Source: High school physics (Chinese)
Problem Sets:
Problem
A non-uniform rod AB of length L = 2.0 m and weight G = 100 N is suspended horizontally by two ropes and is in static equilibrium. One rope at end A makes an angle of 37° with the vertical, and the other rope at end B makes an angle of 53° with the vertical. (Use sin 37° = 0.6).
- Find the tensions, $T_1$ and $T_2$, in the two ropes.
- Find the distance, $x$, from the rod's center of mass C to end A.
[Q1] $T_1 = 80$ N, $T_2 = 60$ N [Q2] $x = 0.72$ m
The rod is in static equilibrium, so the net force and net torque are both zero. We use $\sin 37^\circ = 0.6$, $\cos 37^\circ = 0.8$, $\sin 53^\circ = 0.8$, and $\cos 53^\circ = 0.6$.
For force equilibrium: Sum of horizontal forces: $\sum F_x = T_2 \sin 53^\circ - T_1 \sin 37^\circ = 0 \implies T_2(0.8) = T_1(0.6)$
$$T_1 = \frac{4}{3} T_2$$Sum of vertical forces: $\sum F_y = T_1 \cos 37^\circ + T_2 \cos 53^\circ - G = 0$
$$T_1(0.8) + T_2(0.6) = 100 \text{ N}$$Substituting $T_1$ into the vertical force equation:
$$(\frac{4}{3} T_2)(0.8) + T_2(0.6) = 100$$ $$T_2 (\frac{3.2}{3} + 0.6) = T_2 (\frac{5.0}{3}) = 100 \implies T_2 = 60 \text{ N}$$ $$T_1 = \frac{4}{3} (60 \text{ N}) = 80 \text{ N}$$For torque equilibrium, we take the sum of torques about pivot A:
$$\sum \tau_A = (T_2 \cos 53^\circ) L - Gx = 0$$Solving for x:
$$x = \frac{T_2 L \cos 53^\circ}{G}$$Substituting the values:
$$x = \frac{(60 \text{ N})(2.0 \text{ m})(0.6)}{100 \text{ N}} = 0.72 \text{ m}$$