Turning effect of forces asks whether an object will turn, not only whether it will move. The moment of a force about a pivot is the product of the force and the perpendicular distance from the pivot to the line of action of the force, and every moment is either clockwise or anticlockwise.
The topic covers the moment of a force, the principle of moments and the two conditions for equilibrium, centre of gravity and stability. The idea of a moment underpins every lever, beam, see-saw and stability question in O-Level Paper 1 and Paper 2, so the definitions and formulas must be known exactly.
What these notes cover
- •Moment of a force
- •Clockwise and anticlockwise moments
- •Forces not at right angles to the rod
- •Principle of moments
- •The two conditions for equilibrium
- •A 4-step method for equilibrium problems
- •Worked examples: rules, planks and trapdoors
- •Centre of gravity
- •Centre of gravity of an irregular lamina
- •Stability and the three types of equilibrium
- •How to increase stability
- •Common mistakes and exam technique
Key definitions
- Moment of a force
- The turning effect of the force about a pivot. It is equal to the product of the force and the perpendicular distance from the pivot to the line of action of the force.
- Pivot
- The fixed point about which an object is free to rotate. It is also called the fulcrum or the hinge.
- Line of action of a force
- The straight line drawn along the direction of the force, extended as far as needed in both directions.
- Centre of gravity
- The point through which the whole weight of the body may be taken to act.
- Stability
- A measure of the ability of a body to return to its original position after it has been tilted slightly.
Key formulas
Moment of a force
M = F × d
M is the moment in N m, F is the force in N and d is the perpendicular distance in m from the pivot to the line of action of the force. Convert cm to m first unless the answer is asked for in N cm.
Principle of moments
sum of clockwise moments = sum of anticlockwise moments
Applies only when the body is in equilibrium. Add up all clockwise and all anticlockwise moments, every one taken about the same point.
Zero resultant force
sum of upward forces = sum of downward forces
The other condition for equilibrium. Include the upward force from the pivot or each support, even if it is not drawn in the question diagram.
Weight
W = m × g
W is the weight in N, m is the mass in kg and g is the gravitational field strength. Take g = 10 N/kg unless stated. Never put a mass in kg into a moment calculation.
Perpendicular distance for a force at an angle
d = L × sin θ
L is the distance along the rod from the pivot to where the force is applied and θ is the angle the force makes with the rod. d is smaller than L unless θ = 90°.
Moment of the weight about a pivot
moment = W × d
W is the weight in N and d is the perpendicular distance in m from the pivot to the vertical line through the centre of gravity. For a uniform rod, the weight acts at its midpoint.
Worked example
A uniform plank AB of length 3.0 m and weight 150 N rests horizontally on two supports, one at end A and one at end B. A box of mass 40 kg sits on the plank 1.0 m from A. Take g = 10 N/kg. Find the force exerted on the plank by each support.
- 1Weight of the box: W = mg = 40 × 10 = 400 N, acting downward 1.0 m from A.
- 2The plank is uniform, so its weight of 150 N acts downward at its centre, 1.5 m from A; the supports push upward with RA at A and RB at B, 3.0 m from A.
- 3Taking moments about A removes RA, because RA acts at A and has zero perpendicular distance from A; the two weights give clockwise moments and RB gives an anticlockwise moment.
- 4Sum of clockwise moments about A = 400 × 1.0 + 150 × 1.5 = 400 + 225 = 625 N m.
- 5By the principle of moments, RB × 3.0 = 625, so RB = 625 ÷ 3.0 = 208 N upward (210 N to 2 s.f.).
- 6By zero resultant force, RA + RB = 400 + 150 = 550 N, so RA = 550 − 208 = 342 N upward (340 N to 2 s.f.).
- 7Check by taking moments about B: RA × 3.0 = 400 × 2.0 + 150 × 1.5 = 1025 N m, so RA = 1025 ÷ 3.0 = 342 N, which agrees; A carries more of the load because the box is nearer to A.
Watch out: A force that acts at the point you take moments about has zero moment about that point, but it is still a force and must be included when you apply zero resultant force.
Common mistakes
Using the distance along the rod instead of the perpendicular distance
Wrong: The force acts at the end of a 2.0 m rod, so d = 2.0 m, even when the rod is tilted or the force is applied at an angle.
Right: d is the shortest distance from the pivot to the line of action of the force, measured at 90° to that line. The distance along the rod is correct only when the force is at right angles to the rod.
Forgetting the weight of a uniform beam
Wrong: Only the hanging loads and applied forces produce moments.
Right: If the beam is uniform and its weight or mass is given, its weight acts downward at the midpoint. Its moment is zero only if the pivot is exactly at the midpoint; leave it out only if the beam is light or of negligible weight.
Giving an incomplete statement of the principle of moments
Wrong: Clockwise moments equal anticlockwise moments.
Right: For a body in equilibrium, the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point.
Questions students ask
What is the principle of moments?
When a body is in equilibrium, the sum of clockwise moments about any point is equal to the sum of anticlockwise moments about the same point. It applies only when the body is balanced, and every moment in one equation must be taken about the same point. Because it works about any point, taking moments where an unknown force acts removes that force from the equation.
What are the two conditions for equilibrium in physics?
The resultant force on the body is zero, and the resultant moment about any point is zero. The first means the sum of upward forces equals the sum of downward forces and leftward forces balance rightward forces; the second is the principle of moments. Each condition gives one equation, so together they can find two unknown forces, such as the forces at two supports.
How do you find the centre of gravity of an irregular lamina?
Hang the lamina freely from a pin through a small hole, hang a plumb line from the same pin and mark the line of the string on the lamina once both have stopped swinging. Repeat from a second hole; the centre of gravity is where the lines intersect. A line from a third hole is a check and should pass through the same point.
How do you make an object more stable?
Lower the centre of gravity by placing as much of the mass as possible near the bottom, and widen the base area. Both changes let the body be tilted through a larger angle before the line of action of its weight falls outside its base. Saying an object is stable because it is “heavy” or “big” does not earn the mark.
In the full PDF
- •Labelled diagrams of a pivoted rod showing each force and its perpendicular distance from the pivot
- •Two further worked examples: a metre rule pivoted off-centre and a hinged trapdoor
- •The full plumb-line experiment with apparatus, why the method works and the precautions to quote
- •A table comparing stable, unstable and neutral equilibrium, with worked explanation answers on a traffic cone and a loaded lorry
- •The remaining common mistakes: mixing cm and m, using mass instead of weight, and stopping after the moments equation
- •An exam technique checklist and a five-point summary of the whole topic