Free Turning effect & centre of gravity questions for IGCSE Physics (0625), part of Motion, forces & energy. Each question has the answer and the marking points an examiner would look for.
Generate a full Turning Effect & Centre of Gravity worksheet →All Motion, forces & energy questionsA tall, empty lorry is parked on a slope.
1 A wider wheelbase makes the lorry less likely to topple.
2 The lorry topples when the line of action of its weight passes outside its base.
3 The lorry topples if its centre of gravity is directly above the base.
Which statements are correct?
Answer: B
A uniform plank of length 4.0 m and weight 200 N rests on two supports, A and B, one at each end. A man of weight 600 N stands 1.5 m from A.
What are the upward forces on the plank at A and at B?
| force at A / N | force at B / N | |
|---|---|---|
| A | 400 | 400 |
| B | 325 | 475 |
| C | 475 | 325 |
| D | 375 | 225 |
Answer: C
Fig. 3.1 shows a uniform beam pivoted at its centre. It is in equilibrium.
State the principle of moments.
Calculate the weight W.
The beam weighs 6.0 N. Calculate the force exerted on the beam by the pivot.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | For equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments about the same point Sum of clockwise moments = sum of anticlockwise moments B1 about a point, for an object in equilibrium B1 | 2 |
| (b) | 9 N moment = force × perpendicular distance C1 3 × 30 = W × 10 C1 9 A1 | 3 |
| (c) | 18 N upwards Upward force = total downward force C1 18 A1 | 2 |
Fig. 4.1 shows a uniform plank of length 3.0 m and weight 100 N resting on two supports, P at the left end and Q at the right end. A load of weight 400 N is placed 0.75 m from P.
Define the moment of a force.
State the two conditions for an object to be in equilibrium.
State where the weight of the plank acts.
By taking moments about P, calculate the upward force exerted by support Q.
Use your answer to (d) to calculate the upward force exerted by support P.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | force × perpendicular distance from pivot force × perpendicular distance from the pivot B1 | 1 |
| (b) | no resultant force; no resultant moment resultant force is zero B1 resultant moment (about any point) is zero B1 | 2 |
| (c) | at its centre (midpoint) at its centre of gravity, 1.5 m from P B1 | 1 |
| (d) | 150 N moments about P: 400 × 0.75 + 100 × 1.5 C1 = F_Q × 3.0 C1 F_Q = 150 N A1 | 3 |
| (e) | 350 N 400 + 100 − 150 = 350 N B1 | 1 |
A wheelbarrow is used to lift a load of 600 N. The load is 0.40 m from the wheel axle and the handles are 1.2 m from the axle.
Calculate the moment of the load about the axle.
Calculate the upward force needed at the handles.
Explain why it is easier to lift the load if it is placed closer to the wheel.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 240 N m 240 N m B1 | 1 |
| (b) | 200 N F × 1.2 = 240 C1 200 N A1 | 2 |
| (c) | smaller moment the moment of the load is smaller, so a smaller force is needed B1 | 1 |
A tall vase has a narrow base. A similar vase has a wide base.
Explain, in terms of centre of gravity, why the vase with the wide base is more stable.
Suggest one other way to make the vase more stable.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | centre of gravity stays over the base a wide base means the centre of gravity stays above the base when tilted further B1 the weight then produces a moment that returns it to its original position B2 | 3 |
| (b) | lower the centre of gravity put a heavy weight in the bottom / lower the centre of gravity B1 | 1 |
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