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Question

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=50×102m
Width of slab=10cm=10×102m
Area of slab =50×102×10×102
=5×102m2
Modulus of rigidity of glass =5.6×109Nm2
Tangential stress =F/A=9.0×1045×102
=1.8×106Nm2
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=956×103
=0.16×103m=0.16mm
1755237_1834270_ans_98ca6b097e4b49eaac2174c5a281466d.png

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