Question

1. a 2.6 m uniform beam (mass of 9.0 kg) is attached to a wall by a hinge and supported by a rope. a 13 kg mass hangs from the beam 2.2 m from the hinge. find the tension in the rope that is attached to the beam 1.1 m from the wall.

Explanation:

Step1: Calculate the weight of the beam

The weight of the beam $W_{beam}=mg$, where $m = 9.0$ kg and $g = 9.8$ m/s². So $W_{beam}=9.0\times9.8 = 88.2$ N. Its center - of - mass is at $L_{beam}/2=2.6/2 = 1.3$ m from the hinge.

Step2: Calculate the weight of the hanging mass

The weight of the hanging mass $W_{mass}=m_{mass}g$, where $m_{mass}=13$ kg and $g = 9.8$ m/s². So $W_{mass}=13\times9.8 = 127.4$ N. It is at $L_{mass}=2.2$ m from the hinge.

Step3: Take the torque about the hinge

Let the tension in the rope be $T$. The rope is attached $L_{rope}=1.1$ m from the hinge. The angle between the rope and the beam is $25^{\circ}$. The torque due to the beam's weight is $\tau_{beam}=W_{beam}\times1.3\times\sin(90^{\circ})$, the torque due to the hanging mass is $\tau_{mass}=W_{mass}\times2.2\times\sin(90^{\circ})$, and the torque due to the tension in the rope is $\tau_{T}=T\times1.1\times\sin(25^{\circ})$.
Since the beam is in rotational equilibrium, $\sum\tau = 0$. So $\tau_{T}=\tau_{beam}+\tau_{mass}$.
$T\times1.1\times\sin(25^{\circ})=W_{beam}\times1.3 + W_{mass}\times2.2$.
Substitute $W_{beam}=88.2$ N and $W_{mass}=127.4$ N into the equation:
$T\times1.1\times\sin(25^{\circ})=88.2\times1.3+127.4\times2.2$
$T\times1.1\times0.4226=(88.2\times1.3)+(127.4\times2.2)$
$T\times0.4649 = 114.66+280.28$
$T\times0.4649 = 394.94$

Step4: Solve for the tension $T$

$T=\frac{394.94}{0.4649}\approx849.5$ N

Answer:

$849.5$ N