Assume that if the shear stress in steel exceeds about 4.00 108 n/m2, the steel ruptures. (a) determine the shearing force necessary to shear a steel bolt 1.15 cm in diameter. (b) determine the shearing force necessary to punch a 1.45-cm-diameter hole in a steel plate 0.860 cm thick.

A. Determine the area of a circle with 1.15 cm diameter. 
                    A = πr²

Substituting,
                  A = π(1.15 cm/100 cm)² = 4.15 m²

Multiply this to the given shearing stress. 
              shearing force = (4 x 10^8 N/m²)(4.15 m²)
                 shearing force = 1.66 x 10^9 N

b. Determine the surface area of the cylinder that is made upon boring the hole in steel plate.
                      A = 2πrh

Substituting,
                    A = 2π(0.0145m)(0.0086) 
                     A = 7.84 x 10^-4 m²

Multiplying with shear stress,
                  shear force = 3.13 x 10^5 N

Is this a question of an answer

Since the pully is a simple machine so the mechanical advantage is a 1