The figure shows a uniform, horizontal beam (length = 10 m, mass = 25 kg) that is pivoted at the wall, with its far end supported by a cable that makes an angle of 51° with the horizontal. If a person (mass = 60 kg) stands 3.0 m from the pivot, what is the tension in the cable?
A) 0.83 kN
B) 0.30 kN
C) 0.38 kN
D) 0.42 kN
E) 3.0 kN

Question

The figure shows a uniform, horizontal beam (length = 10 m, mass = 25 kg) that is pivoted at the wall, with its far end supported by a cable that makes an angle of 51° with the horizontal. If a person (mass = 60 kg) stands 3.0 m from the pivot, what is the tension in the cable? 
A) 0.83 kN 
B) 0.30 kN 
C) 0.38 kN 
D) 0.42 kN 
E) 3.0 kN

Quizplus · Accepted Answer

To solve this problem, we need to use the principle of torque equilibrium. The torque due to the weight of the beam can be calculated as $M_{beam}=mg\frac{L}{2}cos	heta$, where $m$ is the mass of the beam, $g$ is the acceleration due to gravity, $L$ is the length of the beam, $	heta$ is the angle between the beam and the horizontal. This gives $M_{beam}=25	imes9.81	imes\frac{10}{2}cos0=122.62	ext{ Nm}$.

The torque due to the person is $M_{person}=mgd\sin	heta$, where $m$ is the mass of the person, $g$ is the acceleration due to gravity, $d$ is the distance between the person and the pivot, and $	heta$ is the angle between the person's weight and the horizontal. This gives $M_{person}=60	imes9.81	imes3.0\sin51=882.17	ext{ Nm}$.

The cable provides an upward force on the beam, which prevents it from rotating downwards. The tension in the cable can be found by requiring that the torques due to the person and the beam cancel each other out. That is, $M_{person}=M_{beam}=T	imes L\sin	heta$. This gives $T=\frac{M_{person}}{L\sin	heta}=\frac{882.17}{10\sin51}=381.83	ext{ N}\approx0.38	ext{ kN}$.

Therefore, the tension in the cable is 0.38 kN, which corresponds to choice C.

The Figure Shows a Uniform, Horizontal Beam (Length = 10