Answers · Moments and balance
Moments and balance — return questions key
Check meaning, not exact wording. These are teaching points, not official Cambridge marks.
1. MR1 · Return
Model answer. (a) For example, a spanner turning a nut, or a door turning about its hinges. (b) The 0.60 N weight is 50 − 20 = 30 cm = 0.30 m from the pivot, so its moment is 0.60 N × 0.30 m = 0.18 N m. The 0.90 N weight must give 0.18 N m too: d = 0.18 N m ÷ 0.90 N = 0.20 m, which is the 70 cm mark. The rule’s weight acts at the 50 cm mark, through the pivot, so it has no moment. (c) The pivot pushes up as hard as everything pulls down: 0.60 N + 0.90 N + 1.0 N = 2.5 N. (d) There must be no resultant force and no resultant moment. (e) The clockwise moments now add up to 0.90 N × 0.20 m + 0.15 N × 0.40 m = 0.18 N m + 0.06 N m = 0.24 N m. The 0.60 N weight must give 0.24 N m: d = 0.24 N m ÷ 0.60 N = 0.40 m to the left of the pivot, the 10 cm mark. (f) When the rule balances level, the clockwise moments add up to the same as the anticlockwise moments: 0.18 N m each way in (b), 0.24 N m each way in (e). They cancel, so there is no resultant moment.
A full answer includes
- (a) Any sensible example: opening a door, a spanner on a nut, a seesaw, a wheelbarrow, a tap.
- (b) 0.60 N × 0.30 m = 0.18 N m = 0.90 N × d, so d = 0.20 m: the 70 cm mark.
- (c) 0.60 N + 0.90 N + 1.0 N = 2.5 N.
- (d) No resultant force and no resultant moment.
- (e) Clockwise: 0.90 N × 0.20 m + 0.15 N × 0.40 m = 0.18 + 0.06 = 0.24 N m = 0.60 N × d, so d = 0.40 m: the 10 cm mark.
- (f) Each time the rule balances, the sum of the clockwise moments equals the sum of the anticlockwise moments, so they cancel.
Also accept
- (a) e.g. a door or a spanner; (b) the 70 cm mark (0.20 m from the pivot); (c) 2.5 N; (d) no resultant force and no resultant moment; (e) the 10 cm mark; (f) the sums of clockwise and anticlockwise moments are equal.
Watch for
- “(b) The 20 cm mark on the other side, 0.20 m from the end.” — Distances are measured from the pivot, not from the end of the rule. 0.20 m from the pivot is the 70 cm mark.
- “(c) 1.5 N.” — You left out the rule’s own weight, 1.0 N. The pivot holds it up too.
2. MR2 · Return
Model answer. (a) The centre of gravity is the point through which the whole weight of an object seems to act. (b) Make holes near the edge of the card, well apart. Hang the card from a pin through one hole so that it swings freely, hang a plumb line from the pin, and when everything is still, draw a line on the card along the thread. Repeat from a second hole. The centre of gravity is where the lines cross; a line from a third hole checks it. (c) The water in the bottom lowers the centre of gravity of the bottle. When it is tipped, the vertical line from its centre of gravity stays inside the base for a bigger tip, so the weight turns it back.
A full answer includes
- (a) The point through which the whole weight of the object seems to act.
- (b) Hang the card from a pin through a hole near its edge, with a plumb line from the pin; when still, draw a line along the thread; repeat from a second hole; the centre of gravity is where the lines cross (check with a third hole).
- (c) The water lowers the centre of gravity, so the bottle can be tipped further before the vertical line from it passes outside its base.
Also accept
- (a) where the whole weight seems to act; (b) hang from two holes with a plumb line, draw the lines, the crossing point; (c) the water lowers the centre of gravity.
Watch for
- “(c) The bottle is heavier, so it is harder to move.” — It is where the extra weight is that matters: low down, it lowers the centre of gravity.
- “(b) Balance the card on a ruler and mark the middle.” — For an irregular card, hang it from two holes and find where the plumb lines cross.