Moments and equilibrium

The Moment of a Force

Key Equations

Moment of a force:

$$M=Fx$$

Variables:

• F = applied force, in N
• x = perpendicular distance from the pivot to the line of action of the force, in m
• M = moment of the force, in N m

SI unit: N m (newton metre)

Rearrangements:

$$F=\frac{M}{x}$$

$$x=\frac{M}{F}$$

ProportionalityThe moment is directly proportional to the force when the perpendicular distance is constant. The moment is also directly proportional to the perpendicular distance when the force is constant.

The moment of a force is the turning effect produced by that force about a chosen pivot or axis of rotation. The size of the moment depends on two things: the magnitude of the force, and the perpendicular distance from the pivot to the line of action of the force. A larger force or a longer perpendicular distance produces a greater turning effect, which is why a long spanner loosens a tight bolt more easily than a short one.

The “perpendicular distance” is measured at right angles to the line along which the force acts, not along the body itself. When the force acts perpendicular to the beam, the perpendicular distance equals the distance along the beam. When the force acts at an angle, the line of action must be extended, and the perpendicular distance is measured from the pivot to this line.

MisconceptionStudents often multiply the force by the distance from the pivot to where the force is applied, ignoring the angle. The correct distance is always perpendicular to the line of action of the force, not along the body.

Exam TipIf a force acts at an angle, draw the line of action and drop a perpendicular from the pivot to it.

Centre of Gravity and the Principle of Moments

The centre of gravity of an extended body is the single point at which the entire weight of the body can be considered to act. For a uniform body with a regular shape, the centre of gravity lies at its geometric centre. Treating the weight as acting at this single point allows the principle of moments to be applied to extended objects such as beams, rulers, and bridges.

The principle of moments states that for a body in rotational equilibrium, the sum of the clockwise moments about any pivot equals the sum of the anticlockwise moments about that same pivot. Combined with the requirement that the resultant force is zero, this gives the two conditions for full equilibrium of an extended body.

When applying the principle of moments, the pivot can be chosen at any point. Choosing the pivot at the line of action of an unknown force eliminates that force from the moment equation, because its perpendicular distance is zero. This is a powerful problem-solving technique for beams supported at two points.

Examiner InsightIn equilibrium problems, always state the pivot chosen and identify each moment as clockwise or anticlockwise. Marks are awarded for correctly identifying perpendicular distances and the direction of each moment.

Exam TipWrite “Taking moments about [point]:” before any moment equation.

Worked Example: Beam in Equilibrium

Scenario

A uniform beam of weight 60 N and length 4.0 m rests horizontally on two supports, one at each end. A 200 N load is placed 1.5 m from the left support. Calculate the upward force from each support.

Step 1 — Identify the forces and distances. The beam’s weight acts at its centre, 2.0 m from each support. The load is 1.5 m from the left support and 2.5 m from the right support. Let $N_{\mathrm{L}}$ and $N_{\mathrm{R}}$ be the support forces.

Step 2 — Take moments about the left support. Choosing the left support eliminates $N_{\mathrm{L}}$ from the equation. Clockwise moments come from the load and the beam’s weight. The anticlockwise moment comes from $N_{\mathrm{R}}$.

$${N}_{R}\times 4.0=(200\times 1.5)+(60\times 2.0)$$

$$4.0 {N}_{R}=300+120=420$$

$${N}_{R}=\frac{420}{4.0}=105 \text{N}$$

Step 3 — Apply translational equilibrium for $N_{\mathrm{L}}$. The total upward force equals the total downward force.

$${N}_{L}+{N}_{R}=200+60$$

$${N}_{L}=260-105=155 \text{N}$$

Answer

$${N}_{L}=155 \text{N}, {N}_{R}=105 \text{N}$$

Interpretation

The left support carries more weight because the load is closer to it. Both conditions for equilibrium are satisfied: zero resultant force and balanced moments.