Question

A mechanic jacks up the front of a car to an angle of 8.0° with the horizontal in order to change the front tires. the car is 3.05 m long and has a mass of 1130 kg. gravitational force acts at the center of mass, which is located 1.12 m from the front end. the rear wheels are 0.40 m from the back end. calculate the magnitude of the torque exerted by the jack.

Answer 1

First draw a free-body diagram. Torque T = Force F x Distance d where force is the component of gravitational force g and d is the lever arm distance to the pivot point. Since the pivot point is at the back tire we subtract that from the length of the car resulting in d = 1.12 - 0.40 = 0.72 meters = d. We are interested in the perpendicular component of the force exerted on the car jack so use sin 8 degrees then T=1130 kg x 9.81 m/s^2 x sin(8 degrees) x0.72 m = 1,110.80 Newton-meters