Question

A $0.05m$ cube has its upper face displaced by $0.2cm$ by a tangential force of $8N$. Calculate the modulus of rigidity (in $\times 10^{4}N{m}^{-2}$ ) of the material of the cube.

Answer 1

Answer: 8

$l=5\times 10^{-2}m,\Delta l=0.2cm$
$=0.2\times 10^{-2}m,F=8N$
Shearing strain $=\frac{\Delta l}{l}=\frac{0.2}{5}=0.04$
Shearing stress $=\frac{F}{l\times l}=\frac{8}{{\left(5\times 10^{-2}\right)}^2}$
$=3200N{m}^{-2}$
Modulus of rigidity, $\eta =\frac{\text{ Shearing stress }}{\text{ Shearing strain }}=\frac{3200}{0.04}$
$=80000N{m}^{-2}=8\times 10^{4}N{m}^{-2}$