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A-Level · Mechanics

M7 Mechanics: Moments and Statics Depth

EDEXCEL 9MA0 · Calculator allowed · about 135 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A worker uses a crowbar to lift the edge of a heavy paving slab. The crowbar is modelled as a light rigid lever, pivoting on a fixed fulcrum stone at the point O.

(a)The worker applies a force of 180 N, perpendicular to the crowbar, at a point 0.9 m from O. Calculate the moment of this force about O.(2)

(b)The slab requires a moment of at least 126 N m about O to lift it. Find the minimum distance from O at which the worker would need to apply a force of 120 N, acting perpendicular to the crowbar, to lift the slab.(2)

(Total for Question 1 is 4 marks)

A light rigid rod AB has length 1 m. It rests horizontally in equilibrium on a smooth pivot at the point C on the rod. Weights of 40 N and 70 N hang from hooks on the rod at distances 0.25 m and 0.5 m from C, on one side of the pivot. A weight of magnitude W newtons hangs from a hook at a distance 0.6 m from C, on the other side of the pivot.

(a)By taking moments about C, show that W = 75.(3)

(b)Find the reaction at the pivot C.(2)

(Total for Question 2 is 5 marks)

A uniform beam AB has length 6 m and weight 210 N. The beam rests horizontally in equilibrium on two smooth supports, one at C, 1 m from A, and one at D, 4 m from A.

(a)By taking moments about C, show that the reaction at D, R_D, is 140 N.(3)

(b)Find the reaction at C, R_C.(3)

(Total for Question 3 is 6 marks)

A non-uniform rod AB has length 4 m and weight 200 N. The rod is held in equilibrium in a horizontal position by two vertical strings, one attached at A and the other attached at B. The tension in the string at A is three times the tension in the string at B.

(a)Show that the tension in the string at B is 50 N.(2)

(b)By taking moments about A, find the distance of the centre of mass of the rod from A.(3)

(c)A particle of weight 40 N is attached to the rod at B. Given that the rod remains in horizontal equilibrium, find the new tension in the string at A.(2)

(Total for Question 4 is 7 marks)

A uniform rod AB has length 2.4 m and weight 100 N. The rod rests with end A on rough horizontal ground and passes over a small smooth peg fixed at the point P on the rod, where AP = 1.6 m. The peg supports the rod from below, exerting a force on the rod perpendicular to the rod. The rod is inclined to the horizontal at an angle θ, where tan(θ) = 3/4, and rests in equilibrium.

(a)By taking moments about A, show that the normal reaction at the peg, N, has magnitude 60 N.(5)

(b)Find the normal reaction at A, N_A.(3)

(c)Find the friction force at A.(2)

(d)Given that the rod is on the point of slipping at A, find the coefficient of friction between the rod and the ground.(2)

(Total for Question 5 is 12 marks)

A shop sign of weight 80 N hangs in equilibrium from a single point on the sign. Two light inextensible strings are attached to this point, with their other ends fixed to a horizontal ceiling. String 1 makes an angle of 50 degrees with the ceiling, and string 2 makes an angle of 65 degrees with the ceiling, on opposite sides of the vertical through the sign. Model the sign as a particle.

(a)Find the tension in each string, giving your answers to 3 significant figures.(7)

(b)The sign is now instead supported by string 1 alone (still at 50 degrees to the ceiling) together with a smooth horizontal strut, fixed to a wall, which pushes horizontally on the sign. Find the tension in string 1 and the force in the strut.(4)

(Total for Question 6 is 11 marks)

A uniform ladder AB has length 6 m and weight 290 N. The ladder rests in limiting equilibrium, on the point of slipping at both ends, with end A on rough horizontal ground and end B against a rough vertical wall, inclined at angle θ to the horizontal. The coefficient of friction between the ladder and the ground equals the coefficient of friction between the ladder and the wall, both equal to μ = 0.4. Let N_A and F_A denote the normal reaction and friction force at A, and N_B and F_B denote the normal reaction and friction force at B.

(a)By resolving horizontally and vertically, and using F_A = μ*N_A and F_B = μ*N_B, show that N_A = 250 N and N_B = 100 N.(5)

(b)By taking moments about A, find the angle θ, giving your answer to 3 significant figures.(5)

(c)Find the magnitude of the resultant contact force exerted on the ladder by the ground at A.(3)

(Total for Question 7 is 13 marks)

A uniform lamina is formed from a rectangle ABCD, where AB = 6 cm and BC = 10 cm, together with a uniform semicircular lamina of the same material and thickness, of diameter 6 cm, attached along the edge CD, so that the semicircle bulges outward from the rectangle. The centre of mass of a uniform semicircular lamina of radius r lies a distance 4r/(3*π) from the centre of its diameter, along the axis of symmetry.

(a)Show that the area of the semicircular portion is 4.5*π cm^2.(2)

(b)Find the distance of the centre of mass of the whole lamina from the edge AB (the edge of the rectangle furthest from the semicircle).(4)

(Total for Question 8 is 6 marks)

A uniform lamina occupies the region of a rectangle OABC, where O is the origin, OA lies along the x-axis with OA = 12 cm, and OC lies along the y-axis with OC = 8 cm. A rectangular hole PQRS is removed from the lamina, where P = (2, 2), Q = (6, 2), R = (6, 5) and S = (2, 5), all coordinates in cm.

(a)Find the coordinates of the centre of mass of the remaining lamina.(6)

(b)The lamina is freely suspended from the point A and hangs in equilibrium. Find the angle that AB makes with the vertical, giving your answer to 3 significant figures.(3)

(Total for Question 9 is 9 marks)

A framework is formed from three uniform rods AB, BC and CA, rigidly joined at their ends to form a triangle with a right angle at B. The rods are made of the same material, with the same mass per unit length, and AB = 6a, BC = 8a.

(a)Show that CA = 10a.(1)

(b)Taking B as the origin, with BC along the x-axis and BA along the y-axis, find the coordinates of the centre of mass of the framework, giving your answer in terms of a.(7)

(c)The framework is freely suspended from A and hangs in equilibrium. Find, to 3 significant figures, the angle that AB makes with the vertical.(4)

(Total for Question 10 is 12 marks)

A uniform triangular lamina ABC has a right angle at B, with AB = 9 cm and BC = 12 cm.

(a)Show that AC = 15 cm.(1)

(b)Taking B as the origin, with BC along the x-axis and BA along the y-axis, find the coordinates of the centre of mass of the lamina.(3)

(c)The lamina is freely suspended from the vertex A and hangs in equilibrium. Find the angle that AB makes with the vertical, giving your answer to 3 significant figures.(5)

(Total for Question 11 is 9 marks)

A uniform solid cuboid rests in equilibrium on a rough plane inclined at an angle θ to the horizontal, with its longer edges horizontal and its rectangular cross-section, of width 1.2 m and height 2 m, lying in a vertical plane containing a line of greatest slope. The coefficient of friction between the cuboid and the plane is μ = 0.4. The angle θ is slowly increased from 0. Take g = 9.8 m/s² where needed.

(a)Given that the cuboid does not slide, find the angle θ at which it is on the point of toppling about its lower edge, giving your answer to 3 significant figures.(4)

(b)Given instead that the cuboid does not topple, find the angle θ at which it is on the point of sliding, giving your answer to 3 significant figures.(4)

(c)Hence determine, with justification, whether the cuboid slides or topples first as θ increases from 0, and state the critical value of θ at which equilibrium first breaks down.(2)

(d)Given that the mass of the cuboid is 45 kg, find the normal reaction between the cuboid and the plane at this critical angle, giving your answer to 3 significant figures.(3)

(Total for Question 12 is 13 marks)

Mark scheme · M7 Mechanics: Moments and Statics Depth

Question 1

(a) M1 moment = force * distance = 180 * 0.9
(a) A1 162 N m cao (units required)
(a) Answer: 162 N m
(b) M1 126 = 120 * d
(b) A1 d = 1.05 m cao
(b) Answer: 1.05 m

Question 2

(a) M1 sets up the moments equation about C for the two sides of the pivot
(a) A1 40 * 0.25 + 70 * 0.5 = W * 0.6 (i.e. 10 + 35 = 0.6W)
(a) A1 W = 75 cso
(a) Answer: W = 75 N (shown)
(b) M1 resolves vertically: R = 40 + 70 + 75
(b) A1 R = 185 N cao
(b) Answer: 185 N

Question 3

(a) M1 takes moments about C, using CD = 3 m and the beam's centre of mass 2 m from C
(a) A1 R_D * 3 = 210 * 2 (correct equation)
(a) A1 R_D = 140 (N) cso
(a) Answer: R_D = 140 N (shown)
(b) M1 resolves vertically: R_C + R_D = 210
(b) A1 R_C = 210 - 140
(b) A1 R_C = 70 (N) cao
(b) Answer: 70 N

Question 4

(a) M1 forms 3T_B + T_B = 200, using T_A = 3T_B and T_A + T_B = 200
(a) A1 T_B = 50 cso
(a) Answer: T_B = 50 N (shown)
(b) M1 takes moments about A: 200x = T_B * 4
(b) A1 200x = 50 * 4 = 200
(b) A1 x = 1 (m) cao
(b) Answer: 1 m from A
(c) M1 takes moments about B: T_A' * 4 = 200 * (4-1) + 40 * 0
(c) A1 T_A' = 150 (N) cao
(c) Answer: 150 N

Question 5

(a) M1 recognises that the moment of N about A is N * AP, since N acts perpendicular to the rod at the point P on the rod
(a) A1 recognises that the moment of the weight about A is W * (AB/2) * cos(θ), since the weight acts vertically at the midpoint of the rod
(a) M1 uses the 3-4-5 triangle from tan(θ) = 3/4 to state cos(θ) = 4/5
(a) A1 forms N * 1.6 = 100 * 1.2 * 4/5 (= 96)
(a) A1 N = 96/1.6 = 60 cso
(a) Answer: N = 60 N (shown)
(b) M1 resolves vertically: N_A + N*cos(θ) = 100
(b) A1 N_A = 100 - 60*(4/5) = 100 - 48
(b) A1 N_A = 52 (N) cao
(b) Answer: 52 N
(c) M1 resolves horizontally: F_A = N*sin(θ)
(c) A1 F_A = 60*(3/5) = 36 (N) cao
(c) Answer: 36 N
(d) M1 μ = F_A / N_A
(d) A1 μ = 36/52 = 9/13 (awrt 0.692) cao
(d) Answer: μ = 9/13 (awrt 0.692)

Question 6

(a) M1 resolves horizontally: T1*cos(50) = T2*cos(65)
(a) A1 correct equation
(a) M1 resolves vertically: T1*sin(50) + T2*sin(65) = 80
(a) A1 correct equation
(a) M1 eliminates one tension using the horizontal equation and solves the resulting equation in one unknown
(a) A1 T2 = awrt 56.7 (N)
(a) A1 T1 = awrt 37.3 (N)
(a) Answer: T1 = 37.3 N (string at 50 degrees); T2 = 56.7 N (string at 65 degrees), both 3sf
(b) M1 resolves vertically: T1'*sin(50) = 80
(b) A1 T1' = 80/sin(50) = awrt 104 (N)
(b) M1 resolves horizontally: strut force = T1'*cos(50)
(b) A1 strut force = awrt 67.1 (N)
(b) Answer: T1' = 104 N; strut force = 67.1 N (both 3sf)

Question 7

(a) M1 resolves vertically: N_A + F_B = 290
(a) M1 resolves horizontally: N_B = F_A
(a) B1 substitutes the limiting friction conditions F_A = μ*N_A and F_B = μ*N_B into both equations
(a) M1 combines the two equations to eliminate a variable, e.g. N_A*(1 + μ²) = 290
(a) A1 N_A = 290/1.16 = 250 (N) and N_B = μ*N_A = 100 (N), both values, cso
(a) Answer: N_A = 250 N; N_B = 100 N (shown)
(b) M1 takes moments about A, identifying moment arms (L/2)cos(θ) for W, L*sin(θ) for N_B and L*cos(θ) for F_B, with L cancelling
(b) A1 (W/2)*cos(θ) = N_B*sin(θ) + F_B*cos(θ)
(b) M1 substitutes W = 290, N_B = 100, F_B = 40 and rearranges to isolate tan(θ)
(b) A1 tan(θ) = 105/100 = 1.05 (oe, e.g. 21/20)
(b) A1 θ = awrt 46.4 (degrees)
(b) Answer: θ = awrt 46.4 degrees (tan(θ) = 1.05)
(c) M1 uses Pythagoras: magnitude = √N_A² + F_A²
(c) A1 = √250² + 100² = √72500
(c) A1 = awrt 269 (N) cao
(c) Answer: awrt 269 N (= 50*√29 N exactly)

Question 8

(a) M1 radius = 3 cm (half of the 6 cm diameter); uses area = 0.5*π*r²
(a) A1 = 0.5*π*9 = 4.5*π cm² cso
(a) Answer: 4.5*π cm² (shown)
(b) M1 identifies the rectangle's area (60 cm²) at distance 5 cm from AB, and the semicircle's area (4.5*π) at distance 10 + 4*3/(3*π) from AB
(b) A1 correct areas and positions listed
(b) M1 combines: xbar = (60*5 + 4.5*π*(10 + 4/π)) / (60 + 4.5*π)
(b) A1 xbar = awrt 6.20 (cm) cao
(b) Answer: awrt 6.20 cm from AB

Question 9

(a) B1 area of the hole = 12 cm², with its centroid at (4, 3.5)
(a) B1 remaining area = 96 - 12 = 84 cm²
(a) M1 xbar = (96*6 - 12*4)/84
(a) A1 xbar = 44/7 (awrt 6.29)
(a) M1 ybar = (96*4 - 12*3.5)/84
(a) A1 ybar = 57/14 (awrt 4.07)
(a) Answer: (44/7, 57/14) cm, i.e. awrt (6.29, 4.07) cm
(b) B1 finds the horizontal and vertical displacement of the centre of mass from A: 40/7 cm and 57/14 cm (or equivalent decimals)
(b) M1 tan(angle) = (40/7) / (57/14), using AB as the local vertical direction
(b) A1 angle = awrt 54.5 (degrees) cao
(b) Answer: awrt 54.5 degrees

Question 10

(a) B1 CA = √(6a)²+(8a)² = √100a² = 10a, using Pythagoras' theorem, cso
(a) Answer: CA = 10a (shown)
(b) B1 masses (lengths) of the three rods are proportional to 6a, 8a, 10a, with total 24a
(b) M1 identifies the midpoint of each rod: AB -> (0,3a), BC -> (4a,0), CA -> (4a,3a)
(b) A1 all three midpoints correct
(b) M1 xbar = (6a*0 + 8a*4a + 10a*4a)/24a
(b) A1 xbar = 3a
(b) M1 ybar = (6a*3a + 8a*0 + 10a*3a)/24a
(b) A1 ybar = 2a
(b) Answer: (3a, 2a)
(c) B1 identifies the horizontal and vertical displacement of G from A: 3a horizontally and 6a-2a=4a vertically
(c) M1 sets up tan(angle) = horizontal/vertical = 3a/4a
(c) A1 = 3/4 (oe)
(c) A1 angle = awrt 36.9 (degrees) cao
(c) Answer: awrt 36.9 degrees

Question 11

(a) B1 AC = √9²+12² = √225 = 15, using Pythagoras' theorem, cso
(a) Answer: AC = 15 cm (shown)
(b) M1 uses the standard result that the centre of mass of a uniform triangular lamina is the average of the coordinates of its vertices
(b) A1 xbar = (0+0+12)/3 = 4
(b) A1 ybar = (9+0+0)/3 = 3
(b) Answer: (4, 3) cm
(c) B1 recognises that AG becomes vertical, and that AB is initially vertical in the lamina's own frame, so the required angle is between AG and AB
(c) M1 finds the horizontal and vertical displacement of G from A
(c) A1 4 cm horizontally and 6 cm vertically (or magnitude 2*√13 cm)
(c) M1 tan(angle) = 4/6 (oe)
(c) A1 angle = awrt 33.7 (degrees) cao
(c) Answer: awrt 33.7 degrees

Question 12

(a) M1 identifies the toppling condition: the vertical line through the centre of mass passes through the lower edge when tan(θ) = half-width/half-height
(a) A1 tan(θ) = 0.6/1 = 0.6, using half-width 0.6 m and half-height 1 m
(a) M1 θ = arctan(0.6)
(a) A1 θ = awrt 31.0 (degrees) cao
(a) Answer: awrt 31.0 degrees (tan(θ) = 0.6 exactly)
(b) M1 resolves perpendicular to the plane: R = Mg*cos(θ)
(b) A1 resolves along the plane at limiting equilibrium: F = Mg*sin(θ)
(b) M1 uses F = μ*R to get tan(θ) = μ = 0.4
(b) A1 θ = awrt 21.8 (degrees) cao
(b) Answer: awrt 21.8 degrees (tan(θ) = 0.4 exactly)
(c) M1 compares the two critical angles, 21.8 and 31.0 degrees, noting θ increases continuously from 0 so equilibrium first breaks at the smaller of the two
(c) A1 concludes the cuboid slides first, since 21.8 < 31.0, at θ = awrt 21.8 degrees
(c) Answer: the cuboid slides first, at θ = awrt 21.8 degrees
(d) M1 weight = Mg = 45*9.8 = 441 N
(d) M1 R = Mg*cos(θ) at θ = 21.8 degrees
(d) A1 R = 441*cos(21.8) = awrt 409 (N) cao
(d) Answer: awrt 409 N

Mechanics: Moments and Statics Depth - Worksheets, Questions and Revision