A uniform beam of length 3.00 m and weight 100 N is mounted on an axle at one end perpendicular to the length of the beam.A rope is attached to the end of the beam at the other end from the axle,and the beam is lifted by the rope so that the beam makes an angle of 30.0 $\degree$ with the horizontal.If the rope is straight up,what magnitude torque does it supply about the axle?
A)300 N·m
B)150 N·m
C)130 N·m
D)75.0 N·m

Question

Quizplus · Accepted Answer

The torque supplied by the rope can be calculated as

τ = Fd sinθ

where F is the tension in the rope, d is the distance from the axle to the point where the rope attaches to the beam, and θ is the angle between the rope and the horizontal.

Since the rope is straight up, F is equal in magnitude to the weight of the beam, 100 N. The distance d can be found using geometry:

d = L sinθ

where L is the length of the beam. Plugging in the given values,

d = 3.00 m sin30.0° = 1.50 m

Substituting into the torque formula,

τ = (100 N) (1.50 m) sin30.0° = 130 N·m

Therefore, the answer is (C) 130 N·m.

A Uniform Beam of Length 3 °\degree° With the Horizontal

A Uniform Beam of Length 3 $\degree$ With the Horizontal