Question

A heavy bank-vault door is opened by the application of a force of 3.0 x 102 N directed perpendicular to the plane of the door at a distance of 0.80 m from the hinges. What is the torque?
120 N·m
240 N·m
300 N·m
360 N·m

Answer 1

Torque is the cross product of the distance from the pivot and the force vector:\[t=r \space x \space F=r \space x \space F sin(\theta) \]So in this problem where the force is perpendicular to the plane of the door you'll get:\[torque=(0.80m) (3.0x10^2N)sin(90)=240N *m\]

Answer 2

Thanks again! :)