A cubical elastomeric pad of side 10 cm is firmly fixed to a horizontal table. A metal block of mass 6 kg is placed on top of the pad. The coefficient of static friction between the block and the pad is 0.4 . A horizontal force is gradually applied to the metal block until it just starts to slip relative to the pad. If the shear modulus of the pad's material is 1.2 10^5 N m ⁻² , the maximum shear displacement of the

Step-by-step solution

The maximum horizontal force that can be applied to the block before it slips is equal to the limiting static friction between the block and the pad. f_ max = M g = 0.4 6 10 = 24 N By Newton's Third Law, this is also the maximum tangential force F exerted on the top surface of the elastomeric pad. The side length of the cubical pad is a = 10 cm = 0.1 m . The area of the top surface is A = a^2 = (0.1)^2 = 0.01 m ^2 . The shear stress on the pad is: Stress = F A = 24 0.01 = 2400 N m ⁻² The shear strain is given by: S