A force of `5 xx 10^(3)` N is applied tangentially to the upper face of a cubical block of steel of side 30 cm. Find the displacement of the upper face relative to the lower one, and the angle of shear. The shear modulus of steel is `8.3 xx 10^(10)` pa.

Explanation

Area A of the upper face = `(0.30)^(2) m^(2)`
The displacement `Deltax` of the upper face relative to the lower one is given by
`Deltax=(yF)/(etaA), :.eta=(F/A)/(Deltax//y)`
`(0.30xx5xx10^(3))/(8.3xx10^(10)xx(0.30)^(2))=2xx10^(-7)` m

`:.` Angle of shear `alpha ` is given by
tan`alpha= (Deltax)/(y)`
`alpha=tan^(-1)((Deltax)/(y))`
= `tan^(-1)((2xx10^(-7))/(0.30))=tan^(-1)(0.67xx10^(-6))`