O-Level Physics · Secondary 3 and 4
Turning Effect of Forces
A force does not only push an object along, it can also make it turn about a pivot. This upper secondary topic covers the moment of a force, the principle of moments for objects in equilibrium, the centre of gravity and stability. The ideas explain why spanners have long handles, how a see-saw balances, why a crane needs a counterweight and why racing cars are built low and wide. Many questions combine moments with the forces work from dynamics.
What the syllabus expects
- Describe the moment of a force as a measure of its turning effect and give everyday examples
- Calculate the moment of a force as force × perpendicular distance from the pivot to the line of action of the force
- State and apply the principle of moments for a body in equilibrium
- State that a body is in equilibrium when there is no resultant force and no resultant moment
- Describe the centre of gravity and find it for a flat plate using a plumb line
- Describe how the position of the centre of gravity and the size of the base affect stability
The moment of a force
The moment of a force about a pivot is its turning effect, calculated as moment = force × perpendicular distance from the pivot to the line of action of the force. The unit is the newton metre (N m). A force of 40 N applied at right angles to a spanner 0.25 m from the nut gives a moment of 40 × 0.25 = 10 N m. A longer spanner gives a larger moment for the same force, which is why stiff nuts are easier to loosen with long handles and why door handles are placed far from the hinges. If the force is not at right angles to the lever, you must use the perpendicular distance to its line of action, which is shorter than the distance along the lever. A force whose line of action passes through the pivot has zero moment.
The principle of moments
The principle of moments states that when a body is in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point. For full equilibrium there must also be no resultant force, so for a balanced beam with only vertical forces, the upward force at the pivot equals the total downward force. Consider a see-saw with a child of weight 400 N sitting 1.5 m to the left of the pivot. Where must a child of weight 600 N sit to balance it? Anticlockwise moment = 400 × 1.5 = 600 N m. Clockwise moment = 600 × d. Setting them equal gives d = 1.0 m to the right of the pivot. The heavier child sits closer to the pivot.
Centre of gravity
The centre of gravity of an object is the point through which its whole weight appears to act. For a uniform ruler or plank it is at the middle, so in moment calculations the beam's own weight is drawn as a single arrow at its centre. To find the centre of gravity of an irregular flat plate (a lamina), hang it freely from a small hole near its edge together with a plumb line, and mark the vertical line of the plumb line on the plate. Repeat from a second hole. The centre of gravity is where the two lines cross, and a third hole can be used as a check. The centre of gravity does not have to lie inside the material. For a ring it is at the empty centre.
Stability
An object is in stable equilibrium if, when tilted slightly, it returns to its original position. It is in unstable equilibrium if a small tilt makes it fall further, like a pencil balanced on its point, and in neutral equilibrium if it stays in its new position, like a ball on a flat floor. A tilted object topples when the line of action of its weight falls outside its base. While the line of action is still inside the base, the weight produces a moment that turns the object back. To make an object more stable, lower its centre of gravity and widen its base. That is why racing cars are low and wide, why Bunsen burners and table lamps have heavy, broad bases, and why heavy loads are best stored low down in a cupboard or vehicle.
Worked example
A uniform metre rule of weight 1.5 N is pivoted at its 30 cm mark. A load of unknown weight W hangs from the 5 cm mark, and the rule balances horizontally. Calculate W and the upward force exerted by the pivot on the rule.
- The rule is uniform, so its weight acts at its centre, the 50 cm mark, which is 50 − 30 = 20 cm = 0.20 m to the right of the pivot.
- Clockwise moment from the rule's weight = 1.5 × 0.20 = 0.30 N m.
- The load is 30 − 5 = 25 cm = 0.25 m to the left of the pivot, so its anticlockwise moment = W × 0.25.
- By the principle of moments, W × 0.25 = 0.30, so W = 0.30 ÷ 0.25 = 1.2 N.
- For no resultant force, the upward pivot force equals the total downward force: 1.2 + 1.5 = 2.7 N.
Answer: W = 1.2 N and the pivot pushes up on the rule with 2.7 N.
Common mistakes
- Using the distance along a lever when the force acts at an angle. The moment needs the perpendicular distance from the pivot to the line of action of the force.
- Forgetting the weight of the beam itself. Unless the question says the beam is light, its weight acts at its centre of gravity and has a moment about the pivot.
- Giving a moment in N cm when distances were in cm but the question expects N m. Convert distances to metres before multiplying.
- Taking moments about a point and then including the force acting at that point. A force through the pivot has zero moment about it, which is why choosing the pivot at an unknown force removes it from the equation.
- Saying an object topples when its centre of gravity is outside the base. The correct condition is that the vertical line of action of the weight falls outside the base.
Exam tips
- When there is an unknown force at a support, take moments about that support. The unknown then has zero moment and drops out of the equation.
- Write one clear line, sum of clockwise moments = sum of anticlockwise moments, and label each term with its force and distance before putting in numbers.
- In stability answers, describe where the line of action of the weight falls relative to the base and whether the moment it produces restores the object or tips it over.
Practice questions
1. Easy: A mechanic applies a force of 25 N at right angles to the end of a spanner 0.40 m long. Calculate the moment about the nut.
Show answer
Moment = force × perpendicular distance = 25 × 0.40 = 10 N m.
2. Medium: A uniform plank 4.0 m long and weighing 200 N is pivoted at its centre. A man weighing 300 N stands 1.2 m to the left of the pivot. Where must a woman weighing 450 N stand for the plank to balance?
Show answer
0.80 m to the right of the pivot. The plank's weight acts through the pivot, so it has no moment. Anticlockwise moment = 300 × 1.2 = 360 N m. Clockwise moment = 450 × d. So d = 360 ÷ 450 = 0.80 m.
3. Harder: A uniform beam 2.0 m long weighing 80 N rests on two supports, A at its left end and B at its right end. A load of 120 N is placed 0.50 m from A. Find the upward force from each support.
Show answer
A gives 130 N and B gives 70 N. Take moments about A to remove its force: B × 2.0 = 120 × 0.50 + 80 × 1.0 = 60 + 80 = 140, so B = 70 N. Total downward force = 120 + 80 = 200 N, so A = 200 − 70 = 130 N. Check about B: 130 × 2.0 = 260 and 120 × 1.5 + 80 × 1.0 = 260.
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