The moment of a force about a pivot, using the perpendicular distance to the line of action

Moments of forces that act at an angle

The principle of moments and the two conditions for a body in equilibrium

Centre of gravity, and free-body diagrams of beams and planks with the weight drawn at the centre of gravity

Beams on two supports, see-saws and levers, cranes, and a beam held by a cable at an angle

Finding the centre of gravity of a rod or a flat sheet (lamina) by balancing and by suspension

Stability and toppling, explained with moments

1. A force can make a body turn as well as move: its turning effect, the moment, is force × perpendicular distance from the pivot to the line of action, measured in N m.

2. For a force at angle θ to a lever of length r: moment = Frsin θ — use the perpendicular component of the force, or the perpendicular distance.

3. Principle of moments: for a body in equilibrium, the sum of clockwise moments about any point = the sum of anticlockwise moments about that point.

4. Equilibrium also needs zero resultant force: upward forces = downward forces.

5. The centre of gravity is the point where the whole weight can be taken to act. For a uniform beam it is at the midpoint.

6. Take moments about the point where an unknown force acts — that force then has no moment and drops out.

7. An object topples when the line of action of its weight passes outside its base.

The line of action is the straight line through the point of application, along the direction of the force, extended as far as you need in both directions.

A moment measures the turning effect of a force. It has a sense of rotation: clockwise or anticlockwise about the chosen point.

A force whose line of action passes through the pivot has zero moment about it, however large the force.

The same force has a larger moment further from the pivot — hence door handles far from hinges and long spanners.

Moments are taken about a stated point. The value depends on which point you choose.

Only the perpendicular distance to the line of action counts. When a force is not at 90° to the lever, that distance is shorter than the length of the lever. For a force F applied at distance r from the pivot, at angle θ to the lever, there are two equivalent methods:

Perpendicular distance: x = rsin θ, so moment = F ×rsin θ.

Perpendicular component: resolve F; the component Fsin θ is at 90° to the lever and has moment (Fsin θ) ×r. The component Fcos θ acts along the lever, through the pivot, so its moment is zero.

Both give Frsin θ. The moment is largest when the force is at 90° to the lever (sin 90° = 1).

WORKED EXAMPLE · QUESTION

A mechanic pulls on the end of a spanner 0.240 m long with a force of 45 N. Calculate the moment about the centre of the nut (a) when the force is at 90° to the spanner and (b) when the same force is at 60° to the spanner, as in the figure.

SOLUTION

(a) The force is perpendicular, so the distance is the full length: moment = 45 ×0.240 = 10.8 N m

(b) Perpendicular distance to the line of action: x = 0.240 ×sin 60° = 0.208 m. Moment = 45 ×0.2078 = 9.35 N m

Check with the component method: 45sin 60° = 39.0 N at 90° to the spanner, and 38.97 ×0.240 = 9.35 N m.

A body in equilibrium is at rest or moving at constant velocity without starting to rotate. Both conditions must hold at the same time.

A reliable method

1. Draw a free-body diagram: every force on the body, at the point where it acts.

2. Choose a point to take moments about — usually where an unknown force acts.

3. For each force, write force × perpendicular distance and label it clockwise or anticlockwise.

4. Equate the totals, solve, then use "up = down" for any remaining unknown force.

WORKED EXAMPLE · QUESTION

A uniform see-saw is pivoted at its centre. Its weight is 150 N. Child A, of weight 320 N, sits 1.50 m to the left of the pivot. Child B, of weight 410 N, sits on the right. Calculate (a) how far from the pivot child B must sit for the see-saw to balance horizontally and (b) the force from the pivot on the see-saw.

SOLUTION

(a) Moments about the pivot, clockwise = anticlockwise: 410 ×x = 320 ×1.50, so x = 480 ÷410 = 1.17 m

(b) Resultant force is zero, so up = down: pivot force = 320 + 410 + 150 = 880 N upwards

The products of force and distance balance, not the forces.

Levers

A lever uses a small force at a large distance from the pivot to balance a large force close to it. The pivot need not be between the two forces: in a wheelbarrow the wheel is the pivot and both the load and the lift are on the same side of it.

WORKED EXAMPLE · QUESTION

A loaded wheelbarrow has a total weight of 620 N. Its centre of gravity is 0.45 m horizontally from the axle of the wheel. The gardener holds the handles 1.40 m horizontally from the axle and lifts vertically so the barrow is just off the ground at the back. Calculate the upward force the gardener applies and the force of the ground on the wheel.

SOLUTION

Step 1 Moments about the axle (the ground force on the wheel acts there): F ×1.40 = 620 ×0.45, so F = 199 N

Step 2 Up = down for the wheel force: R = 620 - 199.3 = 421 N

The handles are about three times as far from the axle as G, so the gardener lifts less than a third of the weight.

Every small part of a body has weight; at G one force — the total weight — has the same turning effect as all those small weights together.

A body is uniform when its mass is evenly spread. The centre of gravity of a uniform rod, beam or plank is at its midpoint; for a uniform rectangular sheet it is where the diagonals cross.

For a non-uniform body (a cricket bat, a tree trunk, a loaded van) the centre of gravity is nearer the heavier end, and it is often found by experiment.

The centre of gravity need not lie inside the material — for a ring or a boomerang it is in the space between.

In a uniform gravitational field the centre of gravity is at the same point as the centre of mass.

Free-body diagrams of beams and planks

Draw the body as a simple shape with the weight as one arrow down from the centre of gravity, and each contact force, labelled, where the contact is. Show only forces on the body, and mark the distances between the lines of action.

With two supports there are two unknown forces. Take moments about one support: its force has no moment there, so the equation has one unknown. Then use up = down for the other support force — or take moments about the second support as an independent check. The support nearer to a heavy load carries the larger force.

WORKED EXAMPLE · QUESTION

A uniform plank of length 4.80 m and weight 360 N rests on two trestles: A at one end and B 3.60 m from A, so 1.20 m of the plank overhangs beyond B. A painter of weight 750 N stands 3.00 m from A. (a) Calculate the forces on the plank from the trestles A and B. (b) The painter walks towards the overhanging end. Determine how far beyond B he can stand before the plank starts to tip.

SOLUTION

(a) Moments about A to remove R_A: anticlockwise R_B ×3.60 = clockwise 360 ×2.40 + 750 ×3.00 = 864 + 2250. So R_B = 3114 ÷3.60 = 865 N. Up = down: R_A = 360 + 750 - 865 = 245 N

(b) On the point of tipping the plank lifts off A, so R_A = 0. Moments about B: 750 ×d = 360 ×(3.60 - 2.40), so d = 432 ÷750 = 0.58 m beyond B

He can reach about halfway along the 1.20 m overhang.

A tower crane's long arm (the jib) carries the load, balanced by a heavy counterweight on a shorter arm on the other side of the tower, so the resultant moment about the tower is small. The further out the load, the smaller the load the crane can safely lift. The tower still has to support the total weight: moments cancel, but forces add.

WORKED EXAMPLE · QUESTION

The jib of a tower crane has a counterweight of 54 kN acting 7.5 m from the axis of the tower. The weight of the jib is 30 kN, acting 4.0 m from the axis on the load side. The load hangs 25 m from the axis. Calculate (a) the load for which the resultant moment about the tower axis is zero and (b) the vertical force from the tower on the jib with this load.

SOLUTION

(a) Moments about the tower axis: anticlockwise 54 ×7.5 = 405 kN m; clockwise 30 ×4.0 + L ×25 = 120 + 25L. So L = (405 - 120) ÷25 = 11.4 kN

(b) Up = down: N = 54 + 30 + 11.4 = 95.4 kN

Without a load the resultant moment is 285 kN m anticlockwise — a crane is designed for a range of moments either way.

A beam hinged to a wall and held by a cable (signs, shelves, drawbridges) has three unknowns: the tension and two components of the hinge force.

Take moments about the hinge first. The hinge force has no moment there, so the tension is the only unknown.

The moment of the tension uses its component perpendicular to the beam, Tsin θ, where θ is the angle between the cable and the beam.

Then resolve: horizontal forces balance (the hinge pushes outwards against Tcos θ), and vertical forces balance.

WORKED EXAMPLE · QUESTION

A uniform horizontal beam of length 1.50 m and weight 80 N is hinged to a wall at one end. A sign of weight 120 N hangs from the beam 1.20 m from the wall. A cable from the far end of the beam to a point higher up the wall makes an angle of 35° with the beam. (a) Calculate the tension in the cable. (b) Determine the magnitude and direction of the force from the hinge on the beam.

SOLUTION

(a) Moments about the hinge: Tsin 35° ×1.50 = 80 ×0.75 + 120 ×1.20 = 204 N m. So T = 204/(1.50 ×sin 35°) = 237 N

(b) Horizontal forces balance: H = Tcos 35° = 237.1 ×cos 35° = 194 N, away from the wall. Vertical forces balance: V = 80 + 120 - Tsin 35° = 200 - 136 = 64 N, upwards. Combine: R = √(194² + 64²) = 205 N at tan⁻¹(64 ÷194) = 18°. 205 N at 18° above the horizontal

A rod or beam — balancing

Balance the rod on a knife-edge (or a narrow pencil) and move it until it stays horizontal. The centre of gravity is directly above the knife-edge, because the only force with a moment about that point would be the weight. Repeat a few times, approaching the balance point from both sides, and take the mean position. Hanging a known mass from the rod and balancing again gives the rod's mass by moments.

WORKED EXAMPLE · QUESTION

A non-uniform rod balances on its own on a knife-edge at the 42.0 cm mark of its scale. A 200 g mass is hung at the 5.0 cm mark, and the rod now balances with the knife-edge at the 30.0 cm mark. Determine the mass of the rod.

SOLUTION

Step 1 The rod's weight acts at its centre of gravity, the 42.0 cm mark. Distances from the new pivot: hung mass 30.0 - 5.0 = 25.0 cm; centre of gravity 42.0 - 30.0 = 12.0 cm.

Step 2 Principle of moments (g cancels on both sides): Mg ×0.120 = 0.200g ×0.250, so M = 0.0500 ÷0.120 = 0.417 kg

Each distance is the difference of two readings, so its uncertainty is the sum of two reading uncertainties — keep both distances large.

A flat sheet (lamina) — suspension

1. Make two or three small holes near the edge of the card, well apart.

2. Hang the card from a pin through one hole so that it can swing freely. Hang a plumb line (a thread with a small weight) from the same pin.

3. When the card is at rest, mark the plumb line on the card with two well-separated dots, then join them with a ruler.

4. Repeat from a second hole. The centre of gravity is where the lines cross. A line from a third hole should pass through the same point — a useful check.

When an object is tilted, it pivots about one edge of its base. Its weight, acting at G, now has a moment about that edge. If the vertical line through G still falls inside the base, the moment of the weight turns the object back: it is stable. If the vertical line through G falls outside the base, the moment turns it further over: it topples.

An object is more stable — it can be tilted further before toppling — when its centre of gravity is low and its base is wide.

Racing cars are low and wide; a lorry with a high load is easier to tip; a desk lamp has a heavy, wide base.

2 Forces and Newton's laws

The "zero resultant force" half of equilibrium is Newton's first law; resolving forces into perpendicular components is the same skill used for slopes and for vectors.

Materials (Unit 1) and practical skills (Unit 3)

Moments give the tension in a wire or the force on a spring that supports a beam; from that force you can find the extension using Hooke's law or the Young modulus. Balancing experiments are also good practice in handling uncertainty.

Magnetic fields (Unit 4)

The forces on the two sides of a current-carrying coil in a magnetic field are equal and opposite but act along different lines, so together they produce a turning effect — the basis of an electric motor.

1. A force of 25 N, at 90° to a door, is applied 0.80 m from the hinges. Calculate its moment about the hinges. What force at 0.20 m from the hinges would have the same moment?

2. A 60 N force acts on the end of a 0.50 m lever at 30° to the lever. Calculate its moment about the pivot.

3. State the two conditions for an extended body to be in equilibrium.

4. A uniform pole 2.0 m long, weight 40 N, rests on supports at its two ends. A 100 N load hangs 0.50 m from the left end. Find the two support forces.

5. Explain why a racing car is built with a low centre of gravity and a wide wheelbase.

1. 25 ×0.80 = 20 N m. Same moment at 0.20 m: 20 ÷0.20 = 100 N — four times the force at a quarter of the distance.

2. Perpendicular distance = 0.50sin 30° = 0.25 m; moment = 60 ×0.25 = 15 N m.

3. The resultant force on it is zero, and the resultant moment about any point is zero (sum of clockwise moments = sum of anticlockwise moments).

4. Moments about the right end: R_L ×2.0 = 40 ×1.0 + 100 ×1.50, so R_L = 95 N. Up = down: R_R = 140 - 95 = 45 N.

5. When the car tilts, it pivots about the outer wheels. A low G and a wide base mean the car must tilt much further before the vertical line through G passes outside the wheels, so the moment of the weight keeps turning it back for longer: it is harder to tip over.