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 $F_{g1}=mg$, where $m = 9.0\ kg$ and $g = 9.8\ m/s^{2}$. So $F_{g1}=9.0\times9.8 = 88.2\ N$. The center - of - mass of the uniform beam is at its mid - point, $L_1=\frac{2.6}{2}=1.3\ m$ from the hinge.

Step2: Calculate the weight of the hanging mass

The weight of the hanging mass $F_{g2}=m_2g$, where $m_2 = 13\ kg$ and $g = 9.8\ m/s^{2}$. So $F_{g2}=13\times9.8=127.4\ N$, and it is at $L_2 = 2.2\ m$ from the hinge.

Step3: Set up the torque equation about the hinge

Let the tension in the rope be $T$. The torque due to the tension in the rope is $\tau_T=T\times1.1\times\sin25^{\circ}$, the torque due to the beam's weight is $\tau_1 = F_{g1}\times1.3$ and the torque due to the hanging mass is $\tau_2=F_{g2}\times2.2$. For the beam to be in rotational equilibrium, $\sum\tau = 0$. Taking counter - clockwise torque as positive, we have $T\times1.1\times\sin25^{\circ}-F_{g1}\times1.3 - F_{g2}\times2.2=0$.

Step4: Solve for the tension $T$

$$
LATEXBLOCK0
$$

Answer:

$T\approx849.5\ N$