Turning forces — answer key

1. MQ1 · Practice

Options as numbered on the pupil sheet: 1 — the fixed point that an object turns about; 2 — the turning effect of a force; 3 — the shortest distance from the pivot to the line of action of the force; 4 — no resultant force and no resultant moment; 5 — no forces act at all.

Model answer. The pivot is the fixed point an object turns about, such as the hinge of a door. The moment of a force measures its turning effect. The perpendicular distance is measured from the pivot to the force’s line of action, at right angles to it. An object in equilibrium has no resultant force and no resultant moment on it; it can still have forces on it.

A full answer includes

• Pivot – the fixed point that an object turns about.
• Moment – the turning effect of a force.
• Perpendicular distance – the shortest distance from the pivot to the line of action of the force.
• Equilibrium – no resultant force and no resultant moment.

Also accept

• pivot – fixed point; moment – turning effect; perpendicular distance – shortest distance to the line of action; equilibrium – no resultant force and no resultant moment.

Watch for

• “Equilibrium means that no forces act on the object.” — A balanced beam has its weights pulling down and the pivot pushing up. In equilibrium the forces and their moments cancel; the forces are still there.
• “The perpendicular distance is the distance along the object to where the force acts.” — Only when the force is at right angles to the object. In general, measure from the pivot to the line of action, at right angles to it.

2. MQ2 · Practice

Model answer. Opening a door turns it about the hinge. Turning a tap turns its handle about the spindle. Lifting the handles of a wheelbarrow turns it about the wheel. Pushing a bicycle pedal turns the crank about its axle. All of these use the moment of a force. Pulling a sledge in a straight line moves it along without turning it, and a book resting on a table is not turning at all.

A full answer includes

• I chose a, b, c and e.
• A door turns about its hinge, a tap about its spindle, a wheelbarrow about its wheel and a pedal about the crank axle.
• Pulling a sledge along in a straight line and a book resting on a table do not turn anything.

Also accept

• a, b, c and e: door, tap, wheelbarrow and pedal.

Watch for

• “A wheelbarrow just lifts the load straight up.” — The barrow turns about its wheel as you lift the handles: the wheel is the pivot.
• “Pulling a sledge uses a moment.” — The sledge moves along in a straight line; nothing turns about a pivot.

3. MQ3 · Practice

Model answer. All four pushes are the same size, so the biggest moment has the biggest perpendicular distance from the hinge. P acts at the handle at right angles to the door, so its perpendicular distance is the full width of the door. Q is at right angles too, but only half as far out, so its moment is half as big. R acts at the handle, but at a slant: its line of action passes closer to the hinge, so its perpendicular distance is shorter than P’s. S acts through the hinge, so its perpendicular distance is zero and it has no moment.

A full answer includes

• I chose P.
• P acts at the handle, at right angles to the door, so its perpendicular distance is the whole width of the door.
• R also acts at the handle, but at a slant, so its perpendicular distance is shorter; S acts through the hinge and has no moment.

Also accept

• P: at the handle and at right angles, so the largest perpendicular distance.

Watch for

• “R, because it acts at the handle too.” — R acts at the handle, but at a slant. Draw its line of action: the perpendicular distance from the hinge to that line is shorter than the width of the door.
• “They are all the same, because the pushes are the same size.” — A moment depends on the distance as well as the force. S, through the hinge, has no moment at all.

4. MQ4 · Practice

Model answer. The push is at right angles to the spanner, so the perpendicular distance is the distance from the nut to the hand, 20 cm = 0.20 m. Moment = force × perpendicular distance = 30 N × 0.20 m = 6.0 N m.

Checked automatically

• Moment: 6.0 N m (within ±0.05 N m)
  • Common wrong answer 600 : That is the moment in N cm. Change 20 cm to 0.20 m first: 30 N × 0.20 m = 6.0 N m.
  • Common wrong answer 150 : You divided. A moment is force × distance: 30 N × 0.20 m.
  • Common wrong answer 1.5 : You divided 30 by 20. Multiply the force by the distance in metres instead.
  • Common wrong answer any value J: A moment is measured in newton metres, N m, not joules. Joules measure energy.

A full answer includes

• The perpendicular distance is 20 cm = 0.20 m.
• Moment = 30 N × 0.20 m = 6.0 N m.

Also accept

• 6.0 N m

Watch for

• “600 N m.” — You used 20 cm without changing it to metres; that gives 600 N cm. In metres the distance is 0.20 m, so the moment is 6.0 N m.
• “150 N m, from 30 ÷ 0.20.” — A moment is the force multiplied by the distance, not divided by it.

5. MQ5 · Practice

Model answer. (1) turning: a moment measures the turning effect of a force. (2) perpendicular: the distance is measured from the pivot to the line of action, at right angles to it. (3) newton metres, N m: force in newtons times distance in metres. (4) anticlockwise: when a beam balances, the clockwise and anticlockwise moments are equal. (5) moment and (6) equilibrium: no resultant force and no resultant moment means the object is in equilibrium.

A full answer includes

• (1) turning
• (2) perpendicular
• (3) newton metres, N m
• (4) anticlockwise
• (5) moment
• (6) equilibrium

Also accept

• turning; perpendicular; newton metres (N m); anticlockwise; moment; equilibrium.

Watch for

• “(3) newtons.” — A moment is a force times a distance, so its unit is the newton metre, N m.
• “(5) force.” — Both are needed: no resultant force and no resultant moment. The second condition is about moments.

6. MQ6 · Practice

Model answer. The plank balances, so the clockwise moment equals the anticlockwise moment about the pivot. The 3.0 N weight gives an anticlockwise moment of 3.0 N × 0.40 m = 1.2 N m. W gives a clockwise moment of W × 0.30 m. So W × 0.30 m = 1.2 N m, and W = 1.2 N m ÷ 0.30 m = 4.0 N. W is closer to the pivot, so it must be bigger than 3.0 N.

Checked automatically

• Weight W: 4.0 N (within ±0.05 N)
  • Common wrong answer 3 : Equal weights would need equal distances. W is closer to the pivot, so it must be bigger: W × 0.30 m = 3.0 N × 0.40 m.
  • Common wrong answer 2.25 : You swapped the distances: 3.0 N is at 0.40 m and W at 0.30 m.
  • Common wrong answer 1.2 : 1.2 N m is the moment of the 3.0 N weight. Divide it by W’s distance, 0.30 m, to find W.

A full answer includes

• Anticlockwise moment = 3.0 N × 0.40 m = 1.2 N m.
• Clockwise moment = W × 0.30 m = 1.2 N m, so W = 1.2 N m ÷ 0.30 m = 4.0 N.

Also accept

• 4.0 N

Watch for

• “W = 3.0 N, because the plank balances.” — Balance needs equal moments, not equal weights. W is closer to the pivot, so it must be bigger.
• “W = 2.25 N.” — You swapped the distances. The 3.0 N weight is 0.40 m from the pivot and W is 0.30 m from it.

7. MQ7 · Practice

Model answer. For the seesaw to balance, the clockwise and anticlockwise moments about the pivot must be equal. The first child’s moment is 450 N × 1.2 m = 540 N m. The friend must give the same moment: 360 N × distance = 540 N m, so the distance is 540 N m ÷ 360 N = 1.5 m. The friend is lighter, so she sits further from the pivot.

Checked automatically

• Distance from the pivot: 1.5 m (within ±0.02 m)
  • Common wrong answer 0.96 : The lighter friend must sit further out than 1.2 m, not closer. Set 360 N × d = 450 N × 1.2 m.
  • Common wrong answer 1.2 : Equal distances balance only equal weights. The friend is lighter, so she must sit further out.
  • Common wrong answer 540 : 540 N m is the first child’s moment. Divide it by the friend’s weight, 360 N.

A full answer includes

• The first child’s moment = 450 N × 1.2 m = 540 N m.
• For balance, 360 N × distance = 540 N m, so the distance = 540 N m ÷ 360 N = 1.5 m.

Also accept

• 1.5 m

Watch for

• “0.96 m.” — You worked out 360 × 1.2 ÷ 450. The lighter friend must sit further out, not closer: 450 × 1.2 = 360 × d.
• “1.2 m, the same distance.” — At the same distance the heavier child’s side would go down. The lighter friend needs a bigger distance.

8. MQ8 · Practice

Model answer. Ayesha is wrong about the force: her hand pushes just as hard on a long spanner as on a short one. What changes is the perpendicular distance from the nut to where she pushes. The turning effect is the moment, force × perpendicular distance, so a longer handle gives a bigger moment for the same push. The nut needs a certain moment to undo it, and with a longer handle a smaller push gives that moment.

A full answer includes

• The force on the spanner is the same push from her hand; a longer handle does not make the force bigger.
• The moment is force × perpendicular distance from the nut, and a longer handle gives a bigger distance.
• So the same push gives a bigger moment, or the same moment needs a smaller push.

Also accept

• Wrong: the force is the same; the longer handle increases the perpendicular distance, so the moment (force × distance) is bigger, or the same moment needs a smaller force.

Watch for

• “A longer spanner is heavier, so it pushes harder.” — The spanner’s own weight is not what undoes the nut. Your push acts further from the nut, so its moment is bigger.
• “The longer spanner makes the force bigger.” — The force is your push, the same in both cases. The distance is bigger, and moment = force × distance.

9. MQ9 · Practice

Model answer. An object is in equilibrium when there is no resultant force and no resultant moment on it. The beam still has forces on it, its weight and the pivot’s push, but they cancel, and so do their moments. Equal forces alone are not enough: forces that are equal and opposite but not in line can still turn an object.

A full answer includes

• I chose c.
• Equilibrium needs both: no resultant force and no resultant moment.
• There are still forces on the beam: the weights pull down and the pivot pushes up.

Also accept

• c: the resultant force is zero and the resultant moment is zero.

Watch for

• “a: no forces act on a beam in equilibrium.” — The beam’s weight pulls it down and the pivot pushes it up. The forces cancel; they are not absent.
• “b: only the forces need to balance.” — Equal and opposite forces not in line, like two hands on a steering wheel, still turn an object. The moments must balance too.

10. MQ10 · Practice

Model answer. (a) The upward 20 N and the downward 20 N are equal and opposite, so there is no resultant force. (b) The forces are not in line. Pushing up on the left and pulling down on the right both turn the wheel clockwise, so their moments add and there is a resultant moment. (c) The wheel is not in equilibrium: no resultant force is not enough, because there is a resultant moment, and the wheel turns.

A full answer includes

• (a) No: 20 N up and 20 N down cancel.
• (b) Yes: both forces turn the wheel the same way (clockwise), so their moments add.
• (c) No: equilibrium needs no resultant force and no resultant moment; there is a resultant moment, so the wheel turns.

Also accept

• (a) no; (b) yes, both turn it the same way; (c) no, because there is a resultant moment.

Watch for

• “(c) Yes, because the forces are equal and opposite.” — Equal and opposite forces cancel only as forces. Here they act on opposite sides of the centre and turn the wheel the same way, so there is a resultant moment.
• “(b) No, the moments cancel because the forces are opposite.” — Up on the left and down on the right both turn the wheel clockwise. Their moments add, they do not cancel.

11. MQ11 · Practice

Model answer. (a) The plank is uniform and the pivot is at its middle, so its weight acts through the pivot and has no moment. The anticlockwise moment is 200 N × 1.5 m = 300 N m. For balance, 300 N × d = 300 N m, so d = 1.0 m. (b) The plank is in equilibrium, so there is no resultant force: the pivot pushes up as hard as everything pulls down, 200 N + 300 N + 50 N = 550 N.

Checked automatically

• (a) Distance of the 300 N child from the pivot: 1.0 m (within ±0.02 m)
  • Common wrong answer 2.25 : The heavier child sits closer, not further: 300 N × d = 200 N × 1.5 m.
  • Common wrong answer 1.5 : Equal distances would balance only equal weights. The heavier child sits closer.
• (b) Upward force of the pivot: 550 N (within ±1 N)
  • Common wrong answer 500 : You left out the plank’s own weight, 50 N.
  • Common wrong answer 250 : The pivot holds up both children and the plank: add all three weights.

A full answer includes

• (a) 200 N × 1.5 m = 300 N m, so 300 N × d = 300 N m and d = 1.0 m.
• The plank’s weight acts at its middle, through the pivot, so it has no moment.
• (b) No resultant force: the pivot pushes up with 200 N + 300 N + 50 N = 550 N.

Also accept

• (a) 1.0 m; (b) 550 N.

Watch for

• “(b) 500 N.” — You left out the plank. Its weight, 50 N, also pulls down and the pivot holds it up too.
• “(b) 0 N, because the plank is balanced.” — Balanced means the forces cancel. The pivot must push up as hard as the total weight pulls down.

12. MQ12 · Practice

Model answer. The plank balances, so the sum of the clockwise moments equals the sum of the anticlockwise moments about the pivot. Anticlockwise: 4.0 N × 0.50 m + 2.0 N × 0.20 m = 2.0 N m + 0.4 N m = 2.4 N m. Clockwise: 3.0 N × 0.40 m + F × 0.60 m = 1.2 N m + F × 0.60 m. Setting them equal, F × 0.60 m = 2.4 N m − 1.2 N m = 1.2 N m, so F = 2.0 N.

Checked automatically

• Weight F: 2.0 N (within ±0.05 N)
  • Common wrong answer 4 : You left out the 3.0 N weight on the right; its moment, 1.2 N m, counts towards the clockwise total.
  • Common wrong answer 3 : Equal total weights on both sides do not make it balance. Add the moments on each side instead.
  • Common wrong answer 1.2 : 1.2 N m is the moment F must give. Divide it by F’s distance, 0.60 m.

A full answer includes

• Sum of anticlockwise moments = 4.0 N × 0.50 m + 2.0 N × 0.20 m = 2.0 + 0.4 = 2.4 N m.
• Sum of clockwise moments = 3.0 N × 0.40 m + F × 0.60 m = 1.2 N m + F × 0.60 m.
• So F × 0.60 m = 1.2 N m and F = 2.0 N.

Also accept

• 2.0 N

Watch for

• “F = 4.0 N.” — You left out the 3.0 N weight on the right. Its moment, 1.2 N m, is part of the clockwise total.
• “F = 3.0 N, so the weights on each side add up to the same.” — Balance needs equal sums of moments, not equal sums of weights.

13. MQ13 · Practice

Model answer. Each child has their own moment about the pivot: 400 N × 1.2 m = 480 N m and 300 N × 0.60 m = 180 N m, so the anticlockwise moments add up to 660 N m. For balance, the adult’s clockwise moment must equal this sum: 600 N × d = 660 N m, so d = 660 N m ÷ 600 N = 1.1 m.

Checked automatically

• Distance of the adult from the pivot: 1.1 m (within ±0.03 m)
  • Common wrong answer 2.1 : Do not add the forces and the distances separately. Work out 400 × 1.2 and 300 × 0.60, then add the moments.
  • Common wrong answer 0.8 : You used only the 400 N child. Add the 300 N child’s moment too.
  • Common wrong answer 660 : 660 N m is the total moment on the left. Divide it by the adult’s weight.

A full answer includes

• Sum of anticlockwise moments = 400 N × 1.2 m + 300 N × 0.60 m = 480 + 180 = 660 N m.
• 600 N × d = 660 N m, so d = 1.1 m.

Also accept

• 1.1 m

Watch for

• “2.1 m, from (400 + 300) × (1.2 + 0.60) ÷ 600.” — Each force has its own distance. Work out each moment, then add the moments.
• “0.80 m.” — You left out the second child. Both children’s moments are on the left.

14. MQ14 · Practice

Options as numbered on the pupil sheet: 1 — Measure the distance from the pivot to each thread; 2 — Repeat with other masses and other positions; 3 — Balance the rule on its own on the pivot; 4 — Work out each weight × distance, and add up the clockwise and the anticlockwise moments; 5 — Slide one loop until the rule rests level; 6 — Hang masses from loops of thread on both sides of the pivot.

Model answer. She first balances the rule on its own, so that its weight acts through the pivot and has no moment. Then she hangs masses on loops on both sides and slides one loop until the rule rests level. With the rule level, she measures each distance from the pivot to a thread, works out each moment as weight × distance, and adds up the clockwise and the anticlockwise moments. The sums are equal. She repeats with other masses and positions to show it is always true.

A full answer includes

• Balance the rule on its own on the pivot.
• Hang masses from loops of thread on both sides of the pivot.
• Slide one loop until the rule rests level.
• Measure the distance from the pivot to each thread.
• Work out each weight × distance, and add up the clockwise and the anticlockwise moments.
• Repeat with other masses and other positions.

Also accept

• balance the rule alone; hang masses on loops; slide one until level; measure the distances; work out and add the moments; repeat.

Watch for

• “Measure the distances first, then make the rule level.” — The distances are read when the rule is level, so they are at right angles to the vertical weights. Level it first.
• “Balancing the rule on its own is not needed.” — If the pivot is not under the rule’s centre of gravity, the rule’s own weight has a moment that the masses must also balance.

15. MQ15 · Practice

Model answer. (a) Balancing the rule alone puts the pivot under its centre of gravity. Then the rule’s weight acts through the pivot and has no moment, so only the masses’ moments need comparing. (b) The weights act vertically. When the rule is level, the distance along it from the pivot to each thread is at right angles to the weight, so it is the perpendicular distance. (c) The sum of the clockwise moments equals the sum of the anticlockwise moments, so they cancel and there is no resultant moment.

A full answer includes

• (a) Then the pivot is under the rule’s centre of gravity, so the rule’s own weight has no moment about the pivot.
• (b) When the rule is level, each distance along it is at right angles to the vertical weights: it is the perpendicular distance.
• (c) The sum of the clockwise moments equals the sum of the anticlockwise moments (within the accuracy of the readings).

Also accept

• (a) its weight then acts through the pivot, so it has no moment; (b) so the distances along the rule are perpendicular to the weights; (c) sum of clockwise moments = sum of anticlockwise moments.

Watch for

• “(b) So that the masses do not slide off.” — The loops stay on the rule either way. Level matters because the distances along the rule are then at right angles to the vertical weights.
• “(c) The rule does not move, so there are no forces on it.” — There are forces: the weights and the pivot’s push. The evidence is that the clockwise and anticlockwise moments add up to the same.

16. MT1 · Apply

Model answer. (a) Looking from the front, the sack hangs on the left, so its weight turns the bar anticlockwise about the hook; the slider hangs on the right and turns it clockwise. (b) The bar is level, so the moments about the hook balance. The slider’s clockwise moment is 20 N × 0.60 m = 12 N m. The sack’s anticlockwise moment must also be 12 N m: W × 0.05 m = 12 N m, so W = 12 N m ÷ 0.05 m = 240 N. The sack is so close to the hook that a small slider balances a big weight. (c) The bar is in equilibrium, so there is no resultant force: the hook’s upward pull equals all the weights on the bar, 240 N + 20 N + the bar’s own 15 N = 275 N. (d) Now two weights turn the bar anticlockwise, and their moments add: 240 N × 0.05 m + 30 N × 0.10 m = 12 N m + 3 N m = 15 N m. The slider must give the same clockwise moment: 20 N × d = 15 N m, so d = 0.75 m. (e) First balance the ruler on its own on the pencil, so its weight acts through the pivot and has no moment. Put a stack of coins on each side, say two stacks on the left and one on the right, and slide one stack until the ruler rests level. Measure the distance from the pencil to the middle of each stack. All the coins weigh the same, so each moment is the number of coins × its distance, in coin cm. Add the moments on each side: the clockwise and anticlockwise totals come out equal, so there is no resultant moment. Repeat with other stacks and positions.

A full answer includes

• (a) Looking from the front, the sack on the left turns the bar anticlockwise and the slider on the right turns it clockwise.
• (b) 20 N × 0.60 m = W × 0.05 m, so W = 12 N m ÷ 0.05 m = 240 N.
• (c) No resultant force: the hook pulls up with 240 N + 20 N + 15 N = 275 N.
• (d) Anticlockwise: 240 N × 0.05 m + 30 N × 0.10 m = 12 + 3 = 15 N m = 20 N × d, so d = 0.75 m.
• (e) Balance the ruler on the pencil on its own first; put stacks of coins on both sides and move one until the ruler is level; for each stack, number of coins × distance from the pencil; the totals on the two sides are equal; repeat with other stacks and positions.

Also accept

• (a) the sack turns it anticlockwise, the slider clockwise; (b) 240 N; (c) 275 N; (d) 0.75 m; (e) balance the ruler alone on the pencil, put coins on both sides and move one stack until level, then the number of coins × distance adds up to the same on both sides; repeat.

Watch for

• “(b) 1.67 N, from 20 × 0.05 ÷ 0.60.” — The sack is much nearer the hook than the slider, so it must be much heavier than the 20 N slider, not lighter.
• “(c) 260 N.” — That leaves out the bar’s own weight. The hook holds up the bar as well as the sack and the slider: add its 15 N.
• “(d) 0.15 m.” — The sack is still on the bar. Its moment, 12 N m, adds to the bag’s 3 N m on the left.