Equilibrium of a rigid body — practice questions

1. Draw a diagram and label forces:
   • Weight $W = 12g$ acting downwards from the centre of the ladder (3 m from the base).
   • Normal reaction from the wall, $R_W$, acting horizontally outwards.
   • Normal reaction from the ground, $R_G$, acting vertically upwards.
   • Frictional force, $F$, acting horizontally inwards along the ground.
2. Resolve forces vertically: The only vertical forces are $R_G$ and $W$. For equilibrium: $\sum F_y = 0 \implies R_G - 12g = 0$ $R_G = 12g = 12 \times 9.8 = 117.6$ N. So, the normal reaction at the ground is 118 N (3 s.f.).
3. Take moments to find an unknown force. A good pivot is the base of the ladder, as this eliminates $R_G$ and $F$ from the moment equation.
   • Pivot: Point on the ground where the ladder rests.
   • Clockwise moment: Caused by the weight $W$. The perpendicular distance from the pivot to the line of action of the weight is $3 \cos \theta$. Moment = $12g \times 3 \cos \theta = 36g \cos \theta$.
   • Anticlockwise moment: Caused by the wall reaction $R_W$. The perpendicular distance is $6 \sin \theta$. Moment = $R_W \times 6 \sin \theta$.
4. Equate moments: $R_W \times 6 \sin \theta = 36g \cos \theta$ $R_W = \frac{36g \cos \theta}{6 \sin \theta} = \frac{6g}{\tan \theta}$ Given $\tan \theta = 2$, we have: $R_W = \frac{6g}{2} = 3g = 3 \times 9.8 = 29.4$ N.
5. Resolve forces horizontally: The only horizontal forces are $F$ and $R_W$. $\sum F_x = 0 \implies F - R_W = 0$ $F = R_W = 29.4$ N.

Answer: The frictional force is 29.4 N and the normal reaction at the ground is 118 N (3 s.f.).

1. Draw a diagram:
   • Rectangular block with base on the ground. Let the bottom corners be A and B.
   • Weight $W = 40g$ acts at the centre of the block (0.25 m from A and B, 0.6 m from the base).
   • Applied force $P$ acts horizontally at height 0.9 m.
   • Normal reaction $R$ acts upwards from the base. When toppling is about to occur about corner B, the entire reaction force is concentrated at B.
   • Frictional force $F$ acts at B, opposing the motion that would be caused by $P$.
2. Identify the condition for toppling: 'On the point of toppling' means the block is about to rotate about the bottom-right corner (let's call it B). At this instant, the normal reaction from the ground at any other point (like corner A) is zero. The entire system pivots about B.
3. Take moments about the pivot point B: This is the most efficient method as it eliminates the unknown forces $R$ and $F$ (which both act through B).
   • Clockwise moment (toppling moment): Caused by the applied force $P$. The perpendicular distance from B to the line of action of $P$ is 0.9 m. Moment = $P \times 0.9$.
   • Anticlockwise moment (restoring moment): Caused by the weight $W$. The perpendicular distance from B to the line of action of $W$ is half the width of the block, which is $0.5 / 2 = 0.25$ m. Moment = $W \times 0.25 = 40g \times 0.25 = 10g$.
4. Equate moments for equilibrium: $0.9 P = 10g$ $P = \frac{10g}{0.9} = \frac{10 \times 9.8}{0.9}$ $P = \frac{98}{0.9} \approx 108.88...$ N.

Answer: The force required to cause toppling is $P = 109$ N (3 s.f.).

(Note: We would also need to check if the block slips first by calculating the required friction $F=P=109$ N and comparing it with the maximum available friction $\mu R = \mu (40g)$. The problem states to assume it does not slip.)