Question

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?

Answer 1

Hi! Have you tried drawing a diagram?

It might help! Then simplify it so you see only the important stuff!

You'll probably want to use \[\large \tau =r\ F\ sin\left(\theta\right)\]
Where
\(\tau\) is the torque,
\(r\) is the distance from where the pivot is to where the force is,
\(F\) is the force exerted, and
\(\theta\) is the angle between the force and radius

Answer 2

You can check to see if my diagram will work with your problem. If you have additional questions, you can probably get help by posting more! What I posted are general guidelines for you!