F(min)/A = 4.0*10^8 N/m^2
A = 6*(0.0556)*(0.005) m^2
= 1.668*10^-3 m^2
is the load-bearing area.
Solve for F(min)
A manufacturer wishes to produce crude benzene models for the chemistry department. The manufacturer plans to do this by punching regular hexagons out of steel. If the steel plate is 0.500 cm thick, What is the minimum shearing force needed to punch regular hexagons with sides of length 5.56 cm? ________N Note: Assume that if the shear stress in steel exceeds 4.0 x 10^8 N/M^2 the steel ruptures.
2 answers
Thanks a lot