A rectangular elastic pad of thickness 5 cm and cross-sectional area 0.1 m ^2 is glued to a fixed inclined plane which makes an angle of 30^ with the horizontal. A block of mass 4 kg is glued to the upper face of the pad. The system is in equilibrium under gravity. If the shear modulus of the pad is 2 10^5 N/m ^2 and the acceleration due to gravity is 10 m/s ^2 , the displacement of the block parallel to the incline

Step-by-step solution

The tangential force causing the shear in the elastic pad is the component of the block's weight parallel to the inclined plane. F_ shear = Mg (30^ ) = 4 10 1 2 = 20 N The shear stress is given by: = F_ shear A = 20 0.1 = 200 N/m ^2 The shear modulus G is the ratio of shear stress to shear strain ( ): G = = G = 200 2 10^5 = 10⁻³ The shear strain is also related to the lateral displacement x and thickness t of the pad by = x t . x = t = 10⁻³ (5 10⁻² m ) = 5 10⁻⁵ m Converting to micrometers: x = 50 10⁻⁶ m = 50 m Answ