A cubical elastomeric block of side 20 cm has its lower face firmly fixed to a rigid floor. A tangential force is applied to its upper face, causing it to be displaced by 0.5 cm relative to the lower face. If the shear modulus of the material is 4 10^4 N m ⁻² , the magnitude of the applied tangential force is _____ N .

Step-by-step solution

The side length of the cubical block is a = 20 cm = 0.2 m . The area of the upper face on which the tangential force is applied is: A = a^2 = (0.2)^2 = 0.04 m ^2 The displacement of the upper face is x = 0.5 cm = 0.005 m . The shear strain is given by the ratio of displacement to the height of the block: Shear strain = x a = 0.005 0.2 = 0.025 The shear modulus is defined as the ratio of shear stress to shear strain: = F / A x / a Rearranging for the tangential force F : F = A ( x a ) Substituting the values: F = (4