Question

A steel plate of face area $4c{m}^2$ and thickness $0.5$cm is fixed rigidly at the lower surface. A tangential force of $10$N is applied on the upper surface. Find the lateral displacement of the upper surface with respect to the lower surface. Rigidity modulus of steel $=8.4\times {10}^{10}$ $N{m}^{-2}$.

Answer 1

$Area,A=4{cm}^2=4\times {10}^{-4}{m}^2$

$Thickness,d=0.5cm=0.5\times {10}^{-2}m$

$Force,F=10N$

$Rigiditymodulus,m=8.4\times {10}^{10}N{m}^{-2}$

$Angulardisplacement=\theta$

$m=\frac{F}{A\theta }=\frac{10}{4\times {10}^{-4}\theta }$

$\theta =\frac{10}{4\times {10}^{-4}\times 8.4\times {10}^{10}}=0.297\times {10}^{-6}$

$Lat.displacement=\theta \times d=0.297\times {10}^{-6}\times 0.5\times {10}^{-2}=1.5\times {10}^{-9}m$