Question

1. a uniform beam (mass = 22 kg) is supported by a cable that is attached to the centre of the beam as shown in the diagram.

a. find the tension in the cable.
b. find the horizontal and vertical forces acting on the hinge.

Explanation:

Step1: Calculate the weight of the beam

The weight of the beam $W_{beam}=mg$, where $m = 22\ kg$ and $g = 9.8\ m/s^{2}$. So $W_{beam}=22\times9.8 = 215.6\ N$. The weight of the object $W_{object}=25\times9.8=245\ N$.

Step2: Take moments about the hinge

Let the length of the beam be $L$. The tension in the cable is $T$. The vertical - component of the tension is $T\sin45^{\circ}$. The weight of the beam acts at its center and the weight of the object acts at the end of the beam. Taking moments about the hinge $\sum\tau = 0$. The weight of the beam creates a clock - wise moment and the weight of the object creates a clock - wise moment, while the vertical component of the tension creates a counter - clockwise moment. So $T\sin45^{\circ}\times\frac{L}{2}=W_{beam}\times\frac{L}{2}+W_{object}\times L$. Canceling out $L$ from both sides, we have $T\sin45^{\circ}\times\frac{1}{2}=W_{beam}\times\frac{1}{2}+W_{object}$.

Step3: Solve for the tension $T$

$T\sin45^{\circ}\times\frac{1}{2}=\frac{215.6}{2}+245$. $T\times\frac{\sqrt{2}}{2}\times\frac{1}{2}=107.8 + 245$. $T\times\frac{\sqrt{2}}{4}=352.8$. $T=\frac{352.8\times4}{\sqrt{2}}=\frac{1411.2}{\sqrt{2}}\approx997.7\ N$.

Step4: Analyze the forces at the hinge

In the horizontal direction, the horizontal component of the tension $T_x=T\cos45^{\circ}$. $T_x=\frac{1411.2}{\sqrt{2}}\times\frac{\sqrt{2}}{2}=705.6\ N$. In the vertical direction, let the vertical force at the hinge be $F_y$. The sum of vertical forces $\sum F_y = 0$. $F_y+T\sin45^{\circ}=W_{beam}+W_{object}$. $F_y=(W_{beam}+W_{object})-T\sin45^{\circ}$. Substituting the values, $F_y=(215.6 + 245)-\frac{1411.2}{\sqrt{2}}\times\frac{\sqrt{2}}{2}=460.6-705.6=- 245\ N$. The negative sign indicates the direction is downward.

Answer:

a. The tension in the cable is approximately $997.7\ N$.
b. The horizontal force acting on the hinge is $705.6\ N$ and the vertical force acting on the hinge is $245\ N$ (downward).