1. **Problem (a):** A see-saw pivoted at the middle is balanced by weights of Richard (60 kg), John (unknown mass), and Philip (15 kg) sitting at distances 2 m, 3 m, and 4 m respectively from the pivot. Find John's mass. 2. **Formula and principle:** For balance, the sum of clockwise moments equals the sum of anticlockwise moments about the pivot. 3. **Moments calculation:** - Richard's moment = $60 \times 2 = 120$ kg·m (one side) - John's moment = $m \times 3$ kg·m (other side) - Philip's moment = $15 \times 4 = 60$ kg·m (same side as John) 4. **Balance equation:** $$ 120 = 3m + 60 $$ 5. **Solve for $m$:** $$ 3m = 120 - 60 = 60 $$ $$ m = \frac{60}{3} = 20 $$ 6. **Answer (a):** John's mass is 20 kg. --- 7. **Problem (b):** A 12 kg body rests on a rough plane inclined at $30^\circ$. Coefficient of friction $\mu = 0.23$. A force $P$ acts along the plane. Find $P$ when the body is about to move down the plane. 8. **Forces on the body:** - Weight component down the plane: $W_{\text{down}} = mg \sin 30^\circ$ - Normal force: $N = mg \cos 30^\circ$ - Friction force up the plane: $F = \mu N$ 9. **Calculate components:** - $mg = 12 \times 9.8 = 117.6$ N - $W_{\text{down}} = 117.6 \times \sin 30^\circ = 117.6 \times 0.5 = 58.8$ N - $N = 117.6 \times \cos 30^\circ = 117.6 \times 0.866 = 101.8$ N - $F = 0.23 \times 101.8 = 23.4$ N 10. **At impending motion down the plane, $P$ acts up the plane to prevent motion:** Sum of forces along plane = 0: $$ P + F = W_{\text{down}} $$ 11. **Solve for $P$:** $$ P = W_{\text{down}} - F = 58.8 - 23.4 = 35.4 $$ 12. **Answer (b):** The force $P$ is 35.4 N to keep the body at the point of moving down the plane.