A uniform bar AB of mass m and length L is fixed to a vertical wall. The bar can rotate about end A in the vertical plane. A supporting string has one end fixed to the well and the other end to the midpoint of the bar making an angle $\theta$ with it as shown in Fig. A block of mass M hangs from end B of the bar. The system is in static equilibrium. If $\theta$ = 30° and M = m, the tension in the string is

 

Let $\overrightarrow{F}$ be the contact (reaction) force exerted by the wall or end A of the bar. Since the frictional force between the wall and the bar is present, force $\overrightarrow{F}$ will not be normal to the wall. Let F_x and F_y be the horizontal and vertical components of F. The horizontal and vertical components of tension T in the string are T_x = T cos $\theta$ and T_y = T sin $\theta$. Figure shows the forces acting on the bar. 

The bar is in translational as well as rotational equilibrium. For translational equilibrium, the net horizontal and vertical force must be zero, i.e. 

$$\begin{gathered} F_x-T_x=0 \Rightarrow F_x-T\cos \theta =0 \left(1\right) \\ and F_y+T_y-mg-Mg=0 \\ or F_y+T \sin \theta -\left(m+M\right) g=0 \left(2\right) \end{gathered}$$

For rotational equilibrium, the net clockwise torque about A = net anti-clockwise torque about A, i.e. ·

                 $Mg\times L+mg\times \frac{L}{2}=T_y\times \frac{L}{2}$

$$\begin{gathered} \Rightarrow \left(M+\frac{m}{2}\right)g=\frac{T \sin \theta }{2} \\ \Rightarrow T=\frac{\left(2M+m\right)g}{\sin \theta } \left(3\right) \end{gathered}$$

Notice that we do not need Eqs. (1) and (2) to find T. Putting M = m and  $\theta$= 30° in Eq. (3), we get T= 6mg.