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?
This can be solved with the following procedure: 150N * sin30 *. 15 = 11.25NM. Remember to use the triangle method for visualization, draw the right triangle with the properties given. Hope this is useful
