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

The moment of a force about a point is the force multiplied by the perpendicular distance from the point to the line of action of the force: $M=F\times d$, in N m. It measures the turning effect and is clockwise or anticlockwise. If the force is at an angle, either resolve it into components perpendicular and parallel to the rod, or find the perpendicular distance. A rigid body is in equilibrium if both conditions hold:

• the resultant force is zero (resolve in two perpendicular directions);
• the resultant moment about any point is zero (total clockwise moment $=$ total anticlockwise moment). The weight of a uniform body acts at its midpoint (centre of mass). A light rod has negligible weight.

For a horizontal beam on supports all forces are vertical, so there are two equations: forces up $=$ forces down, and moments about one point. If the beam rests on two supports, take moments about one support to find the other reaction directly. A beam is on the point of tilting about a support when the reaction at the other support is zero: put that reaction equal to $0$ and take moments about the pivot. Example: $AB=6$ m, mass $15$ kg, supports at $1$ m and $4$ m from $A$. Moments about $D$ (at $4$ m) give $3R_C=147\times1$ so $R_C=49$ N, and $R_D=147-49=98$ N.

When forces act at different angles, resolve into horizontal and vertical components and take moments. A smooth hinge exerts a force of unknown size and direction, so represent it by two perpendicular components $H$ and $V$ and find them by resolving. A light string exerts a tension along its length. For a horizontal rod $AB$ held by a string $BC$ from a wall, the moment of the tension about $A$ is $T\sin\varphi\times AB$, where $\varphi$ is the angle between string and rod; only the component perpendicular to the rod has a moment. The magnitude of the hinge force is $\sqrt{H^2+V^2}$ and its direction is $\tan^{-1}\frac VH$ to the horizontal.

Ladder problems combine moments, resolving, and friction. Friction satisfies $F\le\mu R$, with $F=\mu R$ in limiting equilibrium (about to slip).

• Smooth wall, rough ground: the wall exerts only a horizontal normal reaction $R$; the ground exerts a vertical normal reaction $N$ and a horizontal friction $F$ directed away from the wall. Then $N=W$ and $F=R$, and the moment equation about the foot gives $R$.
• Rough wall and ground: the wall adds a vertical friction force $F_2\le\mu_2N_2$ acting upwards. Resolve horizontally and vertically, then take moments about the foot. A person on the ladder adds a weight at their position. The ladder is on the point of slipping when friction is limiting at every contact.

Ladder $AB$, $5$ m, $20$ kg, $B$ on a smooth wall $4$ m high, rough ground at $A$. Moments about $A$: $R(4)=196(1.5)$ (the weight is $1.5$ m horizontally from $A$), so $R=73.5$ N. Resolving: $N=196$, $F=73.5$. So $\mu\ge\frac{73.5}{196}=\frac38$. Method: (1) draw a large clear diagram with every force, including friction directions; (2) resolve horizontally and vertically; (3) take moments about the point which removes the most unknowns; (4) apply $F\le\mu R$, or $F=\mu R$ if limiting; (5) check the answer is sensible, e.g. a distance up the ladder must be no more than the ladder's length. Use exact fractions or give three significant figures.