Torque: A 1.10-kg wrench is acting on a nut trying to turn it. The length of the wrench lies directly to the east of the nut. A force 150.0 N acts on the wrench at a position 15.0 cm from the center of the nut in a direction 30.0° north of east. What is the magnitude of the torque about the center of the nut?
A) 22.5 N ∙ m
B) 11.3 N ∙ m
C) 19.5 N ∙ m
D) 2250 N ∙ m
E) 1949 N ∙ m

Question

Quizplus · Accepted Answer

First, we need to find the component of the force that is perpendicular to the wrench. This can be found by multiplying the force by the sine of the angle between the force and the wrench:
$F_{perp} = 150.0 N \sin(30.0°) = 75.0 N$
Next, we can find the torque about the center of the nut by multiplying the force perpendicular to the wrench by the length of the wrench:
$	au = F_{perp} \cdot r = (75.0 N)(0.15 m) = 11.3 N \cdot m$
Therefore, the correct answer is B.

Torque: a 1