A shearing force of 9.0×104N is applied on a thin surface of a square glass slab of 50cm side and width10cm. Second thin surface is stick to the floor. How much will the upper surface get displaced?
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Solution
Given, F=9.0×104N, length of s lab =50cm=50×10−2m Width of slab=10cm=10×10−2m ∴ Area of slab =50×10−2×10×10−2 =5×10−2m2 Modulus of rigidity of glass =5.6×109Nm−2 ∴ Tangential stress =F/A=9.0×1045×10−2 =1.8×106Nm−2 Shearing strain, ϕ=lL=l0.50 η= tangential stress shearing strain = tangential stress ×0.50l l= tangential stress ×0.50η =1.8×106×0.55.6×109=956×10−3 =0.16×10−3m=0.16mm