A square lead slab of side 50 cm and thickness 5cm is subjected to a shearing force on its narrow face of magnitude 9×10⁴ N. The lower edge is riveted to the floor. How much is the upper edge displaced if the shear modulus of lead is 5.6×10⁹ Pa?
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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=50cm×10cm=0.5m×0.1m=0.05m2
If ΔL is the displacement of the upper edge of the slab due to tangential force, F, then
η=ΔL/LF/AorΔL=ηAFL
Substituting the given values, we get
ΔL=5.6×109Nm−2×0.05m2(9×104N)(0.5m)=1.6×10−4m=0.16mm
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Explanation:
A=50cm×10cm=0.5m×0.1m=0.05m
2
If ΔL is the displacement of the upper edge of the slab due to tangential force, F, then
η=
ΔL/L
F/A
orΔL=
ηA
FL
Substituting the given values, we get
ΔL=
5.6×10
9
Nm
−2
×0.05m
2
(9×10
4
N)(0.5m)
=1.6×10
−4
m=0.16mm