A square copper slab of side 20 cm and thickness 5 cm is subject to a shearing force (on its narrow face) of 5.0 × 10² N. The lower edge is fixed to the floor. How much will the upper edge be displaced. G = 42 × 10⁹ N m^-2

Let us first calculate the area of the square face.

A = 20 cm × 5 cm = 0.2 m × 0.05 m = 0.01 m²

Stress = ^{5 × 102 N}/_{0.01 m²} = 5 × 10⁴ N m^-2

Shearing strain = ^Δx/_L = ^Stress/_G

∴ Displacement Δx = ^{(Stress × L)}/_G

Δx = ^{(5 × 104 N m-2 × 0.05 m)}/_{42 × 10⁹ N m^-2}

= 5.95 × 10^-8 m