Instead of 90 degrees to the shaft (optimal) You are at 120 which is 30 degrees off.
so your force perpendicular to the shaft is
F cos 30
The torque is F cos 30 * 0.39
so
F cos 30 * 0.390 = 49.0
Help Needed Please...
After changing a tire on your car, you need to tighten one of the lug nuts to the proper specified torque of 49.0 N m. How much force is needed if you're pushing down on the end of the 39.0 cm long torque wrench at an angle of 120° with respect to the wrench shaft?
Please I need help on how to go about solving this question?
1 answer
1 answer