1. A uniform 22 kg beam, 3.0 m long, is attached to a wall on the left by a hinge, and is supported by a cable on its right end. A 13 kg monkey hangs from the beam, 1.0 m from the left end. 25°7 a) Determine the tension in the cable. b) Determine the magnitude of the force due to the hinge on the beam. FT

Torque acting on an object is defined as the product of perpendicular force to the distance from the pivot point , that is , $\tau =rF$ .

Let the hinge point be the pivot point , so the torques are due to the tension force acting in the anti-clockwise direction , which is , $\tau$ acting at a distance $d_{\tau }=3 m$ from the pivot point , the weight force of the beam acting clockwise direction , which is , $F_b=m_{b}g$ acting at a distance $d_b=1.5 m$ from the pivot point and the weight force of the monkey acting clockwise direction , which is , $F_m=m_{m}g$ acting at a distance $d_m=1 m$ from the pivot point . As the system is at equilibrium , the net torque will be zero , that is ,$\begin{array}{rcl}& & T\end{array}\begin{array}{rcl}& & d\end{array}\begin{array}{rcl}& & {}_T\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & m\end{array}\begin{array}{rcl}& & {}_b\end{array}\begin{array}{rcl}& & g\end{array}\begin{array}{rcl}& & d\end{array}\begin{array}{rcl}& & {}_b\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & m\end{array}\begin{array}{rcl}& & {}_m\end{array}\begin{array}{rcl}& & g\end{array}\begin{array}{rcl}& & d\end{array}\begin{array}{rcl}& & {}_m\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 0\end{array}$ , where $m_b=22 kg$ is the mass of beam , $m_m=13 kg$ is the mass of monkey and $g=9.8 m/s^2$ is the acceleration due to gravity .

The horizontal forces acting are the horizontal component of hinge force which is $F_{Hx}=F_H\cos \left(\theta \right)$ and the the horizontal component of tension force , which is , ${\tau }_x=\tau \cos \left(25°\right)$ . As the system is at equilibrium , the relation will be $F_H\cos \left(\theta \right)=T\cos \left(25°\right)$ .