Edexcel International A Level Maths · M1 · Mechanics 1
Statics of a particle
28 free, high-quality past-paper-style questions on Statics of a particle 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 · 221 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
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Two tugs pull a barge across a harbour. The barge is modelled as a particle and all forces act in a horizontal plane, in which i and j are perpendicular unit vectors. The first tug pulls with a force (9i − 4j) kN and the second with a force (−2i + 7j) kN.
(a)
Find the resultant of the two pulls, in the form (pi + qj) kN.
(1)
(b)
Find the magnitude of this resultant.
(2)
(c)
Find the angle between the resultant and the vector i.
(2)
(d)
The barge moves at constant velocity, so it is in equilibrium. Find the total resistance of the water on the barge as a vector.
(1)
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Answer
Mark
Notes
Part (a)
(9 − 2)i + (−4 + 7)j = (7i + 3j) kN
B1
Part (b)
√(72 + 32)
M1
= √58 = 7.62 kN
A1
Part (c)
tan−1(3 ÷ 7)
M1
= 23.2° (anticlockwise from i, towards j)
A1
Part (d)
−(7i + 3j) = (−7i − 3j) kN
B1
Exam tip: Add vectors component by component. A sketch of the resultant tells you which quadrant its angle is in before you press tan−1.
A sledge on horizontal ice is modelled as a particle. Two children pull it with horizontal ropes. One rope has a tension of 20 N and the other a tension of 12 N, and the angle between the two ropes is 70°.
Figure: the two rope tensions acting on the sledge (viewed from above); not to scale.
(a)
Taking the direction of the 20 N rope as the x-direction, find the components of the resultant of the two tensions parallel and perpendicular to this direction.
(2)
(b)
Find the magnitude of the resultant.
(2)
(c)
Find the angle between the resultant and the 20 N rope.
(1)
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Answer
Mark
Notes
Part (a)
parallel: 20 + 12 cos 70° = 24.1 N
M1
perpendicular: 12 sin 70° = 11.3 N
A1
Part (b)
√(24.102 + 11.282)
M1
= 26.6 N
A1
Part (c)
tan−1(11.28 ÷ 24.10) = 25.1°
B1
Common error: Adding the magnitudes, 20 + 12 = 32 N. Forces at an angle must be added as vectors: resolve, add the components, then recombine.
A lantern of mass 1.5 kg hangs from one end of a light string. The other end of the string is fixed. The lantern is held in equilibrium by a horizontal force of magnitude H newtons, so that the string makes an angle of 25° with the downward vertical. The lantern is modelled as a particle.
Figure: the lantern held by the horizontal force, with the string at 25° to the downward vertical; not to scale.
(a)
Find the tension in the string.
(2)
(b)
Find H.
(2)
(c)
Explain why no horizontal force, however large, could hold the lantern with the string horizontal.
(1)
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Answer
Mark
Notes
Part (a)
resolving vertically: T cos 25° = 1.5 × 9.8
M1
T = 16.2 N
A1
Part (b)
resolving horizontally: H = T sin 25° (or H = 14.7 tan 25°)
M1
H = 6.85 (accept 6.86 from rounded T)
A1
Part (c)
a horizontal string would have no vertical component of tension, so nothing would balance the weight
B1
Exam tip: Resolve vertically first when only one unknown force has a vertical component — it gives the tension straight away.
A box of mass 4 kg rests on rough horizontal ground. The coefficient of friction between the box and the ground is 0.45. A horizontal force of 15 N is applied to the box, which remains at rest. The box is modelled as a particle.
(a)
Find the normal reaction on the box.
(1)
(b)
State the magnitude of the friction force on the box.
(1)
(c)
Show that the box is not in limiting equilibrium.
(2)
(d)
Find the greatest horizontal force that can be applied without the box moving.
(1)
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Answer
Mark
Notes
Part (a)
R = 4 × 9.8 = 39.2 N
B1
Part (b)
15 N (it balances the applied force)
B1
Part (c)
maximum friction μR = 0.45 × 39.2 = 17.64 N
M1
15 < 17.64, so friction is below its maximum: not limiting
A1
Part (d)
17.6 N
B1
Common error: Writing friction = μR for a body at rest. μR is only the maximum; the actual friction is whatever balances the other forces.
Figure 1 shows three horizontal forces A, B and C acting on a puck O on smooth horizontal ice. The forces are drawn on a grid of squares in which the side of each square represents 1 N, and i and j are perpendicular unit vectors along the grid lines, as shown. The puck is modelled as a particle.
(a)
Write each of A, B and C in the form (pi + qj) N.
(2)
(b)
Find the resultant of the three forces.
(1)
(c)
A fourth horizontal force D is now applied to the puck so that it is in equilibrium. Find D.
(1)
(d)
Find the magnitude of D and the angle between D and the vector i.
(3)
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Answer
Mark
Notes
Part (a)
from the grid: A = (5i + 2j) N and B = (−2i + 4j) N
B1
C = (−i − 3j) N
B1
Part (b)
(5 − 2 − 1)i + (2 + 4 − 3)j = (2i + 3j) N
B1
Part (c)
D = −(2i + 3j) = (−2i − 3j) N
B1
Part (d)
|D| = √(22 + 32) = √13 = 3.61 N
B1
D points left and down: tan−1(3 ÷ 2) = 56.3° below −i
M1
angle with i = 180° − 56.3° = 123.7°
A1
Exam tip: Read a grid vector as (squares right, squares up) from the point where the force acts, counting left and down as negative. Check that the direction of your answer matches a sketch.
A shop sign of mass 5 kg hangs at a point P, which is held by two light inextensible strings PA and PB. The ends A and B are fixed to a horizontal beam, and N is the point on the beam vertically above P. Figure 1 shows the distances AN, NB and NP. The sign is modelled as a particle at P.
(a)
Show that PA makes an angle α with the horizontal where sin α = 0.6 and cos α = 0.8, and find the angle that PB makes with the horizontal.
(2)
(b)
By resolving horizontally and vertically, find the tension in each string.
(4)
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Answer
Mark
Notes
Part (a)
PA = √(1.22 + 0.92) = 1.5 m, so sin α = 0.9 ÷ 1.5 = 0.6 and cos α = 1.2 ÷ 1.5 = 0.8
B1
for PB, tan β = 0.9 ÷ 0.9 = 1, so β = 45°
B1
Part (b)
horizontally: 0.8TA = TB cos 45°
M1
vertically: 0.6TA + TB sin 45° = 5 × 9.8 = 49
M1
since TB sin 45° = TB cos 45° = 0.8TA: 1.4TA = 49, TA = 35 N
A1
TB = 0.8 × 35 ÷ cos 45° = 28√2 = 39.6 N
A1
Common error: Using the lengths 1.2 m and 0.9 m as if they were forces. The lengths only fix the directions of the strings: take sin and cos from the right-angled triangles, then resolve the tensions.
A filing cabinet of mass 4 kg stands on a rough horizontal floor. A horizontal force of magnitude P newtons is applied to it, and P is slowly increased from zero. Figure 1 shows how the friction force F newtons on the cabinet changes as P increases. The cabinet is modelled as a particle.
(a)
Use the graph to write down the greatest friction force that the floor can exert on the cabinet.
(1)
(b)
Find the coefficient of friction between the cabinet and the floor.
(2)
(c)
State the friction force when P = 12, and explain how you know that the cabinet is not in limiting equilibrium.
(1)
(d)
Files of total mass 2 kg are put into the cabinet. Find the horizontal force needed for the cabinet to be on the point of moving.
(2)
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Answer
Mark
Notes
Part (a)
18 N, where the graph levels off (allow 17.8–18.2)
B1
Part (b)
R = 4 × 9.8 = 39.2 N and 18 = μ × 39.2
M1
μ = 0.459
A1
Part (c)
F = 12 N, which is less than the maximum 18 N, so friction has not reached μR
B1
Part (d)
R = 6 × 9.8 = 58.8 N
M1
P = μR = 0.4592 × 58.8 = 27.0 N
A1
Exam tip: On a graph of friction against applied force, the sloping part shows friction matching the push; the level part is the limiting value μR. Read μR from where the graph turns.
A trolley of mass 25 kg is held at rest on a smooth ramp by a light rope parallel to the ramp, as shown in Figure 1. The ramp is inclined at an angle α to the horizontal, and Figure 1 shows the length of its sloping surface and its height. The trolley is modelled as a particle.
(a)
Find the exact values of sin α and cos α.
(1)
(b)
Find the tension in the rope.
(2)
(c)
Find the normal reaction between the trolley and the ramp.
(1)
(d)
The rope is removed and the trolley is instead held at rest by a horizontal force acting in the vertical plane containing a line of greatest slope. Find the magnitude of this force.
(2)
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Answer
Mark
Notes
Part (a)
sin α = 0.7 ÷ 2.5 = 7/25 and, with horizontal length √(2.52 − 0.72) = 2.4 m, cos α = 24/25
B1
Part (b)
resolving parallel to the ramp: T = 25 × 9.8 × 7/25
M1
T = 68.6 N
A1
Part (c)
R = 245 × 24/25 = 235.2 = 235 N
B1
Part (d)
resolving parallel to the ramp: H cos α = 245 sin α
M1
H = 245 × 7/24 = 71.5 N
A1
Exam tip: When a figure gives lengths rather than an angle, find exact sin and cos from the right-angled triangle; 7, 24, 25 is a Pythagorean triple worth recognising.
A loudspeaker of mass 8 kg hangs in equilibrium from two light inextensible cables attached to the same point on the loudspeaker. The upper ends of the cables are fixed to a horizontal beam. Cable A makes an angle of 35° with the horizontal and cable B makes an angle of 55° with the horizontal, on opposite sides of the loudspeaker. The loudspeaker is modelled as a particle.
Figure: the loudspeaker hanging from cables A and B; not to scale.
(a)
Show that the two cables are perpendicular to each other.
(1)
(b)
By resolving in the direction of cable A, find the tension in cable A.
(2)
(c)
Find the tension in cable B.
(2)
(d)
Each cable can safely carry a tension of at most 150 N. Find the greatest mass of loudspeaker that can be hung in the same way.
(2)
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Answer
Mark
Notes
Part (a)
the angle between the cables is 180° − 35° − 55° = 90°
B1
Part (b)
cable B has no component along A; weight component along A is 8g cos 55° (= 8g sin 35°): TA = 78.4 sin 35°
M1
TA = 45.0 N
A1
Part (c)
resolving along B: TB = 78.4 cos 35° (= 78.4 sin 55°)
M1
TB = 64.2 N
A1
Part (d)
cable B has the larger tension, so it reaches the limit first: mg sin 55° = 150
M1
m = 18.69, so 18.6 kg (rounded down: 18.7 kg would overload cable B)
A1
Exam tip: When two strings are perpendicular, resolve along each string in turn: the other tension then has no component, and each equation has one unknown.
A toy truck of mass 2 kg is held at rest on a smooth ramp inclined at 20° to the horizontal by a light cord. The cord lies in the vertical plane containing a line of greatest slope and makes an angle of 15° with the ramp, pulling up the slope and away from the ramp. The truck is modelled as a particle.
Figure: the toy truck on the smooth ramp, held by the cord at 15° to the ramp; not to scale.
(a)
Find the tension in the cord.
(3)
(b)
Find the normal reaction between the truck and the ramp.
(3)
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Answer
Mark
Notes
Part (a)
resolving parallel to the ramp: T cos 15° = 2g sin 20°
M1
2g sin 20° = 6.704
A1
T = 6.94 N
A1
Part (b)
resolving perpendicular to the ramp: R + T sin 15° = 2g cos 20°
M1
2g cos 20° = 18.42 and T sin 15° = 1.796
A1
R = 16.6 N
A1
Common error: Using R = mg cos α. The cord pulls the truck partly away from the ramp, so it reduces the normal reaction.
A book of mass 0.6 kg rests on a rough board. One end of the board is slowly raised. The book is on the point of sliding when the board makes an angle of 24° with the horizontal. The book is modelled as a particle and no force other than its weight and the contact forces from the board acts on it.
Figure: the book on the raised board at 24° to the horizontal; not to scale.
(a)
Show that the coefficient of friction μ between the book and the board satisfies μ = tan 24°.
(3)
(b)
Find the value of μ.
(1)
(c)
The board is lowered to 15° to the horizontal. Find the friction force on the book.
(1)
(d)
Show that the book is not in limiting equilibrium at 15°.
(2)
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Answer
Mark
Notes
Part (a)
perpendicular to the board: R = 0.6g cos 24°
M1
parallel to the board, limiting: μR = 0.6g sin 24°
M1
dividing: μ = sin 24° ÷ cos 24° = tan 24°
A1
Part (b)
μ = 0.445
B1
Part (c)
F = 0.6 × 9.8 × sin 15° = 1.52 N up the board
B1
Part (d)
R = 0.6g cos 15° = 5.68 N, so μR = 0.445 × 5.68 = 2.53 N
M1
1.52 < 2.53, so friction is less than its maximum
A1
Exam tip: On a rough plane with no other applied force, the mass cancels: the particle is on the point of sliding when μ = tan α.
A crate of mass 15 kg rests on rough horizontal ground. The coefficient of friction between the crate and the ground is 0.4. A worker pushes the crate with a light rod. The thrust in the rod has magnitude P newtons and acts at 20° below the horizontal. The crate is on the point of moving. The crate is modelled as a particle.
Figure: the crate pushed by the rod, with the thrust at 20° below the horizontal; not to scale.
(a)
Find an expression for the normal reaction in terms of P.
(2)
(b)
Find P.
(3)
(c)
Instead, the crate is pulled by a rope at 20° above the horizontal. Find the tension needed for the crate to be on the point of moving.
(2)
(d)
Explain why pulling needs a smaller force than pushing at the same angle.
(1)
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Answer
Mark
Notes
Part (a)
resolving vertically, with the vertical part of the thrust pushing down:
M1
R = 15g + P sin 20° = 147 + P sin 20°
A1
Part (b)
resolving horizontally, limiting friction: P cos 20° = 0.4R
M1
P(cos 20° − 0.4 sin 20°) = 0.4 × 147 = 58.8
M1
P = 73.2
A1
Part (c)
T cos 20° = 0.4(147 − T sin 20°)
M1
T = 58.8 ÷ (cos 20° + 0.4 sin 20°) = 54.6 N
A1
Part (d)
pushing downwards increases the normal reaction and so the maximum friction; pulling upwards reduces both
B1
Common error: Writing R = mg for a body on horizontal ground. Any force with a vertical component changes the normal reaction.
A hanging basket of weight 50 N is held in equilibrium by two light strings attached to the same point on the basket. String A makes an angle of 25° with the upward vertical and string B makes an angle of 45° with the upward vertical, on the other side. The basket is modelled as a particle.
Figure: the basket held by strings A and B, with angles measured from the upward vertical; not to scale.
(a)
The three forces on the basket form a triangle of forces. Find the angle in this triangle opposite the side representing the weight.
(2)
(b)
Find the tension in string A.
(2)
(c)
Find the tension in string B.
(1)
(d)
Verify your answers by resolving horizontally.
(1)
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Answer
Mark
Notes
Part (a)
the angle between the two strings is 25° + 45° = 70°
M1
angle opposite the weight = 180° − 70° = 110°
A1
Part (b)
sine rule: TA ÷ sin 45° = 50 ÷ sin 110°
M1
TA = 37.6 N
A1
Part (c)
TB = 50 sin 25° ÷ sin 110° = 22.5 N
B1
Part (d)
TA sin 25° = 15.9 and TB sin 45° = 15.9, so the horizontal components balance
B1
Exam tip: In a triangle of forces each interior angle is 180° minus the angle between the two forces that meet there. Check the three angles add to 180°.
A small ring on a smooth horizontal table is pulled by three horizontal cords. The ring is modelled as a particle, and i and j are perpendicular unit vectors in the plane of the table. The cords exert forces P = (2pi − 5j) N, Q = (3i + qj) N and S = (−7i + 2qj) N, where p and q are constants. The ring is in equilibrium.
(a)
Find the value of p and the value of q.
(4)
(b)
The cord exerting S is cut. Find the magnitude of the resultant of the two remaining forces.
(2)
(c)
Find the angle between this resultant and the vector i.
(2)
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Answer
Mark
Notes
Part (a)
i-components sum to zero: 2p + 3 − 7 = 0
M1
p = 2
A1
j-components: −5 + q + 2q = 0
M1
q = 5/3
A1
Part (b)
P + Q = −S = (7i − 10/3 j) N
M1
magnitude √(49 + 100/9) = 7.75 N
A1
Part (c)
tan−1((10/3) ÷ 7)
M1
= 25.5°, measured clockwise from i (below i)
A1
Exam tip: Equilibrium in vector form means the i-components and the j-components each add to zero — two equations for two unknowns.
A load of mass 60 kg is held at rest at P by a crane chain while a worker on the ground steadies it with a guide rope. Figure 1 shows the angle that the chain makes with the vertical and the angle that the rope makes with the horizontal. The load is modelled as a particle, and the chain and the rope are modelled as light inextensible strings in one vertical plane.
(a)
The three forces on the load form a triangle of forces. Find the three angles of this triangle.
(2)
(b)
Find the tension in the chain.
(2)
(c)
Find the tension in the guide rope.
(2)
(d)
Without further calculation, explain why the tension in the chain is greater than the weight of the load.
(1)
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Answer
Mark
Notes
Part (a)
the angle between the chain and the rope at P is 25° + 90° + 15° = 130°, so the angle opposite the weight is 180° − 130° = 50°
M1
the angle opposite the rope tension is 25° and the angle opposite the chain tension is 105°
A1
Part (b)
sine rule: T ÷ sin 105° = 588 ÷ sin 50°
M1
T = 741 N
A1
Part (c)
Q = 588 sin 25° ÷ sin 50°
M1
Q = 324 N
A1
Part (d)
the rope pulls the load downwards as well as sideways, so the chain must support the weight and the downward part of the rope tension
B1
Exam tip: Check sine-rule answers by resolving: horizontally T sin 25° = Q cos 15° and vertically T cos 25° = 588 + Q sin 15°.
A block of mass 3 kg rests on a rough plane inclined at an angle α to the horizontal. A force of magnitude P newtons acts on the block up the plane, along a line of greatest slope. Figure 1 shows the friction force F newtons on the block, taken as positive up the plane, for the values of P for which the block stays at rest. At each end of the line the block is on the point of slipping. The block is modelled as a particle.
(a)
Explain why P + F is constant while the block is at rest, and use the graph to find this constant.
(2)
(b)
Find α.
(2)
(c)
Find the coefficient of friction between the block and the plane.
(3)
(d)
Describe what happens to the block when P = 3, giving a reason.
(1)
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Answer
Mark
Notes
Part (a)
resolving parallel to the plane: P + F = 3g sin α, which does not depend on P
B1
from the graph, for example P = 6, F = 12: P + F = 18 N
B1
Part (b)
3 × 9.8 × sin α = 18
M1
α = 37.8°
A1
Part (c)
limiting friction is 12 N (the value of |F| at either end of the line)
B1
R = 29.4 cos α = √(29.42 − 182) = 23.2 N
M1
μ = 12 ÷ 23.25 = 0.516 (accept 0.517 from a rounded angle)
A1
Part (d)
P + μR = 3 + 12 = 15 < 18, so friction cannot hold the block and it slides down the plane
B1
Common error: Assuming friction always acts up a slope. On this graph F changes sign at P = 18: once the push exceeds the weight component, friction acts down the plane.
A smooth ring R of mass 0.4 kg is threaded on a light inextensible string. The ends of the string are fixed at the points A and B on a horizontal ceiling. A horizontal force of magnitude P newtons, acting in the vertical plane containing the string, holds the ring at rest in the position shown in Figure 1. The ring is modelled as a particle.
(a)
Explain why the tension is the same in both parts of the string.
(1)
(b)
Find the tension in the string.
(3)
(c)
Find P.
(2)
(d)
Find the length of the string.
(2)
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Answer
Mark
Notes
Part (a)
the ring is smooth, so it cannot change the tension in the light string passing through it
B1
Part (b)
resolving vertically: T cos 30° + T cos 60° = 0.4 × 9.8
M1
1.366T = 3.92
A1
T = 2.87 N
A1
Part (c)
resolving horizontally: P + T sin 30° = T sin 60°
M1
P = 2.870 × 0.3660 = 1.05
A1
Part (d)
if R is h m below the ceiling, h tan 30° + h tan 60° = 1.2, so h = 0.520
M1
RA = h ÷ cos 30° = 0.6 m and RB = h ÷ cos 60° = 1.04 m, so the length is 1.64 m
A1
Exam tip: A smooth ring, peg or pulley leaves the tension unchanged, so use a single letter T for both parts of the string; the two parts then differ only in direction.
A small mooring ring O lies on a smooth horizontal jetty. Four ropes attached to the ring pull it horizontally with forces of magnitudes 14 N, P N, Q N and 6 N, in the directions shown in Figure 1. The ring is in equilibrium and is modelled as a particle.
(a)
Show that the angle between the 6 N force and the 14 N force is 70°.
(1)
(b)
By resolving in the direction of the 14 N force, form an equation in P and Q.
(2)
(c)
By resolving perpendicular to the 14 N force, form a second equation in P and Q.
(2)
(d)
Find P and Q.
(2)
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Answer
Mark
Notes
Part (a)
360° − 100° − 115° − 75° = 70°
B1
Part (b)
14 + P cos 100° + Q cos 215° + 6 cos 290° = 0
M1
0.1736P + 0.8192Q = 16.05
A1
Part (c)
P sin 100° + Q sin 215° + 6 sin 290° = 0
M1
0.9848P − 0.5736Q = 5.638
A1
Part (d)
solving the two equations simultaneously
M1
P = 15.3, Q = 16.4
A1
Common error: Using the angle between neighbouring ropes as the angle with the reference direction. Measure every direction from the 14 N force: 100°, 215° and 290°.
A block A of mass 3 kg rests on a rough horizontal table. A light inextensible string attached to A passes over a small smooth pulley fixed at the edge of the table, and a particle B of mass 1.2 kg hangs freely from the other end, as shown in Figure 1. The string between A and the pulley is horizontal. A is on the point of slipping towards the pulley. A and B are modelled as particles.
(a)
Find the coefficient of friction between A and the table.
(3)
(b)
B is replaced by a particle of mass m kg, and a force of magnitude 10 N is applied to A at 30° above the horizontal, pulling away from the pulley, as shown dashed in Figure 1. Find the greatest value of m for which A stays at rest.
(3)
(c)
With the 10 N force still acting, the string is cut. Show that A stays at rest.
(2)
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Answer
Mark
Notes
Part (a)
B is in equilibrium, so T = 1.2 × 9.8 = 11.76 N
B1
R = 3 × 9.8 = 29.4 N and friction is limiting: 11.76 = μ × 29.4
M1
μ = 0.4
A1
Part (b)
R = 29.4 − 10 sin 30° = 24.4 N
M1
A about to slip towards the pulley, friction away from it: 9.8m = 10 cos 30° + 0.4 × 24.4
M1
m = (8.660 + 9.76) ÷ 9.8 = 1.88
A1
Part (c)
component of the pull along the table is 10 cos 30° = 8.66 N, and the maximum friction is 0.4 × 24.4 = 9.76 N
M1
8.66 < 9.76, so friction can hold A at rest
A1
Exam tip: Decide which way the particle is about to move before writing F = μR: friction acts against that motion. A pull with an upward component also reduces R.
A lamp of weight 150 N hangs from a point P. P is supported by a light rod BP, freely hinged to a vertical wall at B, and by a light inextensible string PA, where A is on the wall vertically above B. Figure 1 shows the angles that the rod and the string make with the horizontal through P. The lamp is modelled as a particle at P, and the force from the rod on P acts along the rod. The rod and the string lie in a vertical plane perpendicular to the wall.
(a)
Explain why the force in the rod must be a thrust rather than a tension.
(1)
(b)
Find the tension in the string.
(3)
(c)
Find the thrust in the rod.
(2)
(d)
The string breaks if its tension exceeds 250 N. Find the greatest weight of lamp that can be hung at P in the same way.
(1)
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Answer
Mark
Notes
Part (a)
the string pulls P towards the wall, so the rod must push P away from the wall to balance it horizontally: it is in thrust
B1
Part (b)
resolving horizontally: C cos 15° = T cos 50°
M1
resolving vertically: T sin 50° + C sin 15° = 150
M1
T = 160 N (159.9)
A1
(or triangle of forces with angles 40°, 65°, 75° and the sine rule)
—
Part (c)
C = T cos 50° ÷ cos 15° (or 150 sin 40° ÷ sin 65°)
M1
C = 106 N
A1
Part (d)
the tension is proportional to the weight: 150 × 250 ÷ 159.9 = 235 N
B1
Common error: Drawing the force from a rod as a pull. A light rod can push or pull along its length; decide which by asking what is needed to balance the other forces.
A parcel of mass 4 kg is placed on a rough loading ramp inclined at 25° to the horizontal. The coefficient of friction between the parcel and the ramp is 0.3. The parcel is held at rest by a light rope that lies in the vertical plane containing a line of greatest slope. The rope makes an angle of 20° with the ramp, pulling up the slope and away from the ramp. The tension in the rope is T newtons. The parcel is modelled as a particle.
Figure: the parcel on the rough ramp, held by the rope at 20° to the ramp; not to scale.
(a)
Show that the parcel would slide down the ramp if the rope were removed.
(2)
(b)
Find, in terms of T, the normal reaction between the parcel and the ramp.
(2)
(c)
Find the value of T when the parcel is on the point of sliding up the ramp.
(4)
(d)
Find the value of T when the parcel is on the point of sliding down the ramp.
(3)
(e)
State the range of values of T for which the parcel remains at rest.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
compare 4g sin 25° = 16.6 N with μ × 4g cos 25° = 10.7 N (or tan 25° = 0.466 with μ = 0.3)
M1
16.6 > 10.7, so friction cannot hold it: it slides
A1
Part (b)
resolving perpendicular to the ramp, rope component away from the ramp:
M1
R = 4g cos 25° − T sin 20° = 35.53 − T sin 20°
A1
Part (c)
friction acts down the ramp and F = 0.3R
M1
resolving parallel: T cos 20° = 4g sin 25° + 0.3(4g cos 25° − T sin 20°)
M1
T(cos 20° + 0.3 sin 20°) = 27.22
A1
T = 26.1
A1
Part (d)
friction now acts up the ramp: T cos 20° + 0.3R = 4g sin 25°
M1
T(cos 20° − 0.3 sin 20°) = 5.91
A1
T = 7.06
A1
Part (e)
7.06 ≤ T ≤ 26.1 (ft from (c) and (d))
B1
Exam tip: Write down the direction of friction for each limiting case before resolving. The normal reaction contains T, so substitute it into the friction term each time.
A sack of mass 10 kg rests on a rough slope inclined at 15° to the horizontal. The coefficient of friction between the sack and the slope is 0.4. A rope pulls the sack with a force of magnitude P newtons at an angle θabove the slope, in the vertical plane containing a line of greatest slope, so that the sack is on the point of moving up the slope. The sack is modelled as a particle.
Figure: the sack on the rough slope, pulled by the rope at angle θ above the slope; not to scale.
(a)
Show that P(cos θ + 0.4 sin θ) = 98(sin 15° + 0.4 cos 15°).
(5)
(b)
Find the value of P when the rope is parallel to the slope.
(1)
(c)
The value of P is least when tan θ = 0.4. Find this angle and the least value of P.
(3)
(d)
Show that the sack stays in contact with the slope in part (c).
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
resolving perpendicular to the slope: R + P sin θ = 98 cos 15°
M1
R = 98 cos 15° − P sin θ
A1
resolving parallel, friction down the slope: P cos θ = 98 sin 15° + F
M1
limiting: F = 0.4R, substituted
M1
rearranged to the printed result
A1
Part (b)
θ = 0: P = 98(sin 15° + 0.4 cos 15°) = 63.2
B1
Part (c)
θ = tan−1 0.4 = 21.8°
B1
P = 63.23 ÷ (cos 21.8° + 0.4 sin 21.8°)
M1
P = 58.71, so 58.8 (rounded up: 58.7 is not enough to reach limiting equilibrium)
A1
Part (d)
R = 98 cos 15° − 58.7 sin 21.8° = 72.9 N > 0
B1
Exam tip: For a “show that”, keep every line: the two resolved equations, the substitution of F = μR, and the rearrangement into exactly the printed form.
A particle P of mass m kg is held in equilibrium by two light inextensible strings. String AP is fixed at A and makes an angle of 40° with the horizontal. The second string is attached to P, passes over a small smooth pulley at B and has a particle of mass 3 kg hanging freely at its other end. The part BP of this string makes an angle of 20° with the horizontal, and A and B are on opposite sides of P, both above P. All parts of the strings lie in one vertical plane.
Figure: particle P held by string AP and by the string over the pulley at B carrying the 3 kg particle; not to scale.
(a)
Find the tension in the string passing over the pulley, giving a reason.
(2)
(b)
Find the tension in AP.
(3)
(c)
Find m.
(3)
(d)
Find the magnitude and direction of the resultant of the two tensions acting on P.
(1)
(e)
State how you have used the fact that the strings are light.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the 3 kg particle is in equilibrium, so the tension is 3g = 29.4 N
B1
the pulley is smooth, so the tension is the same on both sides, including BP
B1
Part (b)
resolving horizontally for P: TA cos 40° = 29.4 cos 20°
M1
29.4 cos 20° = 27.63
A1
TA = 36.1 N
A1
Part (c)
resolving vertically for P: TA sin 40° + 29.4 sin 20° = mg
M1
23.18 + 10.06 = 9.8m
A1
m = 33.24 ÷ 9.8 = 3.39
A1
Part (d)
it balances the weight of P: 33.2 N vertically upwards
B1
Part (e)
the weight of each string is ignored, so the tension is the same all along each string
B1
Common error: Setting the tension in BP equal to the weight of P. The string over the pulley carries the weight of the hanging 3 kg particle.
A small block of mass 2 kg is on a rough plane inclined at an angle α to the horizontal, where tan α = 5/12. A horizontal force of magnitude 20 N acts on the block, in the vertical plane containing a line of greatest slope, pushing it towards the plane. The block is on the point of sliding up the plane. The block is modelled as a particle and μ is the coefficient of friction.
Figure: the block on the rough plane, pushed by the horizontal force of 20 N; tan α = 5/12; not to scale.
(a)
Write down the exact values of sin α and cos α.
(1)
(b)
Show that the normal reaction is 25.8 N, correct to 3 significant figures.
(3)
(c)
Find the value of μ.
(4)
(d)
The horizontal force is removed. Show that the block remains at rest.
(3)
(e)
State the magnitude and direction of the friction force on the block after the horizontal force is removed.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
sin α = 5/13, cos α = 12/13
B1
Part (b)
resolving perpendicular to the plane: R = 2g cos α + 20 sin α
M1
= 19.6 × 12/13 + 20 × 5/13 = 335.2/13
A1
= 25.78… = 25.8 N
A1
Part (c)
friction acts down the plane and F = μR
M1
resolving parallel: 20 cos α = 2g sin α + μR
M1
μR = (240 − 98)/13 = 10.92
A1
μ = 142/335.2 = 0.424
A1
Part (d)
now R = 2g cos α = 18.09 N, so maximum friction μR = 7.66 N
M1
component of weight down the plane 2g sin α = 7.54 N
M1
7.54 ≤ 7.66, so friction can hold the block: it remains at rest (or μ = 0.424 ≥ tan α = 0.417)
A1
Part (e)
7.54 N up the plane
B1
Exam tip: With tan α given as a fraction, use the exact sin and cos from a 5-12-13 triangle. Exact working avoids rounding errors in the later comparison.
Figure 1 shows a fixed wedge whose two faces are inclined to the horizontal at the angles shown. Particle A, of mass 4 kg, is on the rough face, and the coefficient of friction between A and this face is 0.25. Particle B, of mass m kg, is on the smooth face. A and B are joined by a light inextensible string that passes over a small smooth pulley at the top of the wedge, and each part of the string is parallel to a line of greatest slope of its face. The system is in equilibrium.
(a)
Find, in terms of m, the tension in the string.
(2)
(b)
Find the normal reaction between A and the wedge.
(1)
(c)
Find the least value of m.
(4)
(d)
Find the greatest value of m.
(3)
(e)
Given that m = 2.5, find the magnitude and direction of the friction force on A.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
resolving along the smooth face for B: T = mg sin 50°
M1
T = 7.507m
A1
Part (b)
R = 4 × 9.8 × cos 30° = 33.9 N
B1
Part (c)
least m: A is on the point of slipping down its face, so friction acts up the face
M1
T + 0.25 × 33.95 = 4 × 9.8 × sin 30°
M1
T = 19.6 − 8.487 = 11.11 N
A1
m = 11.11 ÷ 7.507 = 1.48
A1
Part (d)
A is on the point of slipping up its face, so friction acts down the face: T = 19.6 + 8.487 = 28.09 N
M1
7.507m = 28.09
M1
m = 3.74
A1
Part (e)
T = 2.5 × 9.8 × sin 50° = 18.77 N, which is less than 19.6 N
M1
friction = 19.6 − 18.77 = 0.832 N up the face
A1
Exam tip: Treat each particle separately. The smooth pulley passes the same tension to both, so B fixes T in terms of m and the two limiting cases for A give the range of m.
A crate of mass 50 kg rests on a rough horizontal floor. Two horizontal ropes are attached to it. Figure 1 is a plan view, looking down on the floor, showing the tension in each rope and the angle between the ropes. The crate is on the point of slipping. The crate is modelled as a particle.
(a)
Find the magnitude of the resultant of the two tensions.
(3)
(b)
Find the angle between this resultant and rope 1.
(2)
(c)
State the magnitude and direction of the friction force on the crate.
(2)
(d)
Find the coefficient of friction between the crate and the floor.
(2)
(e)
Rope 2 is released, and the tension in rope 1 stays at 120 N. Show that the crate does not move, and state the friction force on it.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
component along rope 1: 120 + 90 cos 60° = 165 N
M1
component perpendicular to rope 1: 90 sin 60° = 77.94 N
A1
magnitude √(1652 + 77.942) = 182 N
A1
Part (b)
tan−1(77.94 ÷ 165)
M1
= 25.3°, on the same side as rope 2
A1
Part (c)
182 N (182.5)
B1
horizontal, directly opposite to the resultant of the two tensions
B1
Part (d)
R = 50 × 9.8 = 490 N and 182.5 = μ × 490
M1
μ = 0.372
A1
Part (e)
R is still 490 N, so the maximum friction is still 182.5 N, and 120 < 182.5
M1
the crate stays at rest with friction 120 N opposite to rope 1
A1
Common error: Taking friction as opposite to one rope. In a horizontal plane friction opposes the resultant of all the other horizontal forces, so find that resultant first.
A particle P of weight 63 N hangs in equilibrium from two light inextensible strings PA and PB, where A and B are fixed points. Figure 1 shows the positions of A, B and P in a vertical plane, with coordinates in metres. The unit vectors i and j are in the directions of the x- and y-axes, with j vertically upwards.
(a)
Find unit vectors in the directions of PA→ and PB→.
(2)
(b)
Find the tension in each string.
(4)
(c)
A steady wind now exerts a horizontal force of 8i N on P. P stays at the same point and both strings remain taut. Find the new tension in each string.
(3)
(d)
The wind force is wi N. Find the greatest value of w for which the string PB remains taut.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
PA→ = (−1.2i + 0.9j), length 1.5 m, unit vector (−0.8i + 0.6j)
B1
PB→ = (0.5i + 1.2j), length 1.3 m, unit vector (5/13 i + 12/13 j)
i: −0.8TA + (5/13)TB + 8 = 0, and the j equation is unchanged
M1
TB = 2.08TA − 20.8, so 2.52TA − 19.2 = 63
M1
TA = 32.6 N, TB = 47.0 N
A1
Part (d)
at the limit TB = 0: TA(−0.8i + 0.6j) + wi − 63j = 0
M1
TA = 63 ÷ 0.6 = 105 N and w = 0.8 × 105 = 84
A1
Exam tip: A tension acts along its string, so write it as T × (unit vector along the string). The figure gives the unit vectors; equilibrium gives one equation for each of i and j.
A crate of mass 25 kg rests on rough horizontal ground, and the coefficient of friction between the crate and the ground is 0.6. The normal reaction and the friction force on the crate can be combined into a single total contact force S, which acts at an angle λ to the vertical, as shown in Figure 1. A force of magnitude P newtons, at the angle shown, pulls the crate, which is on the point of moving. Figure 1 also shows the triangle of forces for the weight W, P and S. The crate is modelled as a particle.
(a)
Show that tan λ = 0.6 and find λ.
(2)
(b)
Find the three angles of the triangle of forces, and hence find P.
(4)
(c)
The direction of the pull can be changed. Explain, using the triangle of forces, why P is least when the pull is perpendicular to S. Find this least value of P and the angle that the pull then makes with the horizontal.
(3)
(d)
Check your answer to part (c) by resolving to find the normal reaction and the friction force.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the friction F and the normal reaction R are the components of S, so tan λ = F ÷ R = μR ÷ R = 0.6, since friction is limiting
B1
λ = 31.0°
B1
Part (b)
angle between the W and S sides = λ = 30.96°, and angle between the W and P sides = 90° − 10° = 80°
B1
third angle = 180° − 80° − 30.96° = 69.04°
B1
sine rule: P ÷ sin 30.96° = 245 ÷ sin 69.04°
M1
P = 135
A1
Part (c)
W is fixed and S must stay at λ to the vertical, so the shortest side that can close the triangle is perpendicular to the line of S
B1
P = 245 sin λ
M1
= 126 at 31.0° above the horizontal
A1
Part (d)
R = 245 − 126.1 sin 31.0° = 180.1 N, F = 126.1 cos 31.0° = 108.1 N and 108.1 ÷ 180.1 = 0.6, so friction is limiting as required
B1
Exam tip: Combining R and F into one contact force turns a four-force problem into a three-force triangle. In limiting equilibrium that force is at angle λ to the normal, where tan λ = μ.
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