A cube has side length 5 cm and modulus of rigidity 10^5 N/m ^2 . The displacement produced by a force of 10 N in the upper face of cube is _____ mm.

Step-by-step solution

Given: Side length of the cube, L = 5 cm = 0.05 m Modulus of rigidity, = 10^5 N/m ^2 Force applied on the upper face, F = 10 N Area of the upper face, A = L^2 = (0.05)^2 = 2.5 10⁻³ m ^2 Shear stress is given by: Stress = F A = 10 2.5 10⁻³ = 4000 N/m ^2 Shear strain is given by: Strain = x L Modulus of rigidity is defined as the ratio of shear stress to shear strain: = Stress Strain = F/A x/L Rearranging for displacement x : x = F L A Substituting the values: x = 10 0.05 2.5 10⁻³ 10^5 x = 0.5 250 = 0.002 m Convertin