Question

A piece of steel is held firmly in the jaws of a vise. A force larger than 3350 N will cause the piece of steel to start to move out of the vise. If the coefficient of friction between the steel and each of the jaws of the vise is 0.825 and each jaw applies an equal force, what is the magnitude of the normal force exerted on the steel by each jaw?

Answer 1

N=2030.30 N

Step-by-step explanation:

Given that

F= 3350 N

Coefficient of friction ,μ =0.825

Lets take N is the normal force

We know that friction force fr

fr= μ N

The total force in left side = 2 fr

Total force in right side =F

So by equating the above two forces

F= 2 fr

3350 = 2 x 0.825 x N

N=2030.30 N