Question

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

Answer 1

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