Question

In Fig. 12-76, a uniform rod of mass m is hinged to a building at its lower end, while its upper end is held in place by a rope attached to the wall. If angle${\mathbf{\text{θ}}}_{\mathbf{1}}\mathbf{=}\mathbf{60}\mathbf{\text{°}}$, what value must angle${\mathit{\theta }}_{\mathbf{2}}$ have so that the tension in the rope is equal to $\mathbf{\text{mg/2}}$?

Short answer

The exact angle ${\theta }_2$for which tension in the rope is$\frac{\text{mg}}{\text{2}}$ is$60°$

Step-by-step solution

Understanding the given information

$\theta =60°$

$\text{T}=\text{mg/2}$

Concept and formula used in the given question

Using the condition for equilibrium, you can find the required angle for the given tension in the rope. The equations are given below.

$\begin{array}{l}\sum F_x=0\\ \sum F_y=0\\ \sum \tau =0\end{array}$

Calculation for the what value must angle θ2 have so that the tension in the rope is equal to  mg/2

We know the condition of equilibrium in which the moment of force, and torque is zero.

From the figure, we can say that

$\begin{array}{l}\left(\frac{L}{2}\right)\times mgsin{\theta }_1-\left(\frac{mg}{2}sin{\theta }_2\right)\times L=0\\ \frac{sin{\theta }_1}{2}-\frac{sin{\theta }_2}{2}=0\\ \frac{sin60}{2}-\frac{sin{\theta }_2}{2}=0\\ sin{\theta }_2=0.8660\\ {\theta }_2={60}^0\end{array}$$\begin{array}{l}\left(\frac{L}{2}\right)\times mgsin{\theta }_1-\left(\frac{mg}{2}sin{\theta }_2\right)\times L=0\\ \frac{sin{\theta }_1}{2}-\frac{sin{\theta }_2}{2}=0\\ \frac{sin60}{2}-\frac{sin{\theta }_2}{2}=0\\ sin{\theta }_2=0.8660\\ {\theta }_2={60}^0\end{array}$