Moments (Cambridge International IGCSE Physics): Exam Questions

Define the term moment.

A uniform seesaw is in equilibrium with a box placed on each side.

The box on the left has an anticlockwise moment of 150 N m about the pivot.

The box on the right has weight W.

Fig. 1.1

State the clockwise moment due to the box on the right.

Extract:

The box on the left has an anticlockwise moment of 150 N m about the pivot. The box on the right has weight W. State the clockwise moment due to the box on the right:

• 150 N m [1 mark] Distance from the box to the pivot, d = 1.5 m

Calculate the weight W of the box on the right side of the seesaw.

Extract:

A uniform seesaw is in equilibrium with a box placed on each side. The box on the left has an anticlockwise moment of 150 N m about the pivot. The box on the right has weight W. The seesaw is in equilibrium. The clockwise and anticlockwise moments are equal - the resultant moment is zero.

The box on the left hand side of the seesaw is now removed. Explain the subsequent motion of the seesaw.

A mechanic uses a spanner to loosen a bolt on a car engine.

(i) The mechanic applies a force of 80 N at a distance of 0.25 m from the centre of the bolt.

Calculate the moment of this force about the centre of the bolt. Include a unit.

moment = ........................................................... [4]

(ii) The force applied by the mechanic produces a turning effect (moment) about the centre of the bolt.

Describe another example of using the turning effect of a force.

The mechanic tries to use the same spanner to loosen another bolt on the car engine. He applies the same force of 80 N at a distance of 0.25 m from the centre of the bolt.

The turning effect produced is not enough to loosen this bolt.

Describe how the mechanic can increase the turning effect without increasing the force.

A lamp is attached to a wall, as shown in Fig. 4.1.

Calculate the moment of the lamp about the pivot. Give the unit.

moment = ..........................................................

Fig. 4.1 shows a metal triangle suspended from a thread.

Complete the sentence. Choose the correct word or phrase from the box.

The metal triangle will come to rest with its centre of gravity directly ................................. the point of suspension.

A student finds the centre of gravity of a shape made of thin card. Fig. 4.2 shows the equipment.

Describe how the student finds the centre of gravity of the card. Choose from these sentences.

1. A line is drawn on the card showing the position of the string.
2. A pin held in a clamp is put through the hole in the card.
3. The centre of gravity is where the lines cross on the card.
4. The process is repeated using holes near the other two edges.

Complete the flow chart. Write the letter for the correct sentence in each box.

Fig. 4.1 shows a truck lifting a heavy load.

(i) The truck is stationary. Identify the quantities that determine the work done as it lifts the load.

Tick the box next to each correct quantity.

square distance

square force

square time

(ii) Draw a ring around the unit for work done from the list.

Identify the quantities that determine the power of the truck.

Tick the box next to each correct quantity.

square energy transferred

square temperature

square time

Extract:

The truck has a pivot near the front wheel. Fig. 4.2 represents the pivot and the vertical forces acting on the truck.

The truck is in equilibrium.

• Load causing anticlockwise moment = 30 000 N
• Distance of 30 000 N from pivot = 1.5 m
• Distance of load from pivot = 1.0 m

State the principle of moments:

total clockwise moments equals total anticlockwise moments

moment = force × perpendicular distance from the pivot

load × 1.0 = 30 000 × 1.5

load = (30 000 × 1.5) / 1.0

load = 45 000 N

The truck has a pivot near the front wheel. Fig. 4.2 represents the pivot and the vertical forces acting on the truck.

The truck is in equilibrium.

Calculate the load.

load = ..................................................... N

Extract:

For a regular object, its centre of gravity is at the centre of the object.

Stability is all about how close a line drawn from an object's centre of gravity (CoG) is to its pivot. If a line from its centre of gravity is further ahead than the pivot, it is unstable.

Fig. 4.3 shows another truck lifting a pile of identical bricks.

(i) On Fig. 4.3, draw a cross to indicate the centre of gravity of the pile of bricks.

(ii) The truck can tilt the pile of bricks backwards, as shown in Fig. 4.4.

Explain how tilting the pile of bricks backwards makes the truck more stable.

A 50 cm rule is balanced at its mid-point. A force of 8.0 N acts at a distance of 10 cm from one end of the rule.

Fig. 2.1 shows the arrangement.

Calculate the moment of the 8.0 N force about the pivot. Give the unit.

moment = ............................ unit = ............................

Extract:

A 50 cm rule is balanced at its mid-point. A force of 8.0 N acts at a distance of 10 cm from one end of the rule.

Fig. 2.1 shows the arrangement.

[diagram: 8.0 N force acting 10 cm from one end, pivot at mid-point, 50 cm rule]

Another force acts at a point 10 cm from the pivot. It makes the rule balance.

On Fig. 2.1, draw an arrow to show the position and direction of this force.

A metre rule is balanced on a pivot by three vertical forces, as shown in Fig. 5.1.

Show that the moment of the 5.0 N force about the pivot is 200 N cm.

Extract:

A metre rule is balanced on a pivot by three vertical forces, as shown in Fig. 5.1.

Diagram of a 100 cm lever with a pivot at 40 cm from the left-hand side. A 5.0 N force acts at 40 cm from the pivot on the left-hand side, and the 2.0 N weight of the rule acts at 10 cm from the pivot on the right-hand side (weight acts from the centre of the rule).

Force = 5.0 N Distance from the pivot = 40 cm Weight of rule = 2.0 N (acts at 10 cm from the pivot on the right-hand side) Distance from pivot to F = 100 - 40 = 60 cm

Moment = force × distance from the pivot

The size of force is calculated by:

• Moment of 5.0 N force = 200 N cm
• Sum of the clockwise moments = sum of the anticlockwise moments
• Moment due to 2.0 N = 2.0 × 10 = 20 N cm
• Moment due to F = F × (100 – 40) = 60F N cm
• 20 + 60F = 200
• 60F = 200 – 20 = 180
• F = 180 / 60
• F = 3.0 N

Extended tier only

Calculate the size of force F.

F = .................................................... N

Complete the statement by writing in the blank spaces.

The moment of a force about a pivot is equal to ............ multiplied by ............

Fig. 3.1 shows a horizontal rod of length 2.4 m and weight 160 N. The weight of the rod acts at its centre. The rod is suspended by two vertical ropes X and Y. The tension in each rope is 80 N.

(i) State the name given to the point at which the weight of the rod acts.

(ii) Calculate the mass of the rod.

mass = ......................................................... [1]

(iii) The rod is in equilibrium.

Using data from Fig. 3.1, explain why.

Fig. 2.1 shows a man pushing down on a lever to lift one end of a heavy log.

State the term used to describe the turning force exerted by the man.

Extract:

Fig. 2.1 shows a man pushing down on a lever to lift one end of a heavy log.

(i) Fig. 2.2 shows the forces acting as the man starts to lift the heavy log.

Calculate the force F, exerted by the lever on the heavy log.

force = ..................................................... N [3]

(ii) Describe how the man can use a smaller force to lift the heavy log.

During a routine check of security camera footage at a zoo, it is discovered that a toucan and a grass snake have been escaping from their enclosures and performing experiments on a uniform 2 m long seesaw to study the principle of moments.

Fig. 1.1

Mass of a toucan = 600 g

Mass of a grass snake = 250 g

Calculate the weight of each animal in newtons. You may assume the acceleration of free fall is 10 m/s².

Extract:

Weight of the toucan, W_t = 6.0 N Weight of the grass snake, W_g = 2.5 N Distance of toucan from pivot, d_t = 40 cm Distance of grass snake from pivot, d_g = 90 cm Convert the distances to metres: d_g = 90 cm = 0.90 m d_t = 40 cm = 0.40 m Recall the equation for calculating a moment: moment = F × d

For the system in Fig. 1.1, calculate the clockwise and anticlockwise moments separately.

clockwise moment = ...................................................... N m

anticlockwise moment = ...................................................... N m

Extract:

Weight of the toucan, W_t = 6.0 N Weight of the grass snake, W_g = 2.5 N Distance of toucan from pivot, d_t = 40 cm Distance of grass snake from pivot, d_g = 90 cm Convert the distances to metres: d_g = 90 cm = 0.90 m AND d_t = 40 cm = 0.40 m Recall the equation for calculating a moment: moment = F × d Calculate the clockwise moment from the snake, moment_CW: moment_CW = 2.5 × 0.90 moment_CW = 2.25 N m = 2.3 N m (2 s.f.) Calculate the anticlockwise moment from the toucan, moment_ACW: moment_ACW = 6.0 × 0.40 moment_ACW = 2.4 N m (2 s.f.)

State whether the system is in equilibrium or not. Explain your reasoning.

Extract:

a uniform 2 m long seesaw

Extended tier only

On a different day, the toucan and the grass snake sit in new positions at different distances from the pivot. The same 2 metre long seesaw is not in equilibrium and has a resultant clockwise moment of 5.2 N m.

A bearded dragon climbs onto on the left side of the see saw. Suggest whether it is possible for the bearded dragon to bring the seesaw into equilibrium. Explain your answer using the principle of moments.

Average weight of a bearded dragon = 4.9 N

A man uses a metal bar to remove an iron nail from a piece of wood, as shown in Fig. 3.1.

(i) The man applies a force of 150 N perpendicular to the bar and at a distance of 0.50 m from the pivot.

Calculate the moment of this force about the pivot. Give the unit.

moment = .......................................................... [4]

(ii) The force applied by the man produces a turning effect (moment) about the pivot.

Describe another example of using the turning effect of a force.

The man tries to use the metal bar to remove another nail from the piece of wood. He applies the same force of 150 N at a distance of 0.50 m from the pivot. The turning effect produced is not enough to remove this nail from the piece of wood.

Describe how the man can increase the turning effect without increasing the force.

Fig. 4.3 shows the force on a pulley from a load M.

The weight of load M is 2.5 N and the weight acts at a distance of 20 cm from the pivot of the pulley wheel.

Calculate the moment of the weight of load M about the pivot.

moment = ................................................ N cm

Extract:

Force 1030 N at a perpendicular distance of 120 cm from the pivot. Expected answer 123 600 N cm; accept 124 000 N cm, 1.24 × 10^{5 N cm or 1.236 × 10}5 N cm. Accept 1236 N m only if the unit N m is written. Award all three marks for the correct answer alone. Do not accept 120 000 N cm.

A builder starts to raise one end of the beam.

He uses a force of 1030 N at a perpendicular distance of 120 cm from the pivot. Fig. 2.2 shows the arrangement.

Calculate the moment of the 1030 N force about the pivot.

moment = ................................................ N cm

Extract:

Expected answer: apply the force further from the pivot (increase the perpendicular distance). Accept "longer lever" or "lift at the end of the beam".

Describe how the builder can use a smaller force to lift the beam.

Extract:

Expected answer, any one of: higher centre of gravity (in Fig. 2.3); smaller base or area in contact with the ground. Accept the comparison stated either way round.

The builder positions the beam as shown in Fig. 2.3.

State why the beam shown in Fig. 2.3 is less stable than the beam shown in Fig. 2.1.

A tower crane has a load W, as shown in Fig. 3.1.

The counterweight has a weight of 80 000 N. This acts at a distance of 5.0 m from the pivot, as shown in Fig. 3.1.

Calculate the moment of the counterweight about the pivot. Give the unit.

moment = ...........................................................

Extract:

The counterweight has a weight of 80 000 N. This acts at a distance of 5.0 m from the pivot, as shown in Fig. 3.1.

Force, F = 80 000 N Distance from pivot, d = 5.0 m

Moment = Force × distance

M = 400 000 N m

The tower crane in Fig. 3.1 balances horizontally when holding the load W.

Calculate the weight of load W.

weight = ....................................................... N

Extended tier only

Fig. 2.1 shows a uniform plank AB of length 2.0 m suspended from two ropes X and Y.

The weight W of the plank is 210 N. The force in rope X is P. The force in rope Y is Q.

State, in terms of P, the moment of force P about B.

Extract:

Fig. 2.1 shows a uniform plank AB of length 2.0 m suspended from two ropes X and Y. The weight W of the plank is 210 N. The force in rope X is P. The force in rope Y is Q. Distance from A to P is 1.0 m, from P to B is 1.0 m, from A to B is 2.0 m, and from P to B is 1.5 m. The moment of force P about B is P × 1.5. The equation for the size of a moment is Moment = force × distance OR M = Fd. The moment of W about B is W × 1.0. In equilibrium, the total upward forces equal the total downward forces: P + Q = W. To rearrange the moment equation to make F the subject: F = M/d. To rearrange P + Q = W to make Q the subject: Q = W – P. W = 210 N.

Extended tier only

Calculate:

(i) the moment of *W *about B

moment = ...........................................................[1]

(ii) the force P

force P = ...........................................................[2]

(iii) the force Q.

force Q = ...........................................................[2]

Fig. 4.1 shows a tractor fitted with a device for breaking up soil in a field.

(i) The tractor has a heavy weight at the front. Explain why the heavy weight is needed.

(ii) Fig. 4.2 represents the weight of the device and its distance from the pivot.

Calculate the moment of the weight of the device about the pivot. State the unit.

moment = ...................................................... [4]

Fig. 4.3 shows a tractor fitted with narrow tyres and the same tractor fitted with wide tyres.

Explain why wide tyres are more suitable for the tractor on soft soil.

In a double-decker bus there are two passenger compartments, one above the other.

Fig. 3.1 shows a double-decker bus on a tilted platform.

The platform is used to test the stability of the bus.

The angle the bus makes with the horizontal is gradually increased until the bus begins to topple to the left.

Explain why the bus begins to topple.

Extract:

In a double-decker bus there are two passenger compartments, one above the other.

Fig. 3.1 shows a double-decker bus on a tilted platform.

The platform is used to test the stability of the bus.

The angle the bus makes with the horizontal is gradually increased until the bus begins to topple to the left.

There are 30 passengers in the upper compartment of a bus and 2 passengers in the bottom compartment.

State how this affects the stability of the bus and the reason for this.

Extended tier only

A bus is travelling along a straight road. The bus and the driver have a combined mass of 16000 kg when there are no passengers in it. The bus has 73 passengers. The average mass of each of the passengers is 65 kg.

(i) Calculate the total mass of the bus, the driver and the 73 passengers.

mass = ......................................................... [2]

(ii) The fully loaded bus accelerates uniformly from rest to a speed of 14 m / s. The time taken to reach a speed of 14 m / s is 20 s.

Calculate the resultant force on the bus during the acceleration.

force = ......................................................... [2]

A university student is constructing some flat-pack furniture.

An Allen key has a hexagonal end and is used to tighten hexagonal bolts. An Allen key can be used with its longer end in the hexagonal bolt, or can be rotated to place the shorter end in the hexagonal bolt. This is shown in Fig. 1.1.

Fig. 1.1

State which orientation, A or B, will allow the bolt to be tightened more easily.

Explain why, referring to moments in your answer.

Extract:

A university student is constructing some flat-pack furniture.

An Allen key has a hexagonal end and is used to tighten hexagonal bolts. An Allen key can be used with its longer end in the hexagonal bolt, or can be rotated to place the shorter end in the hexagonal bolt. This is shown in Fig. 1.1.

The dimensions of the Allen key are shown in Fig. 1.2.

Fig. 1.2

The student building the furniture applies a force of 150 N to the Allen key.

(i) Calculate the moment when the Allen key is in orientation A.

moment in orientation A = ...................................................... N m [4]

(ii) Calculate the moment when the Allen key is in orientation B.

moment in orientation B = ...................................................... N m [3]

Extract:

Weight of furniture = 800 N F_A is 30% of the furniture's weight Perpendicular distance from pivot to furniture, d_A = 0.20 m Perpendicular distance from pivot to force from student, d_B = 0.80 m F_A = 0.3 × 800 = 240 N Recall the equation to calculate the moment: moment = F × d Use the principle of moments: In equilibrium, moment A = moment B

The student constructs the furniture but realises they need to lift it up to place a rug underneath.

The furniture has a weight of 800 N and is too heavy for the student to lift.

Remembering a lecture about the principle of moments, they use a pivot and a plank of wood to lift it. As shown in Fig. 1.3, the furniture exerts a perpendicular force F_A on the plank, which is equal to 30% of the weight of the furniture.

Fig. 1.3

Calculate the minimum force F_B the student must exert to lift the furniture.

Extract:

The student constructs the furniture but realises they need to lift it up to place a rug underneath.

The furniture has a weight of 800 N and is too heavy for the student to lift.

Remembering a lecture about the principle of moments, they use a pivot and a plank of wood to lift it. As shown in Fig. 1.3, the furniture exerts a perpendicular force F_A on the plank, which is equal to 30% of the weight of the furniture.

Calculate the minimum force F_B the student must exert to lift the furniture.

Extended tier only

The calculation in part (c) neglected the weight of the plank of wood.

Assuming the plank of wood has evenly distributed mass, explain how this would affect the size of the force that the student must exert to lift the furniture.

A builder uses a uniform wooden plank as a lever to lift a heavy concrete block.

Fig. 3.1 shows the arrangement.

Fig 3.1

(i) State the equation that links moment, force, and distance.

(ii) State the unit for the moment of a force.

Extract:

A builder uses a uniform wooden plank as a lever to lift a heavy concrete block.

Fig. 3.1 shows the arrangement.

Fig 3.1

The weight of the uniform plank is 80.0 N.

(i) State the distance from the pivot to the centre of gravity of the plank.

(ii) Calculate the moment of the plank's weight about the pivot.

Extract:

A builder uses a uniform wooden plank as a lever to lift a heavy concrete block.

Fig. 3.1 shows the arrangement.

Fig 3.1

The weight of the uniform plank is 80.0 N.

The concrete block has a weight of 600.0 N and is placed at the left end of the plank, 1.00 m from the pivot.

Calculate the minimum downward force, F, the builder must apply to lift the block.