PhysicsCore22 min read

Turning Effects of Force

Moments, couples, centre of gravity and stability

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01

The turning effect of a force

Definition

Moment of a force — The turning effect of a force about a pivot, equal to the force multiplied by the perpendicular distance from the pivot to the line of action of the force.

A force applied to a body that is free to rotate does not simply push it — it turns it. How much turning effect you get depends on two things: how hard you push, and how far from the pivot you push.

This is why a door handle is fitted on the edge furthest from the hinges. Push near the hinge and almost nothing happens; push at the far edge with the same force and the door swings easily. It is also why a long spanner undoes a tight nut that a short one cannot.

The word perpendicular in the definition is not decoration. The distance must be measured at right angles from the pivot to the line of action of the force. Push along a line that passes straight through the pivot and the perpendicular distance is zero, so there is no turning effect at all, however hard you push.

moment = F × dd is the PERPENDICULAR distance from the pivot to the line of action, in metres
moment
turning effectN m
F
forceN
d
perpendicular distancem

Slide the masses and watch the two moments. Balance is reached when the clockwise total equals the anticlockwise total — a small mass far out can balance a large mass close in, which is the whole principle of a lever.

02

The principle of moments

When an object is balanced and not rotating, the turning effects in the two directions must cancel exactly. This is the principle of moments: for a body in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.

The phrase "about any point" is genuinely useful. You may take moments about whichever point you like, and a well-chosen pivot makes a hard problem easy — pick the point where an unknown force acts, and that force disappears from the equation because its distance is zero.

Setting these problems out neatly saves marks. List the clockwise moments on one side, the anticlockwise on the other, set them equal, and solve. Muddling the two directions is the commonest error, and it is always visible in a tidy layout.

Worked example 15 marks

A uniform metre rule is pivoted at its centre. A 2.0 N weight hangs 40 cm from the pivot on the left. Where must a 5.0 N weight be hung on the right to balance it?

  1. The rule is uniform and pivoted at its centre, so its own weight has no moment.Its weight acts through the pivot, giving zero perpendicular distance.
  2. Anticlockwise moment = 2.0 × 0.40 = 0.80 N m.Convert cm to m before multiplying.
  3. For balance, clockwise moment must also be 0.80 N m.Principle of moments.
  4. 5.0 × d = 0.80.
  5. d = 0.16 m = 16 cm from the pivot.Heavier weight, closer in — as expected.

16 cm from the pivot on the right

03

Centre of gravity

Definition

Centre of gravity — The single point at which the entire weight of an object may be taken to act.

Every particle of an object has weight, but for the purposes of a calculation all of that weight can be treated as acting at one point. For a uniform object of regular shape that point is at its geometric centre — the middle of a uniform metre rule, the centre of a uniform sphere.

For an irregular flat shape it can be found experimentally. Suspend the shape freely from a point, hang a plumb line from the same point and mark the vertical. Repeat from a second point. The centre of gravity lies where the two lines cross, because a freely suspended object always hangs with its centre of gravity directly below the point of support.

A third suspension point is normally used as a check. If all three lines meet at one place, the result is reliable.

04

Stability

An object topples when its centre of gravity moves outside its base. That single sentence answers most stability questions, and everything else follows from it.

Tilt an object and its weight acts vertically down through the centre of gravity. While that line still falls inside the base, the weight produces a moment that turns the object back upright — it is stable. Tilt it further, past the point where the line falls outside the base, and the same weight now turns it over instead.

So an object is made more stable in two ways: give it a low centre of gravity and a wide base. Both increase the angle it must be tilted through before the line of the weight escapes the base.

This is why a racing car is low and wide, why a Bunsen burner has a heavy base, and why a double-decker bus is tested for stability with passengers only on the upper deck. It is also why you instinctively spread your feet and crouch on a moving bus.

Type of equilibriumBehaviour when tilted slightlyCentre of gravity
Stablereturns to its original positionrises when tilted
Unstabletopples further awayfalls when tilted
Neutralstays in the new positionstays at the same height

Key points

  1. Moment = force × perpendicular distance from the pivot.
  2. A force acting through the pivot has no turning effect.
  3. In equilibrium, clockwise moments = anticlockwise moments, about any point.
  4. The centre of gravity is where all the weight can be taken to act.
  5. Stability comes from a low centre of gravity and a wide base.

Practice questions

7 questions · 20 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Define the moment of a force and state its SI unit.
Model answer

The turning effect of a force about a pivot, equal to force × perpendicular distance from the pivot to the line of action. SI unit: the newton metre, N m.

Examiner tip. The word "perpendicular" is a mark on its own. So is the unit.

SQ2[2 marks]
Why can a moment not be measured in joules, when N m is the unit of both?
Model answer

They are different quantities. Work is force acting along a displacement; a moment is force acting across a distance from a pivot. Sharing base units does not make them the same thing.

Examiner tip. Write N m for moments and J for energy every time and this question answers itself.

SQ3[2 marks]
State two conditions that must be satisfied for a body to be in complete equilibrium.
Model answer

ΣF = 0 — no resultant force, so it does not accelerate. Σ moments = 0 — no resultant turning effect, so it does not rotate.

Examiner tip. Both conditions, or one mark. Students routinely give only the force condition.

Solved numericals

1 · 4 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
A uniform metre rule is pivoted at the 50 cm mark. A weight of 3.0 N hangs at the 20 cm mark. Calculate the weight that must hang at the 70 cm mark to balance it.

Given. W₁ = 3.0 N at 20 cm, pivot at 50 cm, W₂ at 70 cm

Full working
  1. Left distance = 50 − 20 = 30 cm = 0.30 mdistance from the pivot, not from the end of the rule[1]
  2. Anticlockwise moment = 3.0 × 0.30 = 0.90 N m[1]
  3. Right distance = 70 − 50 = 20 cm = 0.20 m; for balance W₂ × 0.20 = 0.90principle of moments[1]
  4. W₂ = 4.5 Nunit required[1]

4.5 N

Exam questions

3 · 10 marks

Multi-part questions with a full mark scheme.

Q1[4 marks]
A uniform beam of weight 40 N and length 2.0 m rests on a pivot 0.50 m from its left end. A load of weight W hangs from the extreme left end, and the beam is in equilibrium.
  1. State where the weight of the beam acts.
  2. Calculate the value of W.
Mark scheme
  1. At the centre of gravity, which for a uniform beam is its midpoint, 1.0 m from the left end"uniform" is the word that tells you this[1]
  2. Takes moments about the pivot; the beam's weight acts 1.0 − 0.50 = 0.50 m to the right of it[1]
  3. Clockwise = anticlockwise: 40 × 0.50 = W × 0.50the load acts 0.50 m to the left of the pivot[1]
  4. W = 40 N[1]

(a) at the midpoint, 1.0 m from the left end (b) W = 40 N

Examiner tip. Taking moments about the pivot removes the upward reaction force from the equation, because its moment about the pivot is zero. Choosing the pivot well is usually worth more than any algebra.

Q2[3 marks]
Explain, in terms of centre of gravity and base, why a double-decker bus is more likely to topple when passengers stand on the upper deck than when they sit downstairs.
Mark scheme
  1. Passengers upstairs raise the centre of gravity of the bus[1]
  2. A higher centre of gravity means a smaller tilt is needed before the vertical line through it falls outside the base / wheelbasethis is the key reasoning mark[1]
  3. So the bus topples at a smaller angle / is less stable[1]

Examiner tip. Stability answers need all three steps: what moves, what that does to the line through the centre of gravity relative to the base, and the consequence. Two of the three is the usual score.

Q3[3 marks]
A force of 12 N is applied to a spanner at 30° to the handle, at a distance of 0.25 m from the nut. Calculate the moment of the force about the nut.
Mark scheme
  1. Recognises that only the perpendicular component turns the nutor equivalently uses the perpendicular distance[1]
  2. Perpendicular component = 12 sin 30° = 6.0 N[1]
  3. Moment = 6.0 × 0.25 = 1.5 N munit required[1]

1.5 N m

Examiner tip. When a force is applied at an angle, either resolve the force perpendicular to the arm or find the perpendicular distance to the line of action. Both give the same answer; doing neither gives 3.0 N m and no marks.