: R = 20g + 32g + 40gM1
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.

2IALM-M1-6-B-02 M1 · ASEasy6 marks

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.

AB90 N150 NC0.8 m3.2 m
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)
Show mark scheme
AnswerMarkNotes
Part (a)
takes moments about A, with all three relevant termsM1
RB × 3.2 = 150 × 1.6 + 90 × 0.8A1
RB = 97.5 NA1
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 × AEM1
AE = 2.8 mA1

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.

3IALM-M1-6-B-03 M1 · ASEasy5 marks

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.

PQ0.30g N0.50g NR60 cm
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)
Show mark scheme
AnswerMarkNotes
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 equationsB1

Exam tip: Work in metres throughout, even when the question gives centimetres — or in centimetres throughout. Mixing the two is the slip to avoid.

4IALM-M1-6-B-04 M1 · ASEasy6 marks

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.

AB4g N2g NCD0.3 m0.3 m1.8 m
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.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
takes moments about CM1
TD × 1.2 = 4g × 0.6 + 2g × 1.5A1
TD = 4.5g = 44.1 NA1
Part (b)
resolving vertically: TC = 6g − 4.5g = 14.7 N (ft)A1
Part (c)
on the point of turning about D, TC = 0; moments about DM1
4g × 0.6 = mg × 0.3, so m = 8 kgA1

Exam tip: “The wire is about to become slack” translates directly into “tension = 0”. Then take moments about the other wire.

5IALM-M1-6-FB-01 M1 · ASEasy6 marks

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.

4 N A 12 N C 22 N D 4 N B 0.5 m 1.0 m 1.0 m force arrows not to scale
(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.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
taking anticlockwise as positive: 12 × 0.5 + 4 × 2.5 − 22 × 1.5M1
= −17, so the total moment is 17 N m clockwiseA1
Part (b)
4 × 2.5 + 22 × 1.0 − 12 × 2.0 = 8, so 8 N m anticlockwiseB1
Part (c)
resultant of the four forces: 12 + 4 − 4 − 22 = −10, so the extra force is 10 N upwardsB1
its moment about A must cancel 17 N m clockwise: 10 × d = 17M1
d = 1.7 m from AA1

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.

6IALM-M1-6-FB-02 M1 · ASEasy6 marks

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.

F 0.3 m 0.2 m 1.0 m A L G B wheel
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
takes moments about AM1
F × 1.5 = 15g × 0.5 + 60g × 0.3A1
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 × 0.3M1
m = 100 kgA1

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.

7IALM-M1-6-FB-03 M1 · ASEasy6 marks

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.

10 20 30 40 50 60 70 O B A 1.2 kg counterweight sack pan 10 cm marks: distance from O / cm
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
takes moments about O, including the panM1
(m + 0.40)g × 10 = 1.2g × 30A1
m = 3.2 kgA1
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 × 70M1
m = 8.0 kgA1

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.

8IALM-M1-6-FB-04 M1 · ASEasy5 marks

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.

G L 1.1 m 1.5 m 0.9 m F G R L front
(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.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
moments about F: RR × 2.6 = 1300g × 1.1M1
RR = 550g = 5390 NA1
RF = 1300g − 550g = 750g = 7350 NA1
Part (b)
moments about R: RF × 2.6 = 1300g × 1.5 − 80g × 0.9M1
RF = 7080 N (accept 7079 N)A1

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.

9IALM-M1-6-B-05 M1 · ASMedium8 marks

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.

ABbolt30g N65g NC1.2 m4.0 m
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)
Show mark scheme
AnswerMarkNotes
Part (a)
takes moments about AM1
R × 1.2 = 30g × 2.0 + 65g × 4.0A1
R = 266.7g = 2610 N (accept 2600 N)A1
Part (b)
resolving vertically with the force at A unknown in signM1
force = 2613 − 95g = 1682 N, so 1680 N (accept 1700 N)A1
acting vertically downwardsA1
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 clockwiseM1
the support at C has no moment about C, so with no bolt nothing gives an anticlockwise moment and the board turns about CA1

Common error: Assuming every support pushes upwards. A fixing bolt can pull down — if resolving gives a negative upward force, the force acts downwards.

10IALM-M1-6-B-06 M1 · ASMedium8 marks

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.

AB1.25FF54g NG2.8 m
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)
Show mark scheme
AnswerMarkNotes
Part (a)
resolving: 2.25F = 54gM1
force at B = 24g = 235 N, force at A = 30g = 294 NA1
Part (b)
takes moments about A, with AG = xM1
24g × 2.8 = 54g × xA1
x = 1.244…, AG = 1.24 mA1
Part (c)
each force = 27g; moments about AM1
27g × AD = 54g × 1.244…, so AD = 2.488… mA1
BD = 0.311 mA1

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.

11IALM-M1-6-B-07 M1 · ASMedium7 marks

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.

AB18g N72g NCD0.6 m0.9 m3.6 m
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)
Show mark scheme
AnswerMarkNotes
Part (a)
takes moments about C (or D) with the midpoint of CD 1.05 m from CM1
RD × 2.1 = 18g × 1.2 + 72g × 1.05, RD = 454 NA1
RC = 90g − RD = 428 NA1
Part (b)
on the point of tilting about D, the reaction at C is zeroB1
moments about D: 18g × 0.9 = 72g × y, so y = 0.225 m beyond DM1
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 AB1

Exam tip: Write “about to tilt about D, so RC = 0” before any equation: it tells you exactly which point to take moments about.

12IALM-M1-6-B-08 M1 · ASMedium8 marks

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.

AB0.4g NDC0.7 m0.9 m1.5 m
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)
Show mark scheme
AnswerMarkNotes
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 CB1
Part (b)
takes moments about DM1
0.4g × 0.7 = mg × 0.2A1
m = 1.4A1
Part (c)
1.8g = 17.6 NB1
Part (d)
moments about the new balance point E, with AE = eM1
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.

13IALM-M1-6-B-09 M1 · ASMedium9 marks

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.

AB1200g N600g Nx m8 m
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)
Show mark scheme
AnswerMarkNotes
Part (a)
takes moments about AM1
RB × 8 = 600g × 4 + 1200g × xA1
RB = (300 + 150x)g N, with no errors seenA1
Part (b)
resolving vertically: RA = 1800g − RBM1
RA = (1500 − 150x)g NA1
Part (c)
sets up RA = 2RB using the expressionsM1
1500 − 150x = 600 + 300xA1
x = 2A1
Part (d)
a vehicle has its weight spread over two or more axles at different distances from A, not acting at one pointB1

Exam tip: With a “show that”, write the moments equation in full before simplifying, and finish with exactly the printed expression.

14IALM-M1-6-B-10 M1 · ASMedium8 marks

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.

AB18g N12g NC0.5 m4.5 m
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)
Show mark scheme
AnswerMarkNotes
Part (a)
takes moments about C; the barrier's weight acts 1.75 m from CM1
R × 4.0 + 18g × 0.5 = 12g × 1.75A1
R = 3g = 29.4 NA1
Part (b)
resolving vertically: force = 30g − 3gM1
27g = 265 NA1
Part (c)
the force from the post is zeroB1
moments about C: Mg × 0.5 = 12g × 1.75M1
M = 42 kgA1

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.

15IALM-M1-6-FB-05 M1 · ASMedium7 marks

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.

0 10 20 30 40 50 60 70 80 90 100 100 g knife-edge scale / cm
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
the weight of the rule acts at the 50 cm mark, 10 cm from the knife-edgeB1
moments about the knife-edge: 0.1g × 30 = mg × 10M1
m = 0.30A1
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 itM1
0.1g(k − 10) = 0.30g(50 − k) + 0.1g(70 − k)A1
k = 46, the 46 cm markA1
(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.

16IALM-M1-6-FB-06 M1 · ASMedium8 marks

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.

14.0 kg 10.0 kg A B 3.0 m scale readings in kg
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
14.0 + 10.0 = 24 kgB1
Part (b)
takes moments about AM1
10g × 3.0 = 24g × dA1
d = 1.25 mA1
Part (c)
each reading is (24 + 30) ÷ 2 = 27 kgB1
moments about A: 27g × 3.0 = 24g × 1.25 + 30g × APM1
AP = 1.7 mA1
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 itB1

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.

17IALM-M1-6-FB-07 M1 · ASMedium8 marks

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.

books vase wall 0.2 m 0.3 m 0.7 m 0.4 m A C E D B
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
takes moments about C (or D), with all four termsM1
RD × 1.0 = 4g × 0.6 + 9g × 0.3 + 2g × 1.4A1
RD = 7.9g = 77.4 NA1
RC = 15g − 7.9g = 7.1g = 69.6 NA1
Part (b)
on the point of tilting about D, so RC = 0B1
moments about D: Mg × 0.4 = 4g × 0.4 + 9g × 0.7M1
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 momentB1

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.

18IALM-M1-6-FB-08 M1 · ASMedium8 marks

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.

M kg 0.05 kg m kg P O Q R T S 24 cm 16 cm 20 cm 10 cm
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
moments about T: 0.05g × 20 = mg × 10M1
m = 0.1A1
Part (b)
(0.05 + 0.1)g = 1.47 N (ft from (a))B1
Part (c)
moments about O: Mg × 24 = 0.15g × 16M1
M = 0.1A1
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 cmA1

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.

19IALM-M1-6-FB-09 M1 · ASMedium9 marks

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.

0 0.5 1 1.5 2 2.5 3 3.5 4 0 100 200 300 400 500 600 700 800 reaction at C RC / N x / m
(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
AnswerMarkNotes
Part (a)
moments about D: 2.5RC = 300(3.0 − AG) + W(3.0 − x), so the gradient is −W ÷ 2.5M1
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 mA1
Part (c)
at x = 3.75, RC = 0: the plank is on the point of tilting about DB1
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.

20IALM-M1-6-FB-10 M1 · ASMedium7 marks

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.

FA FB 12 kg 20 kg 0.4 m 1.1 m 0.5 m A C D B
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
takes moments about A (or B), with all three weightsM1
FB × 2.0 = 4g × 1.0 + 12g × 0.4 + 20g × 1.5A1
FB = 19.4g = 190A1
FA = 36g − 19.4g = 16.6g = 163A1
Part (b)
each force is 36g ÷ 2 = 18gB1
moments about A: 18g × 2.0 = 4g × 1.0 + 12g × 0.4 + 20g × xM1
x = 1.36 mA1

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.

21IALM-M1-6-B-11 M1 · ASHard13 marks

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.

AB30g N36g NCDG1.0 m1.0 m5.0 m
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.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
reaction at C = 0B1
moments about D: 30g × (4.0 − AG) = 36g × 1.0M1
AG = 2.8 m, with no errors seenA1
Part (b)
resolving vertically: 2R = 66gM1
R = 33gA1
moments about A: 33g × 1.0 + 33g × 4.0 = 30g × 2.8 + 36g × AEM1
AE = 2.25 mA1
Part (c)
moments about C with the boy at AM1
RD × 3.0 = 30g × 1.8 − 36g × 1.0 = 18gA1
RD = 6g = 58.8 N > 0, so the plank is still in contact with D and does not tiltA1
Part (d)
moments about D, including all four forces with a momentM1
400 × 3.0 + 36g × 1.0 = 30g × 1.2 + mg × 4.0A1
m = 30.6A1

Exam tip: To show that a plank does not tilt, find the reaction at the support that would lose contact and show it is positive.

22IALM-M1-6-B-12 M1 · ASHard11 marks

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.

AB80g N70g NCx m1.5 m6.0 m
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
AnswerMarkNotes
Part (a)
takes moments about AM1
TC × 4.5 = 80g × 3.0 + 70g × xA1
TC = (9.8/4.5)(240 + 70x) N, with no errors seenA1
Part (b)
resolving vertically: TA = 150g − TCM1
TA = (9.8/4.5)(435 − 70x) NA1
Part (c)
sets TC ≤ 1200M1
x ≤ 4.44A1
considers the greatest value of TA, which occurs at x = 0M1
TA ≤ 947 N < 1200 N, so the cable at A is always safeA1
0 ≤ x ≤ 4.44A1
Part (d)
at x = 6, TC = 1440 N (1437 N), which exceeds 1200 N, so the cable at C would not be safeB1

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).

23IALM-M1-6-B-13 M1 · ASHard12 marks

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.

ABMg NCDG0.2 m0.4 md m2.0 m
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
AnswerMarkNotes
Part (a)
resolving vertically: each tension = ½(M + 2)gM1
takes moments about CM1
½(M + 2)g × 1.4 + 2g × 0.2 = Mg(d − 0.2)A1
Md = 0.9M + 1.8, with no errors seenA1
Part (b)
tension at C = 0B1
moments about DM1
Mg(1.6 − d) = 6g × 0.4, i.e. M(1.6 − d) = 2.4A1
Part (c)
substitutes Md from (a): 1.6M − 0.9M − 1.8 = 2.4M1
M = 6A1
d = 1.2A1
Part (d)
resolving vertically: T = (M + 6)gM1
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.

24IALM-M1-6-B-14 M1 · ASHard13 marks

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.

AB15g N25g N80g NCD0.6 m1.2 m4.8 m
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
AnswerMarkNotes
Part (a)
takes moments about C (or D), including the bagM1
RD × 3.0 + 15g × 0.6 = 25g × 1.8 + 80g × 1.5A1
RD = 52g = 510 NA1
RC = 120g − 52g = 68g = 666 NA1
Part (b)
about to tilt about D, so the reaction at C is zeroB1
moments about DM1
25g × 1.2 + 15g × 3.6 = 80g × y, so y = 1.05 m beyond DA1
distance from A = 4.65 m, which is less than 4.8 m, so he cannot reach BA1
Part (c)
man at B, reaction at C = 0, moments about D with (15 + k)g at AM1
25g × 1.2 + (15 + k)g × 3.6 = 80g × 1.2A1
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 uniformB1
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 pointB1

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.

25IALM-M1-6-FB-11 M1 · ASHard12 marks

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.

0 100 200 300 400 0 80 160 240 320 400 480 560 640 reaction at A reaction at B reaction / N load W / N
(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
AnswerMarkNotes
Part (a)
when W = 0 the reactions support the beam alone: 240 + 160M1
weight = 400 NA1
Part (b)
with W = 0, takes moments about AM1
160 × 5.0 = 400 × AGA1
AG = 2.0 mA1
Part (c)
moments about A: 5.0RB = 400 × 2.0 + W × AE, so the gradient of the line for B is AE ÷ 5.0M1
gradient from the graph = (280 − 160) ÷ 400 = 0.3 (allow 0.29–0.31)B1
AE = 1.5 mA1
(or: gradient of the line for A = 0.7 = (5.0 − AE) ÷ 5.0)—
Part (d)
resolving vertically: RA + RB = 400 + WB1
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 = 800A1

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.

26IALM-M1-6-FB-12 M1 · ASHard12 marks

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.

ledge person y m Q P 0.5 m 2.5 m 1.0 m 2.0 m A C D B P
(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
AnswerMarkNotes
Part (a)
takes moments about P for PQ aloneM1
N × 2.0 = 10g × 1.0 + 50g × 0.8A1
N = 25g = 245 NA1
ledge: 60g − 25g = 35g = 343 NA1
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 × 3.5M1
RD = 56g = 549 NA1
RC = 60g − 56g = 4g = 39.2 NA1
Part (c)
AB on the point of tilting about D, so RC = 0; moments about D: N × 1.0 = 35g × 1.0M1
N = 35gA1
moments about P for PQ: 35g × 2.0 = 10g × 1.0 + 50g × yM1
y = 1.2A1

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.

27IALM-M1-6-FB-13 M1 · ASHard14 marks

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.

tin painter x m 1.2 m 3.3 m 1.5 m A C D B
(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
AnswerMarkNotes
Part (a)
takes moments about C, with the tin's moment in the opposite senseM1
RD × 3.3 + 10g × 1.2 = 30g × 1.8 + 75g × 1.8A1
RD = 177g ÷ 3.3 = 526 NA1
RC = 115g − 526 = 601 NA1
Part (b)
on the point of tilting about C, RD = 0; moments about CM1
75g(1.2 − x) + 10g × 1.2 = 30g × 1.8A1
x = 0.64A1
on the point of tilting about D, RC = 0: 75g(x − 4.5) = 30g × 1.5 + 10g × 4.5M1
x = 5.7A1
0.64 ≤ x ≤ 5.7A1
Part (c)
painter at A, tin at B, moments about C: RD × 3.3 = 30g × 1.8 + 10g × 4.8 − 75g × 1.2M1
RD × 3.3 = 12g > 0, so the board stays in contact with D and does not tiltA1
on the point of tilting about D: 75g(x − 4.5) + 10g × 1.5 = 30g × 1.5M1
x = 4.9, so the greatest distance is 4.9 mA1

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.

28IALM-M1-6-FB-14 M1 · ASHard12 marks

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.

18 kN 12 kN towards B 2 m x m 12 m A B
(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
AnswerMarkNotes
Part (a)
takes moments about A, with the front wheel at (x + 2)M1
12RB = 24 × 6 + 18x + 12(x + 2)A1
12RB = 168 + 30x, so RB = 14 + 2.5xA1
Part (b)
resolving vertically: RA = 24 + 18 + 12 − (14 + 2.5x)M1
RA = 40 − 2.5x kNA1
Part (c)
14 + 2.5x = 40 − 2.5xM1
x = 5.2A1
Part (d)
14 + 2.5x ≤ 36M1
x ≤ 8.8A1
both wheels on the girder needs 0 ≤ x ≤ 10, so 0 ≤ x ≤ 8.8A1
Part (e)
for x > 10 the front wheel is beyond B, so it is no longer on the girderB1
its 12 kN force then does not act on the girder, so the term 12(x + 2) must be removed from the moments equationB1

Common error: Using x for both wheels. The front wheel is at (x + 2) from A, so its moment about A is 12(x + 2), not 12x.