ExaminerPrep › Question bank › IAL Physics › 1.4 Moments and centre of gravity
Edexcel International A Level Physics · Unit 1 · Mechanics and Materials
40 free, high-quality past-paper-style questions on Moments and centre of gravity for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which of the following is a correct unit for the moment of a force?
| Answer | Mark | Notes |
|---|---|---|
| Moment = force × perpendicular distance, so its unit is newton × metre = N m. | ✓ 1 | Answer A |
Answer: A
Common error: kg m s−2 is the newton itself, the unit of force. A moment needs a distance multiplied in, not divided.
A person pushes on a door with a force of 35 N at right angles to the door. The force acts 78 cm from the line of the hinges.
What is the moment of the force about the hinges?
| Answer | Mark | Notes |
|---|---|---|
| Convert the distance to metres: 0.78 m. Moment = 35 × 0.78 = 27 N m (27.3 N m). | ✓ 1 | Answer B |
Answer: B
Exam tip: Convert every distance to metres before you multiply, so the moment comes out in N m.
Which statement describes the centre of gravity of a body?
| Answer | Mark | Notes |
|---|---|---|
| The centre of gravity is the single point at which the whole weight can be considered to act. It is at the geometric centre only for a uniform, symmetrical body. | ✓ 1 | Answer D |
Answer: D
Common error: The centre of gravity is not always at the middle of an object, and it need not even be inside the material.
A hanging mobile has a light horizontal rod suspended by a thread. An ornament of weight 0.48 N hangs from the rod 9.0 cm to the left of the thread. A second ornament hangs 12 cm to the right of the thread, and the rod is balanced.
What is the weight of the second ornament?
| Answer | Mark | Notes |
|---|---|---|
| Principle of moments about the thread: W × 12 = 0.48 × 9.0, so W = 0.36 N. The units of distance cancel, so centimetres can be used. | ✓ 1 | Answer B |
Answer: B
Exam tip: The ornament further from the pivot must be lighter: check that your answer makes sense before choosing.
A rower pulls the handle of an oar. The oar turns about a pivot (the rowlock) fixed to the side of the boat. The rower's force is 300 N, at right angles to the oar, 0.85 m from the pivot.
Define the moment of a force about a point.
Calculate the moment of the rower's force about the pivot.
The water pushes on the blade with a force at right angles to the oar, acting 1.95 m from the pivot on the other side. The oar is not turning faster or slower. Calculate this force.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| force × perpendicular distance | ✓ 1 | |
| from the point to the line of action of the force | ✓ 1 | |
| Part (b) | ||
| 300 × 0.85 = 255 N m | ✓ 1 | |
| Part (c) | ||
| F × 1.95 = 255 | ✓ 1 | |
| F = 131 N | ✓ 1 | |
Exam tip: A definition of moment without the word “perpendicular” is incomplete. Learn the wording: force × perpendicular distance from the point to the line of action.
A person holds a push-up position with a straight, rigid body. Her weight is 680 N, and her centre of gravity is 0.95 m horizontally from her toes. Her hands are 1.40 m horizontally from her toes. Treat all the forces as vertical.
State the forces acting on her body, giving the direction of each and where it acts.
Calculate the force from the floor on her hands.
Calculate the force from the floor on her toes.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| weight, 680 N downwards, acting at her centre of gravity | ✓ 1 | |
| upward contact forces from the floor at her hands and at her toes | ✓ 1 | |
| Part (b) | ||
| moments about the toes: F × 1.40 = 680 × 0.95 | ✓ 1 | |
| F = 461 N | ✓ 1 | |
| Part (c) | ||
| 680 − 461 = 219 N (allow error carried forward) | ✓ 1 | |
Exam tip: Taking moments about the toes removes the unknown force there, leaving one equation with one unknown.
The figure shows a light rod pivoted at O. Three forces act on the rod at the positions shown on the scale.
What is the resultant moment of these forces about O?
| Answer | Mark | Notes |
|---|---|---|
| Taking anticlockwise as positive: 6.0 × 0.20 − 4.0 × 0.30 + 5.0 × 0.50 = 1.2 − 1.2 + 2.5 = 2.5 N m anticlockwise. The upward 5.0 N force right of O turns the rod anticlockwise, the same way as the 6.0 N force left of O. | ✓ 1 | Answer B |
Answer: B
Common error: Giving every moment the same sign. Decide the sense of each force first: a downward force left of the pivot and an upward force right of the pivot both turn the rod anticlockwise.
The figure shows an L-shaped bracket drawn on a grid of 0.10 m squares. The bracket can turn about P. A vertical force of 20 N acts at its lower end.
What is the moment of the 20 N force about P?
| Answer | Mark | Notes |
|---|---|---|
| The line of action of the force is vertical and lies 3 squares, 0.30 m, horizontally from P. Moment = 20 × 0.30 = 6.0 N m. The vertical part of the bracket does not change this distance. | ✓ 1 | Answer A |
Answer: A
Common error: Using the length of the bracket. The distance in a moment is the perpendicular distance from the pivot to the line of action; for a vertical force that is the horizontal distance.
A uniform L-shaped plate has a small hole at P, and G is its centre of gravity. The plate is hung from a pin through P and allowed to come to rest. In each diagram the dashed line is vertical.
Which diagram shows the plate at rest?
| Answer | Mark | Notes |
|---|---|---|
| At rest the weight has no moment about the pin, so its line of action passes through P: G must be directly below P. Only in diagram D does G lie on the vertical line through P. | ✓ 1 | Answer D |
Answer: D
Exam tip: A freely hanging body comes to rest with its centre of gravity vertically below the point of suspension. This is the idea behind the plumb-line method.
The figure shows a loaded wheelbarrow at rest. G is the centre of gravity of the wheelbarrow and its load.
What vertical force at the handles just lifts the legs off the ground?
| Answer | Mark | Notes |
|---|---|---|
| When the legs just leave the ground, the only other force is at the wheel. Moments about the axle: F × (0.45 + 0.90) = 720 × 0.45, so F = 324 ÷ 1.35 = 240 N. | ✓ 1 | Answer A |
Answer: A
Common error: Measuring the distance of the handles from G. Every distance in the moment equation must be measured from the pivot, the wheel axle.
The figure shows a steelyard, a simple weighing balance. The rod hangs from a pivot P. The load L hangs from a hook to the left of P, and a poise of weight 5.0 N slides along a scale to the right of P. With no load and the poise at the 0 mark, the rod balances horizontally. In the figure the rod is balanced horizontally with a load on the hook.
State the principle of moments.
Use the figure to determine the weight of the load L.
Explain why the weight of the rod is not needed in your calculation in (b).
The scale is to be marked in newtons of load. Calculate the distance between the marks for 10 N and 20 N.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| for a body in equilibrium, the sum of the clockwise moments about any point | ✓ 1 | |
| equals the sum of the anticlockwise moments about the same point | ✓ 1 | |
| Part (b) | ||
| poise at 36 cm: 5.0 × 0.36 = L × 0.060 | ✓ 1 | |
| L = 30 N | ✓ 1 | |
| Part (c) | ||
| the rod balances on its own with the poise at P, so the weight of the rod has no moment about P / its line of action passes through P | ✓ 1 | |
| Part (d) | ||
| an extra 10 N needs an extra moment of 10 × 0.060 = 0.60 N m, so the poise moves 0.60 ÷ 5.0 = 0.12 m | ✓ 1 | |
Exam tip: Read the position of the poise at its centre, and keep both distances in the same unit before you divide.
The figure shows a nutcracker. Its two arms turn about a hinge H. A nut is held between the arms, and forces of 40 N at right angles to the arms squeeze them together. Treat each arm as light and ignore the small angle between the arms.
Calculate the moment of one 40 N force about H.
Calculate the force exerted by each arm on the nut.
Determine the magnitude and direction of the force from the hinge on the upper arm.
Explain why a nut is easier to crack when it is placed closer to the hinge.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 40 × 0.15 = 6.0 N m | ✓ 1 | |
| Part (b) | ||
| F × 0.030 = 6.0 | ✓ 1 | |
| F = 200 N | ✓ 1 | |
| Part (c) | ||
| resultant force on the arm is zero: the nut pushes the upper arm up with 200 N and the hand pushes it down with 40 N | ✓ 1 | |
| so the hinge pushes down on the upper arm with 160 N | ✓ 1 | |
| Part (d) | ||
| the hand gives the same moment about H, so a smaller perpendicular distance from H means a larger force on the nut | ✓ 1 | |
Common error: Forgetting the hinge force. A lever in equilibrium has zero resultant force as well as zero resultant moment, so the pivot usually pushes on it.
A cyclist pushes on a pedal with a force of 400 N. The crank joining the pedal to the axle is 0.170 m long, and the force acts at 50° to the crank.
What is the moment of the force about the axle?
| Answer | Mark | Notes |
|---|---|---|
| The perpendicular distance from the axle to the line of action is 0.170 sin 50° = 0.130 m. Moment = 400 × 0.130 = 52.1 N m. | ✓ 1 | Answer B |
Answer: B
Exam tip: Only the component of the force at right angles to the crank turns it: 400 sin 50°. The component along the crank passes through the axle and has no moment.
A uniform steel girder of length 6.0 m and weight 2.4 kN rests horizontally on two pillars, P and Q, one at each end. A machine of weight 1.8 kN is placed on the girder 1.5 m from P.
What is the force from pillar P on the girder?
| Answer | Mark | Notes |
|---|---|---|
| Take moments about Q: RP × 6.0 = 2.4 × 3.0 + 1.8 × 4.5 = 15.3, so RP = 2.55 kN. (Pillar Q supplies the other 1.65 kN.) | ✓ 1 | Answer C |
Answer: C
Exam tip: Take moments about the pillar whose force you do not want. Its force then has no moment and drops out.
A double-decker bus is tilted sideways on a test platform. Which change would allow the bus to be tilted to a larger angle before it topples?
| Answer | Mark | Notes |
|---|---|---|
| The bus topples when the vertical line through its centre of gravity passes outside the wheels on the lower side. A lower centre of gravity and a wider base mean it must tilt further before that happens. | ✓ 1 | Answer A |
Answer: A
Exam tip: Toppling depends on where the line of action of the weight falls compared with the edge of the base, not on how large the weight is.
A student finds the centre of gravity of an irregular sheet of card. She hangs the card from a pin through a small hole and marks the line of a plumb line hung from the same pin.
Why does the centre of gravity lie on this line?
| Answer | Mark | Notes |
|---|---|---|
| At rest the resultant moment about the pin is zero, so the line of action of the weight passes through the pin: the centre of gravity is vertically below it, on the plumb line. | ✓ 1 | Answer D |
Answer: D
Exam tip: Explain practical methods with the physics behind them — here, zero resultant moment about the pin.
A car park barrier is a uniform pole of length 4.20 m and weight 95 N. It is pivoted 0.50 m from one end, P. A counterweight is fixed at end P so that the pole balances horizontally.
Explain why the weight of the pole acts 1.60 m from the pivot.
Calculate the weight of the counterweight.
Calculate the force from the pivot on the pole.
The operator wants the barrier to fall closed by itself if the motor fails. Suggest how the counterweight should be changed, and explain your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the pole is uniform, so its centre of gravity is at its midpoint, 2.10 m from P, which is 2.10 − 0.50 = 1.60 m from the pivot | ✓ 1 | |
| Part (b) | ||
| W × 0.50 = 95 × 1.60 | ✓ 1 | |
| W = 304 N | ✓ 1 | |
| Part (c) | ||
| 95 + 304 = 399 N upwards | ✓ 1 | |
| Part (d) | ||
| make the counterweight slightly smaller / move it closer to the pivot | ✓ 1 | |
| so the moments about the pivot no longer balance: the moment of the pole's weight is larger and turns the long arm down | ✓ 1 | |
Common error: Measuring the distance of the pole's weight from the end of the pole instead of from the pivot. Every distance in a moment is measured from the pivot.
A student has cut a flat, irregular shape from a sheet of thin plywood. She wants to find its centre of gravity by experiment. She has a clamp and stand, a pin, thread, a small mass and a pencil.
Describe how she should find the centre of gravity of the shape.
Explain why the centre of gravity lies on each line she draws.
Suggest how she could check her result.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| make a small hole near the edge and hang the shape from the pin so that it can swing freely | ✓ 1 | |
| hang a plumb line (thread with the mass) from the pin and, when everything is at rest, mark the line of the thread on the shape | ✓ 1 | |
| repeat with a second hole well away from the first | ✓ 1 | |
| the centre of gravity is where the two lines cross | ✓ 1 | |
| Part (b) | ||
| at rest the resultant moment about the pin is zero, so the line of action of the weight passes through the pin / G is vertically below the pin | ✓ 1 | |
| Part (c) | ||
| hang the shape from a third hole: the new line should pass through the same point / balance the shape on a point placed under G | ✓ 1 | |
Exam tip: In a “describe” question for a practical, give the steps in order and say what is measured or marked at each step.
A forearm is held horizontally, pivoting at the elbow. The biceps muscle is attached 4.0 cm from the elbow. The weight of the forearm, 15 N, acts 15 cm from the elbow, and a ball of weight 40 N is held in the hand 33 cm from the elbow.
The biceps pulls vertically upwards. Calculate the force in the biceps.
In fact the biceps pulls at 12° to the vertical. Calculate the force now needed in the biceps.
Explain why the force in the biceps is much larger than the weight of the ball.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| moments about the elbow: F × 0.040 = 15 × 0.15 + 40 × 0.33 | ✓ 1 | |
| = 15.45 N m | ✓ 1 | |
| F = 386 N | ✓ 1 | |
| Part (b) | ||
| only the perpendicular component turns the forearm: F cos 12° × 0.040 = 15.45 | ✓ 1 | |
| F = 395 N | ✓ 1 | |
| Part (c) | ||
| the biceps acts much closer to the pivot than the ball, so a much larger force is needed for an equal moment | ✓ 1 | |
Exam tip: Here the angle is measured from the vertical, and the forearm is horizontal, so the component perpendicular to the forearm is F cos 12°.
A mobile crane stands on extended legs (outriggers). The weight of the crane, without any load, is 180 kN. Its centre of gravity is 2.0 m horizontally behind the front outriggers. The crane lifts a load whose line of action is 8.0 m horizontally in front of the front outriggers.
State the line about which the crane would tip if the load were too heavy.
Calculate the largest load the crane can lift at this distance without tipping.
The front outriggers are extended further forward. Explain why the crane can now lift a larger load at the same position.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the line through the front outriggers (where they touch the ground) | ✓ 1 | |
| Part (b) | ||
| on the point of tipping the rear outriggers have no force, so moments about the front outriggers: L × 8.0 = 180 × 2.0 | ✓ 1 | |
| L = 45 kN | ✓ 1 | |
| Part (c) | ||
| the perpendicular distance from the tipping line to the crane's weight increases, so the weight has a larger moment | ✓ 1 | |
| and the load is nearer the tipping line, so its moment is smaller for the same load | ✓ 1 | |
Exam tip: “On the point of tipping” means the support on the other side carries no force. State this before taking moments about the tipping edge.
Two people carry a kayak horizontally. The kayak is 4.8 m long and weighs 260 N. Its centre of gravity is 2.3 m from the bow (front). Person X holds the bow; person Y holds the kayak 0.60 m from the stern (back), that is, 4.2 m from the bow.
Calculate the upward force exerted by Y.
Calculate the upward force exerted by X.
A bag of weight 45 N is placed in the kayak 3.6 m from the bow. Calculate the new force exerted by X.
State which person supports most of the weight of the bag, and give a reason.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| moments about the bow: FY × 4.2 = 260 × 2.3 | ✓ 1 | |
| FY = 142 N | ✓ 1 | |
| Part (b) | ||
| 260 − 142 = 118 N | ✓ 1 | |
| Part (c) | ||
| moments about Y: FX × 4.2 = 260 × 1.9 + 45 × 0.60 = 521 | ✓ 1 | |
| FX = 124 N | ✓ 1 | |
| Part (d) | ||
| Y, because the bag is much closer to Y than to X | ✓ 1 | |
Exam tip: For the second force, taking moments about the other support gives an independent check: the two forces must add up to the total weight.
A student balances a uniform rule on a pivot at its centre. A fixed weight W hangs 0.30 m to the left of the pivot. She hangs a weight w to the right of the pivot and moves it until the rule balances, then records its distance d from the pivot. She repeats this for several values of w. The graph shows d against 1/w.
What is the value of W?
| Answer | Mark | Notes |
|---|---|---|
| The rule's weight acts at the pivot and has no moment. Balance: w d = W × 0.30, so d = 0.30W × (1/w). The gradient is 0.60 ÷ 1.00 = 0.60 N m, so W = 0.60 ÷ 0.30 = 2.0 N. | ✓ 1 | Answer D |
Answer: D
Exam tip: Rearrange the physics into the form y = mx before reading the graph: here d = (0.30W) × (1/w), so the gradient is 0.30W, not W.
The figure shows a hammer drawn to scale. The handle is uniform. The weights of the handle and of the head act at the points marked.
How far from end A is the centre of gravity of the whole hammer?
| Answer | Mark | Notes |
|---|---|---|
| The handle's weight acts at its midpoint, 0.17 m from A. Moments about A: (2.4 + 5.6) x = 2.4 × 0.17 + 5.6 × 0.31 = 2.144 N m, so x = 2.144 ÷ 8.0 = 0.27 m (0.268 m). | ✓ 1 | Answer C |
Answer: C
Exam tip: The centre of gravity of a combined object lies between the separate centres of gravity, closer to the heavier part. Use this to check your answer.
The figure shows a uniform diving board. End A is bolted to a concrete block and the board rests on a support B. A diver stands at the free end.
What is the size of the force from the bolt on the board?
| Answer | Mark | Notes |
|---|---|---|
| The board's weight acts at its midpoint, 2.0 m from A, which is 1.0 m beyond B. Moments about B: F × 1.0 = 400 × 1.0 + 600 × 3.0 = 2200 N m, so F = 2200 N, acting downwards. | ✓ 1 | Answer B |
Answer: B
Exam tip: Take moments about the support B, so its unknown force drops out. Both weights turn the board the same way about B, so the bolt must pull down.
The figure shows a wheel and axle that turn together on a fixed bearing. A load hangs from a rope wound round the axle. A rope wound round the wheel is pulled down with an effort E. Friction is negligible.
What value of E holds the load at rest?
| Answer | Mark | Notes |
|---|---|---|
| Moments about the bearing: E × 0.25 = 400 × 0.050, so E = 20 ÷ 0.25 = 80 N. The radius of the wheel is half of its 0.50 m diameter. | ✓ 1 | Answer C |
Answer: C
Common error: Using the diameter of the wheel. Each rope leaves along a tangent, so its perpendicular distance from the axis is a radius.
The figure shows a uniform flat plate in the shape of a letter T, drawn on a grid of 0.10 m squares.
Explain why the centre of gravity of the plate lies on its axis of symmetry.
Determine the height of the centre of gravity of the plate above its base.
The plate is hung freely from a pin through corner P. Calculate the angle between the axis of symmetry and the vertical when the plate is at rest.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the plate is uniform and symmetrical, so the weights on the two sides of the axis are equal and have equal and opposite moments about it | ✓ 1 | |
| Part (b) | ||
| upright: area 0.20 × 0.60 = 0.12 m2, centre 0.30 m up; crossbar: area 0.80 × 0.20 = 0.16 m2, centre 0.70 m up | ✓ 1 | |
| weight is proportional to area, so 0.28 h = 0.12 × 0.30 + 0.16 × 0.70 = 0.148 | ✓ 1 | |
| h = 0.53 m (0.529 m) | ✓ 1 | |
| Part (c) | ||
| at rest G is directly below P; P is 0.40 m from the axis and 0.80 − 0.529 = 0.271 m above G, measured along the axis | ✓ 1 | |
| angle = tan−1(0.40 ÷ 0.271) = 56° (55.8°; allow 55–57°) | ✓ 1 | |
Common error: Measuring some heights from the base and others from the top of the plate. Choose one reference line and measure every distance from it.
The figure shows a uniform gate of weight 300 N hanging from a post on two hinges, T and B. Assume that hinge B supports the whole weight of the gate, so that the force from hinge T on the gate is horizontal.
By taking moments about B, calculate the horizontal force from hinge T on the gate.
State the direction of the force from hinge T on the gate, and explain your answer.
Calculate the magnitude of the resultant force from hinge B on the gate.
A child of weight 250 N stands on the bottom rail at the free end of the gate, 1.20 m from the hinge line. Calculate the new horizontal force from hinge T.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| FT × 1.20 = 300 × 0.60 | ✓ 1 | |
| FT = 150 N | ✓ 1 | |
| Part (b) | ||
| towards the post (horizontally, to the left in the figure) | ✓ 1 | |
| the weight has a clockwise moment about B, so the force at T must give an equal anticlockwise moment about B | ✓ 1 | |
| Part (c) | ||
| horizontal component 150 N away from the post and vertical component 300 N upwards | ✓ 1 | |
| resultant = √(1502 + 3002) = 335 N | ✓ 1 | |
| Part (d) | ||
| FT × 1.20 = 300 × 0.60 + 250 × 1.20 = 480 N m, so FT = 400 N | ✓ 1 | |
Exam tip: Taking moments about B removes both unknown components of the force at B, leaving the force at T as the only unknown.
The figure shows a method for finding the centre of gravity of a person. A uniform board of weight 80 N rests on a knife-edge K at one end and on a digital scale at the other. A person of weight 650 N lies on the board with the soles of their feet level with K. The scale then shows the reading in the figure.
Show that the scale reads 40 N when the person is not on the board.
Calculate the distance of the person's centre of gravity from the soles of their feet.
The person is 1.78 m tall. Express the height of their centre of gravity above the soles as a percentage of their height.
The reading on the scale has an uncertainty of ±2 N. Calculate the percentage uncertainty in the reading.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the board is uniform, so its weight acts at its midpoint, 1.00 m from K | ✓ 1 | |
| moments about K: R × 2.00 = 80 × 1.00, so R = 40 N | ✓ 1 | |
| Part (b) | ||
| moments about K: 370 × 2.00 = 80 × 1.00 + 650 x | ✓ 1 | |
| 650 x = 740 − 80 = 660 | ✓ 1 | |
| x = 1.02 m (1.015 m) | ✓ 1 | |
| Part (c) | ||
| 1.02 ÷ 1.78 × 100 = 57% | ✓ 1 | |
| Part (d) | ||
| 2 ÷ 370 × 100 = 0.54% | ✓ 1 | |
Exam tip: Either include the board's weight as a separate moment or subtract the 40 N board-only reading first: both routes give the same distance.
The figure shows a uniform crate of weight 900 N on level ground. A horizontal force F is applied to the left-hand face, 1.00 m above the ground. Friction stops the crate from sliding.
The force F is slowly increased until the crate starts to tip. State the edge about which the crate tips, and explain why, as it starts to tip, the whole contact force from the ground acts at this edge.
Calculate the value of F at which the crate starts to tip.
Calculate the horizontal force needed to tip the crate if it is applied 0.60 m above the ground instead.
The crate is repacked so that its centre of gravity is lower, but still midway between the left and right faces. Explain whether the force calculated in (b) changes.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| it tips about edge Q | ✓ 1 | |
| the rest of the base is lifting off the ground, so the ground can only push on the crate at Q | ✓ 1 | |
| Part (b) | ||
| moments about Q: F × 1.00 = 900 × 0.40 | ✓ 1 | |
| F = 360 N | ✓ 1 | |
| Part (c) | ||
| F = 900 × 0.40 ÷ 0.60 = 600 N | ✓ 1 | |
| Part (d) | ||
| it does not change: the moment of the weight about Q depends only on the horizontal distance from Q to the line of action of the weight, which is still 0.40 m | ✓ 1 | |
Common error: Using the height of G as the distance for the weight. The weight acts vertically, so its moment about Q uses the horizontal distance from Q to G.
The figure shows a car standing with its front wheels on one weighing platform and its rear wheels on another. The readings of the two platforms are shown.
Calculate the weight of the car.
Calculate the horizontal distance of the centre of gravity of the car from the front axle.
A load of weight 0.90 kN is placed in the boot at the position shown. Calculate the new readings of the two platforms.
Explain why the reading of the front platform decreases when the load is added.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 7.2 + 5.4 = 12.6 kN | ✓ 1 | |
| Part (b) | ||
| moments about the front axle: 12.6 x = 5.4 × 2.70 | ✓ 1 | |
| x = 1.16 m (1.157 m) | ✓ 1 | |
| Part (c) | ||
| moments about the front axle: R × 2.70 = 5.4 × 2.70 + 0.90 × 3.50 = 17.73 kN m | ✓ 1 | |
| rear platform R = 6.57 kN | ✓ 1 | |
| front platform = 13.5 − 6.57 = 6.93 kN | ✓ 1 | |
| Part (d) | ||
| the load is behind the rear axle, so its moment about the rear axle tends to lift the front of the car, and less force is needed from the front platform | ✓ 1 | |
Exam tip: Check your two new readings: they must add up to the new total weight, 12.6 + 0.90 = 13.5 kN.
A uniform trapdoor of length 0.90 m and weight 150 N is hinged along one edge. It is held horizontal by a rope attached to its free edge. The rope makes an angle of 40° with the trapdoor.
What is the tension in the rope?
| Answer | Mark | Notes |
|---|---|---|
| Moments about the hinge: T sin 40° × 0.90 = 150 × 0.45, so T = 67.5 ÷ (0.90 × sin 40°) = 117 N. | ✓ 1 | Answer C |
Answer: C
Common error: Using the whole tension in the moment. Only the component perpendicular to the trapdoor, T sin 40°, turns it about the hinge.
A fallen tree trunk 3.0 m long lies on level ground. A vertical force of 340 N is needed to just lift end A off the ground, with end B still resting on the ground. A vertical force of 510 N is needed to just lift end B, with end A resting on the ground.
How far is the centre of gravity of the trunk from end A?
| Answer | Mark | Notes |
|---|---|---|
| Lifting A, the trunk pivots about B: 340 × 3.0 = 850 × xB, where the weight is 340 + 510 = 850 N. So xB = 1.2 m and the distance from A is 3.0 − 1.2 = 1.8 m. The heavier end, B, needs the larger force. | ✓ 1 | Answer C |
Answer: C
Exam tip: The two lifting forces add up to the weight, because the two moment equations use the same weight and length.
A student determines the weight of a uniform metre rule. She hangs a 0.50 N weight from the 5.0 cm mark and moves a knife-edge under the rule until the rule balances. The knife-edge is then at the 32.4 cm mark. The centre of gravity of the rule is at the 50.0 cm mark.
Calculate the weight of the metre rule.
Each scale reading has an uncertainty of ±0.1 cm, so each distance has an uncertainty of ±0.2 cm. Assuming the 0.50 N weight is exact, calculate the percentage uncertainty in the weight of the rule.
Explain why a knife-edge is used rather than a wide support.
Suggest how she could improve the method to give a more reliable value.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| distances from the knife-edge: 32.4 − 5.0 = 27.4 cm and 50.0 − 32.4 = 17.6 cm | ✓ 1 | |
| W × 17.6 = 0.50 × 27.4 | ✓ 1 | |
| W = 0.78 N (0.778 N) | ✓ 1 | |
| Part (b) | ||
| (0.2 ÷ 27.4) × 100 + (0.2 ÷ 17.6) × 100 = 0.73% + 1.14% | ✓ 1 | |
| = 1.9% | ✓ 1 | |
| Part (c) | ||
| the pivot is at a single, well-defined position, so the distances from it can be measured precisely | ✓ 1 | |
| Part (d) | ||
| repeat with the weight at several different positions and plot a graph / find a mean value | ✓ 1 | |
Exam tip: When two quantities are divided or multiplied, add their percentage uncertainties.
An angler holds a fishing rod horizontally with both hands. The rod is 2.40 m long, weighs 3.0 N, and its centre of gravity is 0.90 m from the butt end. One hand holds the butt end; the other hand is 0.50 m from the butt end. A fish pulls vertically down on the line at the tip of the rod with a force of 12 N.
By taking moments about the butt end, calculate the force from the hand 0.50 m from the butt end.
Determine the magnitude and direction of the force from the hand at the butt end.
Explain why holding the rod with the hands closer together would make it harder to hold.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F × 0.50 = 3.0 × 0.90 + 12 × 2.40 | ✓ 1 | |
| = 31.5 N m | ✓ 1 | |
| F = 63 N upwards | ✓ 1 | |
| Part (b) | ||
| resultant force zero: 63 − 3.0 − 12 = 48 N (allow error carried forward) | ✓ 1 | |
| downwards | ✓ 1 | |
| Part (c) | ||
| the perpendicular distance between the hands is smaller, so larger hand forces are needed to give the same moment | ✓ 1 | |
Common error: Assuming every support force acts upwards. Here the butt-end hand must push down, or the rod would rotate about the front hand.
A drawbridge is a uniform deck of length 6.0 m and weight 9.0 kN, hinged at one end. A chain attached to the other end holds the deck horizontal. The chain makes an angle of 30° with the deck.
Calculate the tension in the chain.
Determine the magnitude and direction of the force from the hinge on the deck.
The deck is raised until it makes an angle of 40° with the horizontal. Calculate the moment of the weight of the deck about the hinge in this position.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| perpendicular component of tension T sin 30° (or perpendicular distance 6.0 sin 30° = 3.0 m) used | ✓ 1 | |
| moments about the hinge: T sin 30° × 6.0 = 9.0 × 3.0 | ✓ 1 | |
| T = 9.0 kN | ✓ 1 | |
| Part (b) | ||
| horizontal component = T cos 30° = 7.8 kN | ✓ 1 | |
| vertical component = 9.0 − T sin 30° = 4.5 kN | ✓ 1 | |
| resultant = √(7.82 + 4.52) = 9.0 kN at 30° above the horizontal, pointing away from the hinge side (towards the free end of the deck) and upwards | ✓ 1 | |
| Part (c) | ||
| perpendicular distance = 3.0 cos 40° = 2.30 m, so moment = 9.0 × 2.30 = 21 kN m (20.7 kN m) | ✓ 1 | |
Common error: Using 3.0 m as the distance once the deck is tilted. The weight acts vertically, so the perpendicular distance from the hinge to its line of action is 3.0 cos 40°.
The figure shows a loaded cabinet standing on a platform, drawn on a grid of 0.10 m squares. G is the centre of gravity of the cabinet and its contents. The platform is slowly tilted so that the right-hand side goes down. The cabinet does not slide.
At what angle of tilt does the cabinet start to topple?
| Answer | Mark | Notes |
|---|---|---|
| The cabinet starts to topple when the vertical line through G passes the lower, right-hand edge of the base. G is 0.20 m horizontally from that edge and 0.90 m above the base, so tan θ = 0.20 ÷ 0.90 and θ = 12.5°. | ✓ 1 | Answer A |
Answer: A
Common error: Assuming G is above the middle of the base. Read G's actual position: here it is closer to the downhill edge, so the cabinet topples at a smaller angle.
The figure shows a uniform bar hinged at its lower end. The bar is held at rest by a horizontal rope attached to its upper end.
What is the tension in the rope?
| Answer | Mark | Notes |
|---|---|---|
| Moments about the hinge. The rope is horizontal, so its perpendicular distance from the hinge is 2.0 sin 30° = 1.0 m. The weight acts at the midpoint, 1.0 cos 30° = 0.866 m horizontally from the hinge. T × 1.0 = 120 × 0.866, so T = 104 N. | ✓ 1 | Answer C |
Answer: C
Common error: Using the length along the bar as the distance. For the horizontal rope the perpendicular distance from the hinge is the height of the top end; for the vertical weight it is a horizontal distance.
A student investigates a uniform beam of length L = 1.00 m. The beam is hinged at end O and held horizontal by a vertical spring balance at its other end. A load of weight W is placed at different distances x from O, and the reading F of the balance is recorded. The graph shows the results.
Show that F = ½Wb + (W/L)x, where Wb is the weight of the beam.
Use the graph to determine W.
Use the graph to determine the weight of the beam.
State the reading of the balance when the load is placed directly above O, and explain your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| moments about O: F L = Wb × L/2 + W x | ✓ 1 | |
| dividing by L gives F = ½Wb + (W/L)x | ✓ 1 | |
| Part (b) | ||
| gradient = (6.5 − 2.5) ÷ (0.90 − 0.10) = 5.0 N m−1 | ✓ 1 | |
| W = gradient × L = 5.0 N (allow 4.8–5.2 N) | ✓ 1 | |
| Part (c) | ||
| intercept on the F axis = 2.0 N | ✓ 1 | |
| Wb = 2 × 2.0 = 4.0 N (allow 3.8–4.2 N) | ✓ 1 | |
| Part (d) | ||
| 2.0 N, because the load then has no moment about O, so the balance only supports the moment of the beam's weight | ✓ 1 | |
Exam tip: Use a large triangle on the best-fit line to find the gradient, and read the intercept only where the x-axis starts at zero.
The figure shows a uniform ladder of weight 200 N leaning against a smooth vertical wall, with its foot on rough horizontal ground. A person of weight 700 N stands on the ladder at the point shown. Because the wall is smooth, the force from the wall on the ladder is horizontal.
Show that the top of the ladder is 4.8 m above the ground.
Calculate the force from the wall on the ladder.
Determine the magnitude and direction of the resultant force from the ground on the ladder.
Explain why the ladder is more likely to slip as the person climbs higher.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| √(5.02 − 1.42) = √23.04 = 4.8 m | ✓ 1 | |
| Part (b) | ||
| horizontal distances from the foot: ladder's weight 0.70 m, person 3.5 ÷ 5.0 × 1.4 = 0.98 m | ✓ 1 | |
| moments about the foot: R × 4.8 = 200 × 0.70 + 700 × 0.98 = 826 N m | ✓ 1 | |
| R = 172 N | ✓ 1 | |
| Part (c) | ||
| vertical component 900 N; horizontal component (friction) 172 N towards the wall | ✓ 1 | |
| resultant = √(9002 + 1722) = 916 N at 79° to the horizontal | ✓ 1 | |
| Part (d) | ||
| the person's horizontal distance from the foot increases, so their moment about the foot increases, the wall force increases and a larger friction force is needed | ✓ 1 | |
Exam tip: Take moments about the foot of the ladder: the friction force and the normal force from the ground both act there, so both drop out.
The figure shows a person standing on one foot while carrying a bucket in the right hand. The person's weight is 650 N and the weight of the bucket and its contents is 150 N. The foot is 0.10 m wide, and the person's own centre of gravity G is directly above the middle of the foot.
Show that the centre of gravity of the person and bucket together is about 0.06 m to the right of G.
Explain, using ideas about moments and centre of gravity, why a person carrying a heavy bucket in one hand leans the upper body to the other side and often holds the other arm out.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (650 + 150) x = 150 × 0.30, so x = 45 ÷ 800 = 0.056 m | ✓ 1 | |
| Part (b) | ||
| the weight of the bucket has a moment about the foot that is not balanced by the person's own weight | ✓ 1 | |
| the combined centre of gravity is 0.056 m to the right of the middle of the foot | ✓ 1 | |
| this is more than half the width of the foot, 0.05 m, so the line of action of the total weight falls outside the base | ✓ 1 | |
| there would be a resultant moment about the right-hand edge of the foot, so the person would topple towards the bucket | ✓ 1 | |
| leaning the upper body to the left moves the person's own centre of gravity to the left, bringing the combined centre of gravity back above the foot | ✓ 1 | |
| the outstretched arm places some of the body's weight far to the left, giving an opposite moment that balances the bucket's moment with less lean | ✓ 1 | |
Exam tip: In an extended explanation, link each step: the force, its moment about the edge of the base, and where the centre of gravity lies compared with the base.
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