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

Question

11.37 .. 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. (Consult Table 11.1.) (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. (b) How large must F be to cause the
cube to deform by 0.250 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)?

Explanation / Answer

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

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

shear modulus copper = 48 * 10^9 Pa

we have shear strain = deformation/length = 0.25 *10^-3 / 6*10^-2 = 0.004166

we have shear modulus = shear stress/shear strain

so shear strain = shear stress/shear modulus

so  0.004166 =(277.777 F )/ ( 48 * 10^9)

so F = 719.886 *10^3 N = 719.886 KN is the force

c)

for same force on lead cube

shear modulus lead = 5.6 GPa = 5.6*10^9 Pa

shear stress = Force/Area = (719.886 *10^3) / (36*10^-4) = 19.9968 *10^7 Pa

so shear modulus = shear stress/shear strain

(5.6*10^9) = (19.9968 *10^7 ) / shear strain

shear strain = 0.03570

so shear strain = deformation/length = deformation / 0.06 = 0.03570

deformation = 0.002142 m = 2.142 mm is the answer

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