M9 Moments — coverage pack

1 specification leaves · notes, questions, answers and worked methods

M9.1 · Understand and use moments in simple static contexts.

The moment of a force about a point is $F d$ , where $d$ is the perpendicular distance from the point to the force's line of action; state whether it is clockwise or anticlockwise when required.
For a rigid body in equilibrium, choose a pivot that removes unknown forces where possible, set total clockwise moments equal to total anticlockwise moments, then use force equilibrium.
A uniform rod's weight acts at its midpoint, while a non-uniform body's weight acts at its stated centre of mass; contact forces act at their contact points.
A common error is to use the distance along a rod instead of the perpendicular distance to the line of action, or to omit a force whose line of action does not pass through the pivot.

1. A $12\,\text{N}$ force acts perpendicular to a spanner at a distance $0.35\,\text{m}$ from a nut. Find the magnitude of its moment about the nut.[2 marks]
Answer
- $4.2\,\text{N m}$
Method: The force is perpendicular, so the perpendicular distance is $0.35\,\text{m}$ . The moment is $F d=12(0.35)=4.2\,\text{N m}$ .

1. A uniform horizontal beam $AB$ has length $4\,\text{m}$ and weight $120\,\text{N}$ . It is supported vertically at $A$ and $B$ . Find the upward force at each support.[4 marks]
Answer
- Upward force at $A=60\,\text{N}$
- Upward force at $B=60\,\text{N}$
Method: The beam's weight acts at its midpoint, $2\,\text{m}$ from $A$ . Taking moments about $A$ , $R_B(4)=120(2)$ , so $R_B=60\,\text{N}$ . Vertical equilibrium gives $R_A+R_B=120$ , hence $R_A=60\,\text{N}$ .

1. A uniform ladder of length $5\,\text{m}$ and weight $240\,\text{N}$ rests with its lower end on rough horizontal ground and its upper end against a smooth vertical wall. The ladder makes an angle of $60^\circ$ with the ground and is in limiting equilibrium. Find the force exerted by the wall and the least coefficient of friction at the ground.[7 marks]
Answer
- Wall force $=40\sqrt3\,\text{N}$
- Least coefficient of friction $=\dfrac{\sqrt3}{6}$
Method: Let the horizontal wall force be $S$ . Taking moments about the foot of the ladder, $S(5\sin60^\circ)=240(2.5\cos60^\circ)$ , so $S=40\sqrt3\,\text{N}$ . Horizontal equilibrium gives friction $F=S=40\sqrt3\,\text{N}$ , and vertical equilibrium gives ground reaction $R=240\,\text{N}$ . At limiting equilibrium $F=\mu R$ , so $\mu=40\sqrt3/240=\sqrt3/6$ .