1. **State the problem:** We have a trap door hinged at point A with width 100 cm. It has a weight of 30 N acting downwards at the centre of gravity 45 cm from A. An object weighing 40 N is placed on the trap door acting downwards at point B, 25 cm from A. We apply a vertical upward force at point C (100 cm from A) just enough to lift the door. We want to find the magnitude of this force at C. 2. **Diagram forces and distances:** - Weight of trap door $W_1=30\,N$ at $45\,cm$ from A. - Weight of object $W_2=40\,N$ at $25\,cm$ from A. - Unknown upward force $F_C$ at $100\,cm$ from A. 3. **Apply conditions for equilibrium:** Since the door is just lifted, the hinge at A experiences no vertical reaction force. The door is on the verge of turning upwards about A, so the sum of moments about A must be zero. 4. **Write the moment equilibrium equation about A:** Taking clockwise moments as positive: $$ \text{Sum of moments about } A = 0 = F_C \times 100 - W_1 \times 45 - W_2 \times 25 $$ 5. **Substitute known values:** $$ 0 = F_C \times 100 - 30 \times 45 - 40 \times 25 $$ $$ 0 = 100F_C - 1350 - 1000 $$ $$ 100F_C = 2350 $$ 6. **Solve for $F_C$:** $$ F_C = \frac{2350}{100} = 23.5\,N $$ **Final answer:** The magnitude of the force applied vertically at C to just lift the door is **23.5 N**.