91Ó°ÊÓ

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

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:

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

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,

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

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,

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

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,$F_y=55.5\text{N}$ .