MomentsAQA A-Level Maths: Revision notes

The moment of a force about a point measures its turning effect: $\text{moment}=\text{force}\times\text{perpendicular distance from the point to the line of action}.$ It is a derived quantity: it is built from the base quantities force ($\text{kg m s}^{-2}$) and length, so its unit is the newton metre, N m (equivalent to $\text{kg m}^2\text{s}^{-2}$). In a plane a moment is clockwise or anticlockwise; choose one direction as positive. Example: a child of weight $294$ N sits $1.5$ m from a pivot, so the moment about the pivot is $294\times1.5=441$ N m. If a force $F$ acts at angle $\theta$ to the rod, use the perpendicular component: moment $=Fd\sin\theta$ where $d$ is the distance along the rod.

A rigid body (such as a rod) is in equilibrium when two conditions hold:

1. the resultant force is zero, so upward forces equal downward forces;
2. the resultant moment about any point is zero, so total clockwise moment equals total anticlockwise moment (principle of moments). Because the second condition holds about any point, choose the point where an unknown force acts, so that its moment is zero and it drops out of the equation. Example: a light plank balanced on a pivot, with $30$ kg at $1.5$ m on one side and $m$ kg at $2$ m on the other: $m(9.8)(2)=30(9.8)(1.5)$, so $m=22.5$ kg.

A uniform rod has its mass spread evenly, so its weight acts at its midpoint. A light rod has negligible mass, so its weight is ignored. A non-uniform rod has its centre of mass elsewhere, and its position is found from moments. Example: a $4$ m rod of mass $15$ kg is held by vertical strings at each end, with tension $88.2$ N at $B$. Vertically $T_A=147-88.2=58.8$ N. Moments about $A$: $147x=88.2(4)$, so the centre of mass is $x=2.4$ m from $A$ (not at the midpoint, so the rod is non-uniform). In modelling, a person or object on a rod is treated as a particle whose weight acts at a point.

A rod resting horizontally on two supports has a reaction at each, acting vertically upwards. Use both equilibrium conditions: moments about one support, then resolve vertically. Example: a uniform beam $AB$, $5$ m, mass $50$ kg, supports at $C$ ($AC=1$ m) and $D$ ($DB=1$ m), a $60$ kg man at $E$ with $EB=0.5$ m. Moments about $C$: $3R_D=490(1.5)+588(3.5)=2793$, so $R_D=931$ N. Vertically $R_C=1078-931=147$ N. Supports are modelled as smooth, so reactions are vertical, and the beam is rigid.

A rod is about to tilt about a support when the reaction at the other support becomes zero. Then take moments about the support it tilts about. Example: the uniform $50$ kg beam above, with a child of mass $m$ at $A$ and nothing else: about to tilt about $C$, so $R_D=0$ and $m(9.8)(1)=50(9.8)(1.5)$, giving $m=75$ kg. A greater mass at $A$ would make the beam tilt. For a diving board bolted at $A$ and supported at $P$ ($AP=1$ m, $3$ m long, $24$ kg), a $60$ kg diver at $B$ requires a downward bolt force $X$ with $X(1)=588(2)+235.2(0.5)$, so $X=1294$ N. Increasing the diver's distance from $P$ increases $X$. To find the greatest load or distance, set the reaction (or force) to its limiting value and solve.