A square lead slab of side $ 50 \mathrm{~cm} $ and thickness $ 10 \mathrm{~cm} $ is subjected to a shearing force (on its narrow face) of $ 9 \times 10^{4} \mathrm{~N} $. The lower edge is riveted to the floor. How much will the upper edge be displaced? (Shear modulus of lead = $ 5.6 \times 10^{9} \mathrm{~N} \mathrm{~m}^{-2} $ )

Question

A square lead slab of side $ 50 \mathrm{~cm} $ and thickness $ 10 \mathrm{~cm} $ is subjected to a shearing force (on its narrow face) of $ 9 \times 10^{4} \mathrm{~N} $. The lower edge is riveted to the floor. How much will the upper edge be displaced? (Shear modulus of lead = $ 5.6 \times 10^{9} \mathrm{~N} \mathrm{~m}^{-2} $ ), $ 0.16 \mathrm{~mm} $, $ 1.6 \mathrm{~mm} $, $ 0.16 \mathrm{~cm} $, $ 1.6 \mathrm{~cm} $

MARKS App · Accepted Answer

The lead slab is fixed and force is applied parallel to the narrow face as shown in the figure. Area of the face parallel to which this force is applied is A = 50 c m × 10 c m = 0.5 m × 0.1 m = 0.05 m 2 If Δ L is the displacement of the upper edge of the slab due to tangential force F, then η = F A Δ L L or Δ L = F l η A Substituting the given values, we get Δ L = 9 × 10 4 N ( 0.5 m ) 5.6 × 10 9 N m 2 × 0.05 m 2 = 1.6 × 10 – 4 m = 0.16 mm

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