Question

If couple forces of F = 36 N are applied to the handle of the machinist’s vise, determine the compressive force developed in the block. Neglect friction at the bearing A. The guide at B is smooth. The single square-threaded screw has a mean radius of 6.5 mm and a lead of 8.2 mm, and the coefficient of static friction is mu_s = 0.27.

Answer 1

The compressive force developed in the block is 28.571 N.

Step-by-step explanation:

To determine the compressive force developed in the block, we can analyze the forces acting on the vise. The applied couple forces of F = 36 N will create a torque which can be translated into an axial compressive force.

The torque is given by the product of the force and the radius of the screw:

T = F * r.

The axial compressive force is then equal to the torque divided by the lead of the screw:

F_compressive = T / lead.

Substituting the given values, T = 36 N * 6.5 mm and lead = 8.2 mm, we find the axial compressive force F_compressive = (36 N * 6.5 mm) / (8.2 mm) = 28.571 N.