Source: High school physics (Chinese)
Problem Sets:
Problem
A trapdoor on a ceiling is a square board with a side length of $1.00$ m and a weight of $100$ N. One of its sides is connected to the ceiling with hinges, while the opposite side is secured by a latch. It is known that the door's center of gravity is located on the same horizontal plane as the door's center but is displaced by $10$ cm towards the hinged side.
Force on the latch: $F_l = 40$ N. Force on the hinges: $F_h = 60$ N.
This is a static equilibrium problem. Let $F_h$ be the force from the hinges and $F_l$ be the force from the latch. Both act upwards, opposing the downward weight $G$. The door has length $L = 1.00$ m and weight $G = 100$ N. The center of gravity (CG) is displaced by $d = 10$ cm $= 0.10$ m from the geometric center towards the hinges.
We apply the condition for rotational equilibrium by taking torques about the hinge axis. This eliminates the hinge force $F_h$ from the equation. The latch is at a distance $L$ from the hinge. The CG is at a distance $x_{CG} = L/2 - d$ from the hinge. The sum of torques is zero:
$$\sum \tau_{\text{hinge}} = F_l \cdot L - G \cdot x_{CG} = 0$$ $$F_l = G \frac{x_{CG}}{L} = G \frac{L/2 - d}{L}$$Substitute the values:
$$F_l = (100 \text{ N}) \frac{1.00 \text{ m}/2 - 0.10 \text{ m}}{1.00 \text{ m}} = 100 \frac{0.50 - 0.10}{1.00} \text{ N}$$We apply the condition for translational equilibrium. The sum of vertical forces is zero:
$$\sum F_y = F_h + F_l - G = 0$$ $$F_h = G - F_l$$Substitute the value of $F_l$:
$$F_h = 100 \text{ N} - 40 \text{ N}$$