Question

A copper cube measures 6.00 cm on each side. The bottom face is held in place by very strong glue to a flat horizontal surface, while a horizontal force F is applied to the upper face parallel to one of the edges.

A.) Show that the glue exerts a force F on the bottom face that is equal but opposite to the force on the top face, with a small statement.

B.) How large must F be to cause the cube to deform by 0.190 mm ?

C.) If the same experiment were performed on a lead cube of the same size as the copper one, by what distance would it deform for the same force as in part B?

Answer 1

a) For the block to not move the net force on the block has to zero so the glue exerts a force F on the bottom face that is equal but opposite to the force on the top face

(b) Area of the side = 6*6 *10^-4 = 36*10^-4 m^2

shear stress = Force/Area = F/(36*10^-4) = 277.777 F

shear modulus copper = 48 * 10^9 Pa

shear strain = deformation/length = 0.19 *10^-3 / 6*10^-2 = 0.0031667

shear modulus = shear stress/shear strain

48*10^9 = 277.777F/0.0031667

F = 5.47*10^5 N = 547 kN
(c) lead shear modulus = 5.6*10^9 Pa

G = (F/A)/(DL/L)

5.6*10^9 = (547000/(36*10^-4)) (0.06/DL)

deformation DL = 0.00163 m

deformation DL = 1.63 mm