Question

A rigid, uniform, horizontal bar of mass and length is supported by two identical massless strings. (Figure 1) Both strings are vertical. String A is attached at a distance d < L/2 from the left end of the bar and is connected to the ceiling; string B is attached to the left end of the bar and is connected to the floor. A small block of mass is supported against gravity by the bar at a distance from the left end of the bar, as shown in the figure. Throughout this problem positive torque is that which spins an object counterclockwise. Use g for the magnitude of the acceleration due to gravity.

Part A

Find \(T_{A}\), the tension in string \(A\). Express the tension in string \(\mathbf{A}\) in terms of \(g, m_{1}, L, d, m_{2}\), and \(x\).

Part B

Find \(T_{B}\), the magnitude of the tension in string \(\mathrm{B}\).

Express the magnitude of the tension in string \(\mathrm{B}\) in terms of \(T_{A}, m_{1}, m_{2}\), and \(g\).

Answer 1

1) Diagram:

The net torque about string B should be zero at the equilibrium state of rotation.

$$ \begin{aligned} \sum \tau &=0 \\ -m_{2} g(x)-m_{1} g\left(\frac{L}{2}\right)+T_{A} d &=0 \\ T_{A} &=\frac{m_{2} g(x)+m_{1} g\left(\frac{L}{2}\right)}{d} \\ &=\frac{2 m_{2} g x+m_{1} g L}{2 d} \end{aligned} $$

2) The net torque about string A should be zero at the equilibrium state of rotation.

$$ \begin{aligned} \sum \tau &=0 \\ -m_{2} g(x-d)-m_{1} g\left(\frac{L}{2}-d\right)+T_{B} d &=0 \\ T_{B} &=\frac{m_{2} g(x-d)+m_{1} g\left(\frac{L}{2}-d\right)}{d} \\ &=\frac{m_{2} g x-m_{2} g d+m_{1} g \frac{L}{2}-m_{1} g d}{d} \\ &=\frac{2 m_{2} g x+m_{1} g L-2\left(m_{2} g+m_{1} g\right) d}{2 d} \\ &=\frac{2 m_{2} g x+m_{1} g L}{2 d}-\frac{2\left(m_{2} g+m_{1} g\right) d}{2 d} \\ &=T_{A}-\left(m_{2} g+m_{1} g\right) \end{aligned} $$