Two parallel and opposite forces, each of magnitude 4000 N, are applied tangentially to the upper and lower faces of a cubical metal block 25 cm on a side. Find the displacement of the upper surface relative to the lower surface (in x 10^-5 cm ). The shear modulus for the metal is 80 Gpa.

We use the approximate from $\mathrm{S}=\mathrm{F}/\left(\mathrm{A\varphi }\right)$

Putting $\mathrm{S}=8\times {10}^{10}\mathrm{N}/{\mathrm{m}}^2,\mathrm{F}=4000\mathrm{N}\text{ and }\mathrm{A}=(0.25\mathrm{m}{)}^2$

$=6.25\times {10}^{-2}{\mathrm{m}}^2$ in the above equation and solving for $\mathrm{\varphi }$, we get

$\mathrm{\varphi }=\frac{\left(4000\mathrm{N}\right)}{\left(6.25\times {10}^{-2}{\mathrm{m}}^2\right)\left(8\times {10}^{10}\mathrm{N}/{\mathrm{m}}^2\right)}=8.0\times {10}^{-7}\mathrm{rad}$

The displacement of the upper surface is given by $\mathrm{d}=\mathrm{L\varphi }$,

where L is an edge of the cube.

$\therefore \mathrm{d}=\left(8.0\times {10}^{-7}\right)\left(25\mathrm{cm}\right)=2.0\times {10}^{-5}\mathrm{cm}$