The lower surface of a steel plate having a face area of 4 cm ^2 and a thickness of 0.5 cm is fixed rigidly. A tangential force of 10 N is applied to the upper surface. Determine the lateral displacement of the upper surface relative to the lower surface. The rigidity modulus of steel is 8.4 10¹⁰ N/m ^2 .

Step-by-step solution

Given, tangential force F = 10 N Face area A = 4 cm ^2 = 4 10⁻⁴ m ^2 Thickness L = 0.5 cm = 0.5 10⁻² m Rigidity modulus = 8.4 10¹⁰ N/m ^2 The formula for rigidity modulus is given by: = Shear Stress Shear Strain = F / A x / L Rearranging for lateral displacement x : x = F L A Substituting the given values: x = 10 0.5 10⁻² 4 10⁻⁴ 8.4 10¹⁰ x = 5 10⁻² 33.6 10^6 x = 5 33.6 10⁻⁸ x 1.488 10⁻⁹ m x 1.5 10⁻⁹ m