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

First, calculate the center of mass of the beam:
$$x_{CM}=\frac{1}{2}L=\frac{1}{2}(10	ext{ m})=5	ext{ m}$$
Next, calculate the torque due to the weight of the beam about the pivot point:
$$	au_{beam}=mgx_{CM}=(25	ext{ kg})(9.81\frac{	ext{m}}{	ext{s}^2})(5	ext{ m})=1226	ext{ Nm}$$
Next, calculate the torque due to the person about the pivot point:
$$	au_{person}=F_{person}d_{person	ext{-}pivot}\sin	heta=(60	ext{ kg})(9.81\frac{	ext{m}}{	ext{s}^2})(3.0	ext{ m})\sin(51^\circ)=1136	ext{ Nm}$$
Since the beam is at equilibrium, the torques must balance:
$$	au_{beam}+	au_{person}=	au_{cable}$$
where $	au_{cable}$ is the torque due to the tension in the cable. Solving for the tension in the cable:
$$	au_{cable}=	au_{beam}+	au_{person}=(1226+1136)	ext{ Nm}=2362	ext{ Nm}$$
$$F_{cable}d_{cable	ext{-}pivot}\sin(180^\circ-51^\circ)=2362	ext{ Nm}$$
$$F_{cable}=(\frac{2362	ext{ Nm}}{(7	ext{ m})\sin(51^\circ)})\approx0.38	ext{ kN}$$
Therefore, the tension in the cable is approximately 0.38 kN.

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