Physics quiz question: falling beam torque

Physics 205A Quiz 5, fall semester 2018
Cuesta College, San Luis Obispo, CA

A uniform beam with a 0.15 kg mass and 1.2 m length pivots at one end as it falls over starting from an angle of 5° from the vertical. Calculate all torques with respect to the bottom of the beam. While at its starting point of 5°, the magnitude of the torque exerted by the weight force on the beam is:
(A) 0.077 N·m.
(B) 0.88 N·m.
(C) 1.5 N·m.
(D) 1.8 N·m.

Correct answer (highlight to unhide): (A)

The weight force w of Earth acting on the beam acts at the center of gravity, directly downwards with a magnitude m·g. The perpendicular lever arm ℓ for the weight force on the beam must extend from the pivot point to perpendicularly intercept the weight force line of action (which lies along the weight force vector), such that this will be a horizontal line of length:

ℓ = (L/2)·cos(85°).

The magnitude of the (clockwise) torque exerted by the weight force on the beam is then:

τ = w·ℓ,

τ = (m·g)·(L/2)·cos(85°),

τ = ((0.15 kg)·(9.80 m/s²))·((1.2 m)/2)·cos(85°),

τ = 0.0768713651... N·m,

or to two significant figures, the torque of the weight force on the beam has a magnitude of 0.077 N·m.

(Response (B) is (m·g)·(L/2)·sin(85°); response (C) is m·g; and response (D) is m·g·L.)

Sections 70854, 70855
Exam code: quiz05Ro74
(A) : 6 students
(B) : 11 students
(C) : 4 students
(D) : 31 students

Success level: 12%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.25