IMAT Physics · Dynamics, Work and Energy

Moments, Couples, Equilibrium and Centre of Mass

A force's turning effect is the force times its perpendicular distance from the pivot; a body is in equilibrium when both the forces and the turning effects cancel.

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Why this matters

Four of the older papers asked about moments: balancing a pivoted uniform bar (2011) and a suspended beam (2013), choosing the diagram that shows a couple (2015), and finding the support forces under a loaded beam (2016). The ministry papers have not asked yet.

Concept 1 of 3: The moment of a force, and couples

A door is easy to open when you push at the handle and very hard when you push next to the hinge. The turning effect of a force depends on how hard you push and how far from the pivot, measured at right angles to the force. Two equal and opposite forces that do not act along the same line cannot move a body along, but they can spin it: that is a couple.

Definition

Moment (torque) of a force about a point = force × perpendicular distance from the point to the line of action of the force. Unit: N m.
A force whose line passes through the pivot has zero moment.
If the force acts at angle $\theta$ to a lever of length $r$ , the moment is $Fr\sin\theta$ .
A couple is two equal, opposite, parallel forces not in the same line. Its resultant force is zero, but its torque is $F \times d$ , where $d$ is the perpendicular distance between the two lines. The torque of a couple is the same about any point.

Moment and couple

M = F d_{\perp} \qquad \tau_{\text{couple}} = F d

$d_{\perp}$ perpendicular distance from the pivot to the line of the force, in m
$d$ perpendicular distance between the two forces of a couple, in m

Worked example

A mechanic pulls on the end of a spanner 0.25 m long with a force of 40 N. Find the moment about the nut when she pulls at right angles to the spanner, and when she pulls at 30° to the spanner.

Practice this conceptself-check · 3 quick reps

A couple has zero resultant force but a non-zero torque

Two equal and opposite forces on different lines cancel as forces, so the body does not accelerate along, but they still turn it. The torque is one force times the distance between the lines, not zero and not twice that.

Concept 2 of 3: Equilibrium and the principle of moments

A see-saw balances when the turning effect pushing one way equals the turning effect pushing the other way. A beam resting on supports must also not move up or down, so the supports must push up with exactly the total weight. Two conditions, two equations: that is enough to find two unknown forces.

Definition

A body is in equilibrium when:

the resultant force on it is zero (up forces = down forces, left = right), and
the resultant moment about any point is zero.

Principle of moments: in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point.

The weight of a uniform beam acts at its middle.
Taking moments about a point where an unknown force acts removes that force from the equation.

Principle of moments

\sum M_{\text{clockwise}} = \sum M_{\text{anticlockwise}} \qquad \sum F_{\text{up}} = \sum F_{\text{down}}

$M$ moment about the chosen point, in N m

Worked example

A uniform beam 4.0 m long and of weight 100 N rests on supports at its ends A and B. A 300 N load sits 1.0 m from A. Find the upward force from each support.

Practice this conceptself-check · 3 quick reps

Include the beam's own weight, at its centre

Unless the beam is called light or of negligible mass, its weight acts at its midpoint and has a moment like any other load. Leaving it out, or placing it at the pivot, gives the wrong balance.

Concept 3 of 3: Centre of mass and stability

For many purposes a whole body behaves as if all its mass were at one point, the centre of mass, and its weight acts there. For a uniform body it is at the geometric centre. For two masses on a light rod it sits nearer the heavier one, just where you would balance the rod on a finger.

Definition

The centre of mass is the point where the whole mass can be taken to act; in a uniform gravitational field it is the same as the centre of gravity, where the weight acts.
A uniform rod, plate or sphere: at its geometric centre.
Two point masses on a line: $x_{\text{cm}} = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}$ , closer to the heavier mass.
A body hung from one point comes to rest with its centre of mass directly below that point.
A standing body topples when its centre of mass moves outside its base. A low centre of mass and a wide base make it more stable.

Centre of mass of two masses

x_{\text{cm}} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}

$x_1, x_2$ positions of the two masses along the line
$m_1, m_2$ the two masses

Worked example

A 2.0 kg mass sits at

x = 0

and a 3.0 kg mass at

x = 1.0\ \text{m}

on a light rod. Where is the centre of mass?

Practice this conceptself-check · 3 quick reps

The centre of mass is nearer the heavier mass

The heavier mass pulls the balance point towards itself. Using the mass ratio the wrong way round puts the centre of mass near the light end, and that reversed answer is usually one of the options.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The moment of a force, and couples
Moment and couple
$M = F d_{\perp} \qquad \tau_{\text{couple}} = F d$
Equilibrium and the principle of moments
Principle of moments
$\sum M_{\text{clockwise}} = \sum M_{\text{anticlockwise}} \qquad \sum F_{\text{up}} = \sum F_{\text{down}}$
Centre of mass and stability
Centre of mass of two masses
$x_{\text{cm}} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$

Watch out for (3)

A couple has zero resultant force but a non-zero torque
→ The moment of a force, and couples
Include the beam's own weight, at its centre
→ Equilibrium and the principle of moments
The centre of mass is nearer the heavier mass
→ Centre of mass and stability

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