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.
a) (sum\tau = 0)
(tsin45\times\frac{1}{2}=\frac{215.6 + 245}{2})
(t = 997.7n
ightarrow t = 1.0\times10^{3}n)
b)

Explanation:

Step1: Analyze torque equilibrium for tension

Taking the hinge as the pivot - point, the sum of torques $\sum\tau = 0$. The weight of the beam acts at its center and the weight of the attached mass acts at the end of the beam. The vertical component of the tension in the cable creates a counter - clockwise torque. Let the length of the beam be $L$. The weight of the beam $F_{g1}=mg$, where $m = 22$ kg and $g = 9.8$ m/s², so $F_{g1}=22\times9.8 = 215.6$ N. The weight of the attached mass $F_{g2}=M g$, where $M = 25$ kg and $g = 9.8$ m/s², so $F_{g2}=25\times9.8=245$ N. The vertical component of the tension $T$ is $T\sin45^{\circ}$, and its moment arm is $\frac{L}{2}$. The sum of the clock - wise torques due to the weights is $F_{g1}\times\frac{L}{2}+F_{g2}\times L$. So, $T\sin45^{\circ}\times\frac{L}{2}=F_{g1}\times\frac{L}{2}+F_{g2}\times L$. Canceling out $L$ from both sides, we get $T\sin45^{\circ}\times\frac{1}{2}=\frac{F_{g1}}{2}+F_{g2}$.
$T\times\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{215.6}{2}+245$
$T\times\frac{\sqrt{2}}{4}=107.8 + 245$
$T\times\frac{\sqrt{2}}{4}=352.8$
$T=\frac{352.8\times4}{\sqrt{2}}\approx997.7\approx1.0\times 10^{3}$ N

Step2: Analyze horizontal and vertical forces for hinge forces

For the horizontal force on the hinge $F_x$, since there are no other horizontal forces in the system (assuming no external horizontal loads), the horizontal component of the tension in the cable must be balanced by the horizontal force at the hinge. $F_x=T\cos45^{\circ}$. Substituting $T = 997.7$ N, $F_x=997.7\times\frac{\sqrt{2}}{2}\approx700$ N.
For the vertical force on the hinge $F_y$, the sum of vertical forces in the system is zero, $\sum F_y = 0$. The upward vertical component of the tension is $T\sin45^{\circ}$, and the downward forces are $F_{g1}$ and $F_{g2}$. So $F_y=F_{g1}+F_{g2}-T\sin45^{\circ}$. $F_y=(215.6 + 245)-997.7\times\frac{\sqrt{2}}{2}\approx0$ N (due to the equilibrium condition).

Answer:

a. The tension in the cable $T\approx1.0\times 10^{3}$ N
b. The horizontal force on the hinge $F_x\approx700$ N, the vertical force on the hinge $F_y\approx0$ N