Physics quiz question: cable-held boom

Physics 205A Quiz 5, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 8.14

[3.0 points.] A uniform boom of length 1.20 m is held at an angle of 30.0° above the horizontal. A cable, at angle of 30.0° below the horizontal, is attached to the middle of the boom, and fastened to a wall. The boom is fastened to a hinge attached to the wall. Which is the lesser magnitude force acting on the boom?
(A) Tension.
(B) Weight.
(C) (Tension and weight have the same magnitude.)
(D) (Not enough information is given to determine this.)

Correct answer: (C)

The magnitude of the counterclockwise torque of the weight on the boom is:

tau_weight = r*F_perp = (0.60 m)*w*sin(60.0°).

This must be equal to the magnitude of the clockwise torque of the cable tension on the boom:

tau_tension = r*F_perp = (0.60 m)*T*sin(60.0°).

From inspection, the magnitudes of the weight and tension forces must be equal.

Student responses
Sections 70854, 70855
(A) : 11 students
(B) : 14 students
(C) : 15 students
(D) : 1 student

"Difficulty level": 36%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.12