Magnitude of the torque

Question

A mechanic uses a 12 in wrench to turn a bolt. The wrench makes a 30° angle with the horizontal. If the mechanic applies a vertical force of 14 lb on the wrench handle, what is the magnitude of the torque at point P? Express the answer in foot-pounds. (Round your answer to two decimal places.)

Answer 1

To calculate the torque (τ) produced by the force applied by the mechanic, we use the formula for torque:

τ = r⋅F⋅sin⁡(θ)

where:

1. r is the length of the wrench (12 inches, which we will convert to feet),
2. F is the force applied (14 lb),
3. θ is the angle between the force and the lever arm.

First, convert the length of the wrench from inches to feet: r = 1 foot

The force F is given as 14 lb.

The angle θ given in the problem is 30°, which represents the angle the wrench makes with the horizontal. Since we want the angle between the force and the lever arm, the angle θ we use in our formula is 60° (90° - 30° = 60°).

Now, we can substitute these values into the torque formula: τ = 1 foot ⋅ 14 lb ⋅ (√3)/2

Calculating this, we get: τ = 7 foot-pounds

So, the magnitude of the torque at point P is: τ = 12.12 foot-pounds

Thus, the torque is 12.12 foot-pounds when rounded to two decimal places.

Hope this was helpful. Credits to Roger R. (see below comment thread) for helping me revise the solution for this problem.