Moments and equilibrium of rigid bodiesEdexcel A-Level Maths: Revision notes

The moment of a force about a point measures its turning effect: $\text{moment}=\text{force}\times\text{perpendicular distance from the point to the line of action of the force}.$ The unit is N m. Taking moments about a point removes any force that passes through that point. If the force is not perpendicular to the rod, use the component perpendicular to the rod, $F\sin\theta$ when $\theta$ is the angle between the force and the rod. Example: a 30 N force at $60^\circ$ to a 2 m rod, acting at the end, has a moment about the other end of $30\times2\sin60^\circ=52.0$ N m. Moments are clockwise or anticlockwise; choose one as positive and keep to it.

A rigid body in equilibrium has two conditions:

• the resultant force is zero (resolve in two perpendicular directions)
• the total moment about any point is zero (clockwise moments $=$ anticlockwise moments). Choose the pivot to remove unknown forces. Taking moments about the point where two unknown forces meet leaves one unknown in the equation. The weight of a uniform body acts at its midpoint (centre of mass). For a non-uniform rod the centre of mass is at an unknown distance that you find with moments.

A uniform beam on two supports has two vertical reactions. Moments about one support give the other reaction directly. Example: beam 20 kg, 6 m, supports at $A$ and 4 m from $A$ at $C$. About $C$: $4R_A=196\times1$, so $R_A=49$ N, and vertically $R_C=196-49=147$ N. A beam is about to tilt about a support when the reaction at the other support is zero. A load of $m$ at $B$ gives $mg(2)=196(1)$, so $m=10$ kg. Strings or supports at the ends: the same method works. For a non-uniform rod with tensions 50 N and 67.6 N on a rod of weight 117.6 N, moments about $A$ give the distance of the centre of mass as $x=\frac{67.6\times4}{117.6}=2.30$ m.

When a rod is held by a hinge and a string, the hinge force has an unknown size and direction. Take moments about the hinge to eliminate it, then resolve horizontally and vertically to find its components. Example: rod 8 kg, 3 m, string at $40^\circ$ to the rod. Moments about $A$: $3T\sin40^\circ=78.4\times1.5$, so $T=61.0$ N. Horizontally $X=T\cos40^\circ=46.7$ N and vertically $Y=78.4-T\sin40^\circ=39.2$ N, so the hinge force is $\sqrt{X^2+Y^2}=61.0$ N at $40^\circ$ above the horizontal. Include the moment of each force through the perpendicular distance or resolve each force first.

A ladder $AB$ rests with $A$ on rough ground and $B$ against a smooth wall. The forces are: weight $W$ at the midpoint, wall reaction $R$ (horizontal, perpendicular to the wall), ground reaction $S$ (vertical) and friction $F$ (horizontal, towards the wall). Resolve: $S=W$ (plus any load) and $F=R$. Moments about the foot: $R\times\ell\sin\theta=W\times\frac{\ell}{2}\cos\theta$ for a ladder of length $\ell$ at angle $\theta$ to the ground. Example: 15 kg, 5 m, $60^\circ$: $R=\frac{147\times2.5\cos60^\circ}{5\sin60^\circ}=42.4$ N, $S=147$ N and $F=42.4$ N. Limiting equilibrium: $F=\mu S$. A person of 60 kg at $x$ m from the foot adds $588x\cos60^\circ$ to the moments, and with $\mu=0.4$ the ladder slips at $x=3.71$ m.