Question

# A copper cube measures 6.00 cm on each side. The bottom face is held in place...

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?

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

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