Mechanics: Moments (Part 2) - Worksheets, Questions and Revision

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A-Level · Mechanics

M3B Mechanics: Moments (Part 2)

EDEXCEL 9MA0 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A gardener uses a wheelbarrow to move garden waste. When lifting the wheelbarrow, it is modelled as pivoting about the axle of its single front wheel. The load of garden waste has weight 250 N and acts at a horizontal distance of 0.4 m from the wheel axle. The gardener applies a vertical lifting force at the handles, which are a horizontal distance of 1 m from the wheel axle. The wheelbarrow's own frame is light, so its weight may be ignored.

(a)Calculate the moment of the 250 N load about the wheel axle, stating the appropriate units.(2)

(b)Find the minimum vertical lifting force the gardener must apply at the handles for the wheelbarrow to begin to lift off the ground (other than at the wheel).(2)

(Total for Question 1 is 4 marks)

A cyclist applies a force of 400 N to a bicycle pedal, at an angle of 60 degrees to the crank arm, which has length 0.17 m. The bottom bracket of the bicycle acts as a fixed pivot for the crank.

(a)Calculate the perpendicular distance from the pivot to the line of action of the 400 N force, giving your answer to 3 significant figures.(2)

(b)Hence calculate the moment (turning effect) of the force about the pivot, stating the appropriate units, giving your answer to 3 significant figures.(2)

(c)The cyclist repeats the push, this time applying a force of the same magnitude perpendicular to the crank (that is, at 90 degrees to the crank). Find the percentage increase in the moment compared to part (b), giving your answer to 3 significant figures.(2)

(Total for Question 2 is 6 marks)

A market stallholder's uniform wooden display plank AB has length 3 m and weight 120 N. The plank rests horizontally in equilibrium on two smooth trestle supports, one at C, 0.5 m from A, and one at D, 0.5 m from B. Two display boxes rest on the plank; the table below gives each box's weight and its distance from A.

Item	Weight (N)	Distance from A (m)
Box P	50	1.0
Box Q	30	2.0

(a)By taking moments about C, show that the reaction at D, R_D, is 95 N.(4)

(b)Find the reaction at C, R_C.(2)

(c)Box Q is now repositioned along the plank (its weight is unchanged) so that the reactions at C and D become equal. Find the new distance of Box Q from A, giving your answer to 3 significant figures.(3)

(Total for Question 3 is 9 marks)

A uniform beam AB has length 2 m and weight 50 N. The beam rests horizontally on a pivot at a point M, 0.6 m from A (M is not the midpoint of the beam). A vertical force of magnitude P is applied downward at A, and a counterweight of weight 80 N hangs from B. The beam remains horizontal and in equilibrium.

(a)Explain why the weight of the beam does contribute a moment about M in this situation, unlike in a case where the pivot is at the beam's midpoint.(1)

(b)By taking moments about M, find P.(4)

(c)Find the reaction at the pivot M.(2)

(d)The counterweight is repositioned so that P can be reduced to 150 N, with the beam's weight and the pivot position unchanged. Find the new distance of the counterweight from the pivot M.(3)

(Total for Question 4 is 10 marks)

A non-uniform curtain pole AB has length 2.4 m. It is suspended horizontally by two vertical wires, one attached at A and the other attached at a point 1.6 m from A. The centre of mass of the pole is 1.0 m from A. The tension in the wire at A is 36 N; let T newtons be the tension in the other wire.

(a)By resolving vertically and taking moments about A, form two equations connecting the weight of the pole, W, and T, and hence find the values of W and T.(5)

(b)Verify your answer to part (a) by taking moments about the point where the second wire is attached.(2)

(c)State, giving a reason, whether the tension in the wire at A would be larger or smaller than 36 N if the pole were uniform instead of non-uniform (with the same length, weight and wire positions).(2)

(Total for Question 5 is 9 marks)

A uniform plank footbridge PQ over a garden pond has length 5 m and weight 300 N. It rests horizontally on two supports, one at P and one at R, where PR = 4 m (so R is 1 m from Q). A garden ornament of weight 60 N sits on the plank at a point 3 m from P. A student's attempt to find the reaction at R by taking moments about P is shown below:

Moments about P: R x 4 = 300 x 2.5 + 60 x 4
R = (750 + 240) / 4
R = 247.5 N

(a)Identify the error in the student's working, and state the correct distance that should have been used.(2)

(b)Hence find the correct value of R.(3)

(c)Find the reaction at P.(2)

(Total for Question 6 is 7 marks)

A shop awning arm AB has length 1.6 m and weight 30 N. It is smoothly hinged to the shop wall at A and held horizontal by a rigid strut CB, where C is a fixed point on the wall vertically below A, the strut making an angle of 42 degrees with the arm. The strut exerts a force of magnitude S on the arm at B, directed along CB. A hanging sign of weight 22 N is attached to the arm at B.

(a)By taking moments about A, find the force S in the strut, giving your answer to 3 significant figures.(4)

(b)Find the magnitude of the horizontal and vertical components of the force exerted on the arm by the hinge at A.(4)

(c)Hence find the magnitude of the resultant force exerted by the hinge on the arm, and the angle this resultant makes with the arm, both to 3 significant figures.(3)

(Total for Question 7 is 11 marks)

A garden centre displays a uniform wooden potting bench, modelled as a rigid plank AB of length 2.4 m and weight 180 N, resting on two smooth support posts, one at C, 0.3 m from A, and one at D, 1.8 m from A. A decorative stone urn of weight 40 N sits at end A. A sack of compost of weight w newtons is placed at end B. The bench remains horizontal and rests only on the two posts; it will tip about D (lifting off the post at C) if w becomes too large.

(a)Explain briefly why, as w increases, the bench would tip about D rather than about C.(1)

(b)By taking moments about C, show that R_D = 100 + 1.4w (in newtons), where R_D is the reaction at post D.(4)

(c)Find, in terms of w, an expression for R_C.(2)

(d)Find the maximum value of w for which the bench remains in equilibrium without tipping.(3)

(Total for Question 8 is 10 marks)

A uniform ladder AB has length 5 m and weight 220 N. It rests with end A on rough horizontal ground, where the coefficient of friction is 0.3, and end B against a smooth vertical wall, making an angle of 65 degrees with the ground. A steady horizontal wind force of magnitude P acts at the midpoint of the ladder, blowing in the direction from the wall towards A (tending to push the ladder away from the wall). The ladder remains in equilibrium and is on the point of slipping at A.

(a)By resolving vertically, find the normal reaction at A, N_A. (The wind force is horizontal, so does not affect this.)(2)

(b)Hence write down the friction force at A, given that the ladder is on the point of slipping there.(2)

(c)By taking moments about A, show that N_B = 110/tan(65 degrees) - 0.5P, where N_B is the normal reaction at the wall (both terms in newtons).(4)

(d)By resolving horizontally, find the value of P, giving your answer to 3 significant figures.(3)

(Total for Question 9 is 11 marks)

A shop's hanging flower-basket bracket consists of a uniform rod AB of length 0.9 m and weight 18 N, smoothly hinged to the shop wall at A. The rod is held horizontal by a light string attached to the rod at B and to a point on the wall above A, the string making an angle θ with the rod. A basket of weight 45 N hangs from B. The string is rated to withstand a maximum tension of 150 N before it risks breaking.

(a)By taking moments about A, show that the tension in the string is given by T = 54/sin(θ) (in newtons).(4)

(b)Given that the string must not exceed its rated maximum tension of 150 N, find the minimum angle θ, to 1 decimal place, at which the string may be attached.(3)

(c)The basket is replaced with a heavier one of weight 70 N (the rod's weight is unchanged). Find the new minimum safe angle θ, to 1 decimal place.(3)

(Total for Question 10 is 10 marks)

A uniform ladder AB has length 4 m and weight 180 N. The foot A rests on rough horizontal ground, where the coefficient of friction is 0.28, and the ladder is on the point of slipping at A. The top B rests against a wall which is also rough, at an angle of 60 degrees to the ground.

(a)By resolving horizontally, and using the fact that the ladder is on the point of slipping at A (so the friction there is limiting, F_A = 0.28 N_A), show that N_B = 0.28 N_A, where N_B is the normal reaction at the wall.(3)

(b)By resolving vertically, write down an equation connecting N_A, F_B (the friction force at the wall) and the weight of the ladder.(2)

(c)By taking moments about A, and using your equations from parts (a) and (b), find the values of N_A, N_B and F_B, giving each to 3 significant figures.(4)

(d)Hence find the minimum coefficient of friction between the ladder and the wall required for the ladder to remain in equilibrium in this position, giving your answer to 3 significant figures.(2)

(Total for Question 11 is 11 marks)

A uniform ladder AB has length 5 m and weight 180 N. The foot A rests on rough horizontal ground, where the coefficient of friction between the ladder and the ground is 0.4, and the top B rests against a smooth vertical wall. The ladder makes an angle of 59 degrees with the horizontal ground. A decorator of weight w newtons climbs the ladder to a point 4 m from A.

(a)By resolving vertically, find N_A, the normal reaction at the ground, in terms of w.(2)

(b)By taking moments about A, show that the normal reaction at the wall is given by N_B = 54.1 + 0.481w, to 3 significant figures, where N_B is in newtons.(4)

(c)Given that the ladder is on the point of slipping when w takes its maximum possible value, find this maximum value of w, giving your answer to 3 significant figures.(4)

(d)Comment on whether it would be safe for a decorator of weight 800 N to climb to this same point on the ladder.(2)

(Total for Question 12 is 12 marks)

A uniform ladder of length L and weight W rests with one end A on rough horizontal ground, where the coefficient of friction is μ, and the other end B against a smooth vertical wall. The ladder makes an angle θ with the horizontal ground, and is on the point of slipping.

(a)Show that μ = 1/(2 tan(θ)).(5)

(b)Explain what happens to the minimum coefficient of friction required as θ approaches 90 degrees, and as θ approaches 0 degrees, and explain briefly why each result makes physical sense.(3)

(c)A particular ladder requires a minimum coefficient of friction of 0.35 to remain in equilibrium against a smooth wall. Find, to 1 decimal place, the smallest angle θ at which this ladder can safely be placed.(3)

(Total for Question 13 is 11 marks)

Mark scheme · M3B Mechanics: Moments (Part 2)

Question 1

(a) M1 moment = 250 x 0.4 oe
(a) A1 100 N m cao (units required)
(a) Answer: 100 N m
(b) M1 takes moments about the wheel: F x 1 = 250 x 0.4 oe
(b) A1 F = 100 N cao
(b) Answer: 100 N

Question 2

(a) M1 d = 0.17 x sin(60 degrees) oe
(a) A1 awrt 0.147 (m)
(a) Answer: 0.147 m (3 s.f.)
(b) M1 moment = 400 x their (a), or 400 x 0.17 x sin(60 degrees)
(b) A1 awrt 58.9 N m (units required), ft their (a)
(b) Answer: 58.9 N m (3 s.f.)
(c) M1 new moment = 400 x 0.17 (= 68 N m)
(c) A1 awrt 15.5% increase, ft their (b)
(c) Answer: 15.5% increase (3 s.f.)

Question 3

(a) M1 moments equation about C attempted, with all four terms present (plank weight, Box P, Box Q, R_D)
(a) A1 correct unsimplified equation: R_D x 2.0 = 120 x 1.0 + 50 x 0.5 + 30 x 1.5 oe
(a) dM1 rearranges to make R_D the subject
(a) A1 R_D = 95 N cso
(a) Answer: R_D = 95 N
(b) M1 R_C = (120 + 50 + 30) - 95 (vertical equilibrium), ft their (a)
(b) A1 R_C = 105 N cao
(b) Answer: R_C = 105 N
(c) M1 recognises that equal reactions require R_C = R_D = 100 N (half the total weight of 200 N)
(c) M1 forms a moments equation about C with Box Q's distance d as the unknown: 100 x 2.0 = 120 x 1.0 + 50 x 0.5 + 30 x (d - 0.5) oe
(c) A1 d = awrt 2.33 m cao
(c) Answer: 2.33 m from A (3 s.f.)

Question 4

(a) B1 because M is not the midpoint (centre of mass) of the beam, so the weight's line of action does not pass through M, giving a nonzero perpendicular distance and hence a nonzero moment about M
(a) Answer: The weight's line of action does not pass through M, since M is not the centre of mass, so it has a nonzero moment about M
(b) M1 identifies the distance of each force from M (P at 0.6 m from M; the beam's weight at 0.4 m from M; the counterweight at 1.4 m from M)
(b) A1 correct unsimplified equation: P x 0.6 = 50 x 0.4 + 80 x 1.4 oe
(b) dM1 rearranges to make P the subject
(b) A1 P = 220 N cao
(b) Answer: P = 220 N
(c) M1 R = P + 50 + 80 (vertical equilibrium), ft their P
(c) A1 R = 350 N cao
(c) Answer: 350 N
(d) M1 new moments equation: 150 x 0.6 = 50 x 0.4 + 80 x d
(d) A1 90 = 20 + 80d oe
(d) A1 d = 0.875 m cao
(d) Answer: 0.875 m from M

Question 5

(a) M1 vertical equilibrium equation formed: 36 + T = W oe
(a) M1 moments equation about A formed: T x 1.6 = W x 1.0 oe
(a) dM1 eliminates one variable (e.g. substitutes W = 36 + T) and solves
(a) A1 T = 60 N cao
(a) A1 W = 96 N cao
(a) Answer: W = 96 N, T = 60 N
(b) M1 alternative moments equation formed: 36 x 1.6 = W x (1.6 - 1.0), using their W
(b) A1 both sides equal 57.6 (N m), confirming the values found in part (a)
(b) Answer: Consistent: both sides equal 57.6 N m
(c) B1 correct statement: smaller
(c) B1 valid reason: a uniform pole's centre of mass would be at its midpoint (1.2 m from A), further from A than the given 1.0 m, so more weight is supported by the wire nearer the centre of mass, leaving less tension needed at A
(c) Answer: Smaller than 36 N

Question 6

(a) B1 identifies that the student used 4 m (the distance PR) as the lever arm for the ornament's weight, instead of the ornament's own distance from P
(a) B1 states the correct distance is 3 m (the ornament's actual position from P), so the correct moment term is 60 x 3, not 60 x 4
(a) Answer: The ornament's distance from P should be 3 m, not 4 m
(b) M1 correct moments equation: R x 4 = 300 x 2.5 + 60 x 3
(b) A1 R x 4 = 930 oe
(b) A1 R = 232.5 N cao
(b) Answer: R = 232.5 N
(c) M1 R_P = 300 + 60 - 232.5 (vertical equilibrium), ft their (b)
(c) A1 R_P = 127.5 N cao
(c) Answer: 127.5 N

Question 7

(a) M1 moments equation about A with all three terms: S sin(42 degrees) x 1.6 = 30 x 0.8 + 22 x 1.6 oe
(a) A1 correct unsimplified equation
(a) dM1 rearranges to make S the subject
(a) A1 awrt 55.3 N
(a) Answer: 55.3 N (3 s.f.)
(b) M1 vertical equilibrium: Y + S sin(42 degrees) = 30 + 22 oe
(b) A1 substitutes S sin(42 degrees) = 37 (exact, from part a) to get Y = 15 N cao
(b) M1 horizontal equilibrium: X = S cos(42 degrees), ft their S
(b) A1 awrt 41.1 N
(b) Answer: Vertical: 15 N; Horizontal: 41.1 N (3 s.f.)
(c) M1 magnitude = √X² + Y², using their X and Y
(c) A1 awrt 43.7 N (ft)
(c) A1 angle: awrt 20.1 degrees (ft), e.g. from tan(angle) = Y/X
(c) Answer: 43.7 N, at 20.1 degrees to the arm

Question 8

(a) B1 because the sack (at B) and part of the plank's own weight act beyond D, so increasing w increases the tendency to rotate the bench about D, lifting end A (and the reaction at C) off post C
(a) Answer: The bench would tip about D, lifting off at C
(b) M1 moments equation about C attempted, with all four terms present
(b) A1 correct unsimplified equation: R_D x 1.5 = 180 x 0.9 + w x 2.1 - 40 x 0.3 oe
(b) dM1 rearranges to make R_D the subject
(b) A1 R_D = 100 + 1.4w cso
(b) Answer: R_D = 100 + 1.4w
(c) M1 R_C + R_D = 180 + 40 + w (vertical equilibrium), ft their R_D
(c) A1 R_C = 120 - 0.4w cao
(c) Answer: R_C = 120 - 0.4w
(d) M1 sets R_C ≥ 0 and solves: 120 - 0.4w ≥ 0
(d) A1 w ≤ 300
(d) B1 maximum value of w is 300 N, ft their inequality
(d) Answer: 300 N

Question 9

(a) M1 vertical equilibrium: N_A = 220 (the wall's reaction and the wind force are both horizontal)
(a) A1 N_A = 220 N cao
(a) Answer: 220 N
(b) M1 limiting friction: F_A = 0.3 x N_A, using their N_A
(b) A1 F_A = 66 N cao
(b) Answer: 66 N
(c) M1 moments equation about A with all three terms: N_B x 5 sin(65 degrees) + P x 2.5 sin(65 degrees) = 220 x 2.5 cos(65 degrees) oe
(c) A1 correct unsimplified equation
(c) dM1 divides by 5 sin(65 degrees) and rearranges to isolate N_B
(c) A1 N_B = 110/tan(65 degrees) - 0.5P cso
(c) Answer: N_B = 110/tan(65 degrees) - 0.5P
(d) M1 horizontal equilibrium: F_A = N_B + P, using their F_A and their expression for N_B
(d) M1 substitutes and rearranges: 66 = 110/tan(65 degrees) - 0.5P + P
(d) A1 awrt 29.4 (N)
(d) Answer: 29.4 N (3 s.f.)

Question 10

(a) M1 moments equation about A: T sin(θ) x 0.9 = 18 x 0.45 + 45 x 0.9 oe
(a) A1 correct unsimplified equation
(a) dM1 divides by 0.9 and rearranges to make T the subject
(a) A1 T = 54/sin(θ) cso
(a) Answer: T = 54/sin(θ)
(b) M1 sets T ≤ 150 and forms the inequality 54/sin(θ) ≤ 150
(b) A1 sin(θ) ≥ 0.36 oe
(b) A1 θ ≥ 21.1 degrees (1 d.p.) cao
(b) Answer: 21.1 degrees (1 d.p.)
(c) M1 new constant found: T sin(θ) x 0.9 = 18 x 0.45 + 70 x 0.9, i.e. T sin(θ) = 79
(c) A1 sin(θ) ≥ 79/150 = 0.5267 oe
(c) A1 θ ≥ 31.8 degrees (1 d.p.) cao
(c) Answer: 31.8 degrees (1 d.p.)

Question 11

(a) M1 horizontal equilibrium: N_B = F_A
(a) A1 F_A = 0.28 N_A stated (limiting friction)
(a) A1 N_B = 0.28 N_A cso
(a) Answer: N_B = 0.28 N_A
(b) M1 vertical equilibrium attempted
(b) A1 N_A + F_B = 180 oe
(b) Answer: N_A + F_B = 180
(c) M1 moments equation about A formed: N_B x 4 sin(60 degrees) + F_B x 4 cos(60 degrees) = 180 x 2 cos(60 degrees) oe
(c) A1 correct unsimplified equation, e.g. N_B tan(60 degrees) + F_B = 90 (after dividing by 4 cos(60 degrees))
(c) dM1 substitutes N_B = 0.28 N_A and F_B = 180 - N_A, and solves for N_A
(c) A1 N_A = awrt 175 N, N_B = awrt 48.9 N, F_B = awrt 5.25 N (ft)
(c) Answer: N_A = 175 N, N_B = 48.9 N, F_B = 5.25 N (all 3 s.f.)
(d) M1 μ = F_B / N_B (using their F_B and N_B)
(d) A1 awrt 0.107
(d) Answer: 0.107 (3 s.f.)

Question 12

(a) M1 vertical equilibrium: N_A = 180 + w (the wall is smooth, so its reaction is horizontal)
(a) A1 N_A = 180 + w oe
(a) Answer: N_A = 180 + w
(b) M1 moments equation about A with all three terms: N_B x 5 sin(59 degrees) = 180 x 2.5 cos(59 degrees) + w x 4 cos(59 degrees) oe
(b) A1 correct unsimplified equation
(b) dM1 divides by 5 sin(59 degrees) and rearranges
(b) A1 N_B = 54.1 + 0.481w cso (awrt)
(b) Answer: N_B = 54.1 + 0.481w
(c) M1 limiting friction: F_A = 0.4 x N_A = 0.4(180 + w)
(c) M1 horizontal equilibrium F_A = N_B, set equal to the expression from part (b), using unrounded coefficients for accuracy
(c) dM1 rearranges to make w the subject
(c) A1 awrt 222 (N)
(c) Answer: w = 222 N (3 s.f.)
(d) B1 no, it would not be safe / the ladder would slip
(d) B1 valid reason: 800 N exceeds the maximum weight of 222 N found in part (c), ft their (c)
(d) Answer: No, unsafe, since 800 N > 222 N