torque wrench directions
Please please help! Torque question.?
A 95 N force exerted at the end of a 0.7 m long torque wrench gives rise to a torque of 15 N ∙ m. What is the angle (assumed to be less than 90°) between the wrench handle and the direction of the applied force?
The torque applied to the wrench is the force exerted at the end with a perpendicular component at an unknown angle that when multiplied by 0.7 m will provide 15 Nm of torque. In this case, the perpendicular component is the sine of angle. (I wish there was a way to draw to make this easier to explain.) Therefore,
15 Nm = (95 N) (sin(theta)) (0.7 m)
Solving for theta we get,
sin(theta) = 15 Nm / (95 N) (0.7 m) [Notice that the units cancel out as they should.]
sin(theta) = 15 / 66.5 = 0.2256
theta = arcsin(0.2256) = 13.04 degrees.