Questions on moments and equilibrium

Question 1

Define the moment of a force about a point, and state its SI unit.

Accepted Answer

Moment = force × perpendicular distance from the pivot to the line of action of the force (1). Unit: newton metre, N m (1).

Question 2

A force of 15 N acts at a perpendicular distance of 0.40 m from a pivot. Calculate the moment of the force.

Accepted Answer

Moment = 15 × 0.40 (1) Moment = 6.0 N m (1)

Question 3

State the principle of moments for a body in equilibrium.

Accepted Answer

For a body in equilibrium, the sum of the clockwise moments about any point (1) equals the sum of the anticlockwise moments about that point (1).

Question 4

A cyclist presses vertically downwards on a pedal with a force of 300 N at the moment when the crank is horizontal. The crank is 0.17 m long. Calculate the moment of the force about the crank axle.

Accepted Answer

Moment = 300 × 0.17 (1) = 51 N m (1).

Question 5

State the two conditions that must both be satisfied for a rigid body to be in equilibrium.

Accepted Answer

The resultant force on the body is zero (1). The resultant moment about any point is zero (1).

Question 6

Explain why a force has no turning effect about a pivot when its line of action passes through the pivot.

Accepted Answer

The perpendicular distance from the pivot to the line of action is zero, so the moment F × d is zero (1).

Question 7

A uniform metre rule is pivoted at its centre. A 2.0 N weight hangs 0.20 m to the left of the pivot. Calculate the distance from the pivot at which a 4.0 N weight must hang on the right-hand side to balance the rule.

Accepted Answer

Clockwise moment = anticlockwise moment: 2.0 × 0.20 = 4.0 × d (1) d = (2.0 × 0.20)/4.0 (1) d = 0.10 m to the right of the pivot (1)

Question 8

Two children sit on a see-saw pivoted at its centre. One of weight 300 N sits 1.2 m from the pivot. Calculate the distance from the pivot at which a child of weight 450 N must sit on the other side to balance the see-saw.

Accepted Answer

Clockwise moment = anticlockwise moment: 300 × 1.2 = 450 × d (1) d = (300 × 1.2)/450 (1) d = 0.80 m (1)

Question 9

A uniform beam of weight 120 N and length 4.0 m rests horizontally on two supports, one at each end (A on the left, B on the right). A load of 200 N is placed 1.0 m from A. Calculate the upward force provided by each support.

Accepted Answer

Taking moments about A: R_B × 4.0 = 120 × 2.0 + 200 × 1.0 (1) R_B = 110 N (1) Vertically: R_A = (120 + 200) − 110 (1) R_A = 210 N (1)

Question 10

Two antiparallel forces of 8.0 N act on a bar, separated by a perpendicular distance of 0.25 m. State what this pair of forces is called and calculate its turning effect.

Accepted Answer

This is a couple (1) Its moment (torque) = one force × the separation = 8.0 × 0.25 (1) Moment = 2.0 N m (1)

Question 11

A mechanic pulls on the end of a spanner of length 240 mm with a force of 90 N. The force acts at 55° to the line of the spanner. Calculate the moment of the force about the nut.

Accepted Answer

Perpendicular distance = 0.240 sin 55° (240 mm converted to m) (1). Moment = 90 × 0.240 × sin 55° (1) = 17.7 N m (1).

Question 12

A uniform metre rule has a mass of 120 g. It balances horizontally on a knife-edge pivot at its 35.0 cm mark when an object is hung from the 5.0 cm mark. Taking g = 9.81 N kg⁻¹, calculate the mass of the object.

Accepted Answer

The rule is uniform, so its weight 1.18 N acts at the 50.0 cm mark, 0.150 m from the pivot (1). Moments about the pivot: W × 0.300 = 1.18 × 0.150 (1), so W = 0.59 N (1). Mass = W/g = 60 g (0.060 kg) (1).

Question 13

The blades of a pair of bolt cutters are 45 mm from the pivot and the handles are gripped 0.55 m from the pivot. A student squeezes the handles with a force of 120 N. Calculate the force the blades exert on a bolt.

Accepted Answer

Moments about the pivot: F × 0.045 = 120 × 0.55 (1), so F = 66/0.045 (1) = 1467 N ≈ 1.5 kN (1).

Question 14

Distinguish between the centre of mass and the centre of gravity of an object, and state the condition under which the two lie at the same point.

Accepted Answer

The centre of mass is the single point at which the whole mass of the body may be taken to act (1). The centre of gravity is the single point at which the whole weight of the body may be taken to act (1). The two coincide wherever the gravitational field is uniform over the body, which holds for every object met in this course; they separate only for a body large enough for g to differ measurably from one end of it to the other (1).

Question 15

A uniform rod of weight 40 N and length 2.0 m is hinged at one end and held horizontal by a vertical cable attached at the other end. Calculate the tension in the cable and the vertical force exerted by the hinge on the rod.

Accepted Answer

Moments about the hinge: T × 2.0 = 40 × 1.0 (1) T = 20 N (1) Vertically: hinge force = 40 − 20 (1) Hinge force = 20 N upwards (1)

Question 16

A non-uniform beam balances on a pivot placed 0.40 m from its left-hand end when a 2.0 N weight hangs from that end. The centre of mass of the beam is 0.65 m from the left-hand end. Calculate the weight of the beam.

Accepted Answer

The centre of mass is (0.65 − 0.40) = 0.25 m from the pivot (1) Moments about the pivot: 2.0 × 0.40 = W × 0.25 (1) W = 0.80 / 0.25 (1) W = 3.2 N (1)

Question 17

A door handle is fitted on the edge of the door furthest from the hinge. Explain, in terms of moments, why this makes the door easier to open.

Accepted Answer

The moment of the applied force equals force × perpendicular distance from the hinge (1). Placing the handle far from the hinge maximises that distance (1), so a smaller force produces the moment needed to open the door (1).

Question 18

A uniform scaffold plank of weight 150 N and length 3.0 m rests horizontally on two supports, one at each end, A and B. Support B is damaged and must not carry a force greater than 600 N. A decorator of weight 720 N steps on at A and walks towards B. Deduce whether the decorator can walk all the way to B and, if not, determine the greatest distance from A they can safely reach.

Accepted Answer

Taking moments about A with the decorator a distance x from A: R_B × 3.0 = 150 × 1.5 + 720x (1). With x = 3.0 m, R_B = 795 N, which exceeds 600 N, so the decorator cannot reach B (1). Setting R_B = 600 N: 600 × 3.0 = 225 + 720x (1), so x = 2.2 m from A (1).

Question 19

A uniform horizontal bar of weight 35 N and length 1.2 m is hinged to a wall at one end. A shop sign of weight 90 N hangs from the bar 1.0 m from the hinge. The bar is held by a chain joining its outer end to the wall above the hinge; the chain makes an angle of 30° with the bar. Calculate the tension in the chain.

Accepted Answer

The component of the tension perpendicular to the bar is T sin 30°, acting 1.2 m from the hinge (1). Moments about the hinge: T sin 30° × 1.2 = 35 × 0.60 + 90 × 1.0 (1), so 0.60T = 111 N m (1), giving T = 185 N (1).

Question 20

You are given a metre rule, a knife-edge pivot and a single 50 g mass. Describe how you would use them to determine the mass of the rule, using the principle of moments. State the measurements you would take and show how the mass of the rule is calculated from them.

Accepted Answer

Place the rule on the pivot away from its centre and position the 50 g mass on the shorter side, adjusting until the rule balances horizontally (1). Measure the distance d₁ from the pivot to the centre of the 50 g mass, and the distance d₂ from the pivot to the 50.0 cm mark (1). The weight of a uniform rule acts at its centre of mass, the 50.0 cm mark (1). By the principle of moments: 0.050g × d₁ = mg × d₂, so m = 0.050 d₁/d₂ (g cancels) (1). Repeat with the pivot at different positions and average the values of m to reduce random error (1).

Question 21

An irregular sheet of card is to have its centre of gravity found using a pin held in a clamp, and a plumb line. Describe the method, and explain why the card must be free to swing on the pin.

Accepted Answer

Hang the card from a small hole near its edge so that it swings freely on the pin, and let it settle (1). Hang the plumb line from the same pin and rule the vertical line it marks across the card; the centre of gravity lies somewhere on that line (1). Repeat from a second hole well away from the first: the centre of gravity is where the two lines cross, and a third hole is a check that its line passes through the same crossing (1). The card must swing freely because it can only come to rest with its centre of gravity vertically below the pin; in any other position the weight has a moment about the pin and turns the card further (1).

Question 22

A uniform beam AB of length 6.0 m and weight 400 N rests horizontally on two supports, P and Q. Three loads hang from the beam. The beam's own weight is not marked on the diagram. Determine the magnitude of the force exerted on the beam by support Q, and then the magnitude of the force exerted by support P.

Accepted Answer

Taking moments about P removes the unknown force at P from the equation, since its line of action passes through that point (1)
The 300 N load hangs 0.50 m on the far side of P, so its moment about P acts in the opposite sense to the moments of the other three downward forces, and the uniform beam's 400 N weight acts at the 3.0 m mark, 2.0 m from P (1)
Q × 4.0 + 300 × 0.50 = 600 × 1.0 + 400 × 2.0 + 100 × 3.5 = 1750 N m, so 4.0Q = 1600 and Q = 400 N (1)
The beam is also in vertical equilibrium, so P = (300 + 600 + 400 + 100) − 400 = 1000 N (1)