Question: In Figure 12-40, one end of a uniform beam of weight is hinged to a wall; the other end is supported by a wire that makes angles $\mathit{\theta }\mathbf{}\mathbf{=}{\mathbf{30}}^{\mathbf{o}}$ with both wall and beam. (a) Find the tension in the wire and the (b) Find horizontal and (c) Find vertical components of the force of the hinge on the beam.

1. The weight of the beam is, W = 222 N.
2. The angle between the wall and the cable, $\theta ={30}^0$

You use the concept of torque and Newton’s second law. By writing the equation for net torque and forces along the x and y directions for equilibrium, you can solve for the tension and force components.

You know that the wire makes an angle of 30⁰ at a vertical, and the beam makes an angle of 60⁰ at a vertical.

We can write the equation for net torque about the hinge as:

$$\begin{gathered} \sum Torque=0 \\ TL\sin 30°-W\left(\frac{L}{2}\right)\sin 60°=0 \end{gathered}$$

Rearranging for T, we get,

$T=\frac{W\sin 60°}{2\sin 30°}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} T=\frac{222\sin 60°}{2\sin 30°} \\ =192\text{N} \end{gathered}$$

Thus, the tension in the wire is 192 N.

We can write the equation for net force in the x direction as,

$$\begin{gathered} \sum F_x=0 \\ {F}_x-T\sin 30°=0 \\ {F}_x=T\sin 30° \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} F_x=192\sin {30}^0 \\ =96.1N \end{gathered}$$

Thus, the horizontal component of the force of the hinge on the beam, $F_x=96.1\text{N}$.

We can write the equation for net force in the y direction as,

$$\begin{gathered} \sum F_y=0 \\ {F}_y+T\cos 30°-W=0 \\ {F}_y=W-T\cos 30° \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} F_y=222-192\cos 30° \\ =55.5\text{N} \end{gathered}$$

Thus, the vertical component of the force of the hinge on the beam,$F_y=55.5\text{N}$ .