Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 8.19
A women pushes perpendicularly at the top edge of a fold-down Murphy bed frame (2.1 m length, 91 kg mass) to keep it stationary. Approximate the bed frame as a uniform beam raised at an angle of 55° from the horizontal. (Calculate all torques with respect to the pivot, located at the floor edge.) The magnitude of the force applied by the woman is _________ the magnitude of the weight of the bed frame.
(A) less than.
(B) equal to.
(C) greater than.
(D) (Not enough information is given.)
Correct answer (highlight to unhide): (A)
The force of the woman is already acting perpendicular to the bed, so the perpendicular lever arm rwoman is simply L = 2.1 m.
The weight of the bed acts at its center of gravity, straight downwards. The perpendicular lever arm for the weight force extends from the pivot to perpendicularly intercept the weight force line of action, such that this will be a horizontal line of length:
rw = (L/2)·cos(55°).
Since the bed is (implied to be) in static equilibrium and does not rotate, then the net torque on it is also equal to zero (Στ = 0), such that the clockwise torque of the woman on the bed must equal the counterclockwise torque of weight acting on the bed:
(cw) τwoman = (ccw) τw.
Substituting in the perpendicular lever arms for these torques:
Fwoman·rwoman = w·rw,
Fwoman·(L) = w·((L/2)·cos(55°)),
Fpost = w·(1/2)·cos(25°),
Fpost = w·(0.29).
Because the force of the woman on the bed has a longer perpendicular lever arm than the perpendicular lever arm of the weight on the bed, then the force of the post on the beam has a smaller magnitude than the weight of the bed.
Sections 70854, 70855, 73320
Exam code: quiz05mRp4
(A) : 25 students
(B) : 22 students
(C) : 17 students
(D) : 0 students
Success level: 39%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.79