Moments and equilibrium of rigid bodies

A moment measures the turning effect of a force about a point. A small force far from a pivot can have the same turning effect as a large force close to it.

For a force of magnitude $F$ whose line of action is a perpendicular distance $d$ from a point $O$,

$$\boxed{M_O=Fd}.$$

Moment is measured in newton metres, $\mathrm{N\,m}$. It has a clockwise or anticlockwise sense.

Prerequisites

You should be able to:

• identify weight, tension and reaction forces using free body diagrams;
• resolve forces using sine and cosine;
• apply force equilibrium from Newton’s laws;
• use $W=mg$, with $g=9.8\ \mathrm{m\,s^{-2}}$ unless told otherwise.

Force, line of action and perpendicular distance

The line of action is the infinite straight line through the force arrow. The distance $d$ in $M=Fd$ is the shortest distance from the point to this line, so it must meet the line of action at $90^\circ$.

$$\begin{array}{c} \text{line of action}\quad \longrightarrow F\\[-2pt] \hspace{22mm}\rule{18mm}{0.3pt}\\[-2pt] \hspace{16mm}\underbrace{\qquad}_{d\text{ perpendicular}}\\[-2pt] \hspace{8mm}O \end{array}$$

If the line of action passes through $O$, then $d=0$ and the moment about $O$ is zero. The force itself need not be zero.

Do not use the distance to the force arrow

The required distance is from the pivot to the line of action, not to the point where the force is drawn. It is also not automatically the length of a rod.

Worked example 1: a simple lever

A vertical force of $35\ \mathrm N$ acts downwards at the end of a horizontal spanner, $0.24\ \mathrm m$ from a nut. Find its moment about the nut.

The force is perpendicular to the spanner, so the perpendicular distance is $0.24\ \mathrm m$:

$$M=35(0.24)=8.4\ \mathrm{N\,m}.$$

The force acts downwards to the right of the nut, so the sense is clockwise. Hence the moment is

$$\boxed{8.4\ \mathrm{N\,m}\text{ clockwise}}.$$

Self-check 1

A $16\ \mathrm N$ force acts vertically upwards $0.75\ \mathrm m$ to the right of a pivot. Find its moment about the pivot.

Answer

$$M=16(0.75)=\boxed{12\ \mathrm{N\,m}\text{ anticlockwise}}.$$

Angled forces: two equivalent methods

Suppose a force $F$ acts at angle $\theta$ to a rod, at distance $r$ from $O$. The turning effect comes only from the component perpendicular to the rod:

$$F_\perp=F\sin\theta.$$

Therefore

$$\boxed{M_O=rF\sin\theta}.$$

This is the same as using $M=Fd$, because the perpendicular distance from $O$ to the line of action is $d=r\sin\theta$.

You may therefore use either:

1. force multiplied by perpendicular distance, $F(r\sin\theta)$; or
2. perpendicular component multiplied by distance along the rod, $(F\sin\theta)r$.

Do not resolve the force and also use the shortened perpendicular distance. That would include the factor $\sin\theta$ twice.

Worked example 2: force at an angle

A force of $50\ \mathrm N$ is applied to a straight bar at a point $0.8\ \mathrm m$ from a pivot. The angle between the force and the bar is $30^\circ$. Find the magnitude of the moment.

Using the perpendicular component,

$$F_\perp=50\sin30^\circ=25\ \mathrm N.$$

Hence

$$M=0.8(25)=\boxed{20\ \mathrm{N\,m}}.$$

Equivalently, $d=0.8\sin30^\circ=0.4\ \mathrm m$, giving $M=50(0.4)=20\ \mathrm{N\,m}$.

Self-check 2

A $40\ \mathrm N$ force acts at $60^\circ$ to a rod, $0.5\ \mathrm m$ from a pivot. Find the moment magnitude exactly and to $3$ significant figures.

Answer

$$M=0.5(40)\sin60^\circ =20\left(\frac{\sqrt3}{2}\right) =\boxed{10\sqrt3\ \mathrm{N\,m}} \approx\boxed{17.3\ \mathrm{N\,m}}.$$

The diagram is needed to decide whether the sense is clockwise or anticlockwise.

The principle of moments

For a rigid body in equilibrium, there is no linear acceleration and no turning acceleration. Thus both conditions must hold:

$$\boxed{\sum \mathbf F=\mathbf 0} \qquad\text{and}\qquad \boxed{\sum M_O=0}.$$

The second equation may be taken about any point $O$. Equivalently,

$$\boxed{\text{total clockwise moments} =\text{total anticlockwise moments}}.$$

Force equilibrium alone is not enough for an extended rigid body. Two equal and opposite forces on different lines of action have zero resultant force but still produce a turning effect.

Choose the point strategically

Take moments about a point through which an unknown force acts. Its moment is zero, so that unknown disappears from the moment equation. This is usually the main simplification.

Uniform rods and beams

The weight of a uniform rod acts at its midpoint. More generally, the whole weight of a rigid body is modelled as acting through its centre of mass. A non-uniform rod need not have its centre of mass at the midpoint.

A support reaction acts at the point of contact unless the model says otherwise. A hinge or pivot can exert force in more than one direction, but every component through the pivot has zero moment about that pivot.

Worked example 3: a supported uniform beam

A uniform horizontal beam $AB$ has length $6\ \mathrm m$ and mass $20\ \mathrm{kg}$. It is supported at $A$ and $B$. A particle of mass $30\ \mathrm{kg}$ rests $2\ \mathrm m$ from $A$. Find the upward reactions $R_A$ and $R_B$.

The beam’s weight, $20g$, acts at its midpoint, $3\ \mathrm m$ from $A$. The particle’s weight is $30g$.

Take moments about $A$. This removes $R_A$:

$$R_B(6)=20g(3)+30g(2).$$

Therefore

$$R_B=\frac{60g+60g}{6}=20g=196\ \mathrm N.$$

Now use vertical force equilibrium:

$$R_A+R_B=20g+30g=50g.$$

So

$$R_A=50g-20g=30g=294\ \mathrm N.$$

Hence

$$\boxed{R_A=294\ \mathrm N,\qquad R_B=196\ \mathrm N}.$$

Check: the total upward force is $490\ \mathrm N$, equal to the total weight. The heavier load is nearer $A$, so $R_A>R_B$, which is physically sensible.

Worked example 4: finding a load’s position

A uniform plank $AB$ has length $5\ \mathrm m$ and weight $120\ \mathrm N$. It is supported at both ends. A child of weight $300\ \mathrm N$ stands $x\ \mathrm m$ from $A$. The reaction at $B$ is $180\ \mathrm N$. Find $x$.

Take moments about $A$:

$$180(5)=120(2.5)+300x.$$

Thus

$$900=300+300x,$$

so

$$\boxed{x=2\ \mathrm m}.$$

The reaction at $A$ is not needed. If required, vertical equilibrium gives

$$R_A+180=120+300,$$

and hence $R_A=240\ \mathrm N$.

Self-check 3

A uniform horizontal rod $AB$ has length $4\ \mathrm m$ and mass $10\ \mathrm{kg}$. It is supported at $A$ and $B$. A mass of $6\ \mathrm{kg}$ is attached $1\ \mathrm m$ from $B$. Find both reactions.

Answer

The rod’s weight acts $2\ \mathrm m$ from $A$, and the attached mass is $3\ \mathrm m$ from $A$. Taking moments about $A$,

$$4R_B=10g(2)+6g(3)=38g,$$

so

$$R_B=9.5g=93.1\ \mathrm N.$$

Vertical equilibrium gives

$$R_A+R_B=16g,$$

so

$$R_A=6.5g=\boxed{63.7\ \mathrm N}.$$

Therefore $\boxed{R_A=63.7\ \mathrm N}$ and $\boxed{R_B=93.1\ \mathrm N}$.

A rod held by a cable

Worked example 5: combining moments with resolution

A uniform horizontal rod $AB$ has length $4\ \mathrm m$ and weight $200\ \mathrm N$. It is hinged at $A$. A cable attached at $B$ makes an angle of $30^\circ$ above the rod and holds it in equilibrium. Find the cable tension $T$ and the vertical component of the hinge reaction.

The hinge reaction passes through $A$, so take moments about $A$. Only the vertical component $T\sin30^\circ$ of the tension produces a moment:

$$(T\sin30^\circ)(4)=200(2).$$

Therefore

$$4T\left(\frac12\right)=400,$$

giving

$$\boxed{T=200\ \mathrm N}.$$

Let the vertical hinge component be $Y$ upwards. Vertical equilibrium gives

$$Y+T\sin30^\circ-200=0.$$

Hence

$$Y+100-200=0,$$

so

$$\boxed{Y=100\ \mathrm N\text{ upwards}}.$$

The horizontal hinge component would balance $T\cos30^\circ$. Moments found $T$, but force equilibrium is still needed to find the hinge reaction.

A reliable method for statics problems

1. Isolate the whole rigid body and draw every external force.
2. Replace each mass by its weight $mg$ at the correct position.
3. Mark all relevant distances and angles.
4. Choose a moment centre that removes the least convenient unknown force.
5. Assign clockwise and anticlockwise senses consistently.
6. Write one moment equation, using perpendicular distances or perpendicular components.
7. Use horizontal and vertical force equilibrium for remaining unknowns.
8. Check units, signs, total force, and whether each reaction has a physically possible direction.

Common misconceptions

• Using mass instead of weight: a $12\ \mathrm{kg}$ mass produces a force $12g\ \mathrm N$, not $12\ \mathrm N$.
• Using a sloping distance: $M=Fd$ requires the shortest distance to the line of action.
• Forgetting the rod’s weight: a uniform rod’s weight acts at its midpoint, unless its weight is said to be negligible.
• Assuming equal reactions: supports share a load equally only in a symmetric arrangement.
• Taking moments of only some forces: include every force whose line of action does not pass through the chosen point.
• Using moments without force balance: $\sum M=0$ prevents turning, while $\sum\mathbf F=\mathbf0$ prevents translation. Equilibrium needs both.
• Writing $\mathrm{Nm}$ carelessly: moment has unit $\mathrm{N\,m}$. It is not a force and should not be given in newtons.

Mixed self-check

A uniform horizontal beam $AB$ has length $8\ \mathrm m$ and mass $25\ \mathrm{kg}$. It is hinged at $A$ and supported at $B$ by a vertical cable. A crate of mass $40\ \mathrm{kg}$ is placed $6\ \mathrm m$ from $A$.

1. Find the tension in the cable.
2. Find the vertical component of the hinge reaction.
3. Explain why no horizontal calculation is needed.

Answer

The beam’s weight $25g$ acts $4\ \mathrm m$ from $A$. Let the cable tension be $T$. Taking moments about $A$,

$$8T=25g(4)+40g(6).$$

Therefore

$$T=\frac{100g+240g}{8}=42.5g =\boxed{416.5\ \mathrm N}.$$

Let the vertical hinge component be $Y$, positive upwards. Vertical equilibrium gives

$$Y+42.5g-25g-40g=0,$$

so

$$Y=22.5g=\boxed{220.5\ \mathrm N\text{ upwards}}.$$

Every applied force is vertical, so horizontal equilibrium gives a zero horizontal hinge component immediately.

Next steps

Moments problems often combine several earlier ideas. Revisit resolving forces for angled tensions, friction and limiting equilibrium for rough contacts, and forces and free body diagrams if support reactions are hard to identify. Then use the broader mechanics overview to connect statics with dynamics and connected particles.