Source: High school physics (Chinese)
Problem Sets:
Problem
A rope has one end fixed at point A on a vertical wall. The other end supports an object of mass $m$. A horizontal wooden rod BC holds the rope away from the wall, with point B on the rope and point C on the wall. The system is in static equilibrium. The segment of the rope AB makes an angle $\theta$ with the horizontal rod. The weights of the rope and the rod are negligible.
- Find the tension in the rope segment AB.
- Find the compressive force in the rod BC.
[Q1] $T_{AB} = \frac{mg}{\sin\theta}$ [Q2] $F_{BC} = mg \cot\theta$
We analyze the forces acting on the junction point B, where the rope and rod meet. Let $T_{AB}$ be the tension in rope AB, $F_{BC}$ be the compressive force from rod BC, and $W=mg$ be the weight of the mass. Since the system is in static equilibrium, the vector sum of forces at point B is zero ($\sum \vec{F} = 0$).
We resolve the forces into horizontal (x) and vertical (y) components.
For vertical equilibrium ($\sum F_y = 0$): The upward vertical component of the tension $T_{AB}$ must balance the downward weight of the mass.
$$T_{AB} \sin\theta - mg = 0$$Solving for $T_{AB}$:
$$T_{AB} = \frac{mg}{\sin\theta}$$For horizontal equilibrium ($\sum F_x = 0$): The horizontal force from the rod $F_{BC}$ must balance the horizontal component of the tension $T_{AB}$.
$$F_{BC} - T_{AB} \cos\theta = 0$$Solving for $F_{BC}$ and substituting the expression for $T_{AB}$:
$$F_{BC} = T_{AB} \cos\theta = \left(\frac{mg}{\sin\theta}\right) \cos\theta = mg \cot\theta$$