Question

A rubber cube of side $5\text{cm}$ has one side fixed while a tangential force equal to $1800\text{N}$ is applied to the opposite face. Find the shearing strain and the lateral displacement of the strained face. Modulus of rigidity for rubber is $2.4\times {10}^6{\text{N/m}}^2$.

• A. $0.3\text{radian},1.5\text{cm}$
• B. $1.2\text{radian},1.225\text{cm}$
• C. $0.72\text{radian},1.96\text{cm}$
• D. $0.98\text{radian},2.25\text{cm}$

Answer 1

The correct option is A $0.3\text{radian},1.5\text{cm}$

Given,
$L=5\times {10}^{-2}\text{m}$
Tangential Force $F=1800\text{N}$
Modulus of rigidity $\eta =2.4\times {10}^6{\text{N/m}}^2$

As we know,
Modulus of rigidity $\eta =\frac{\text{Tangential Stress}}{\text{Shear strain}}=\frac{\left(\frac{F}{A}\right)}{\theta }$
$\Rightarrow \text{Shear strain}\left(\theta \right)=\frac{F}{A\times \eta }=\frac{1800}{(5\times 5\times {10}^{-4})\times 2.4\times {10}^6}$
$=\frac{180}{25\times 24}=\frac{3}{10}=0.3\text{radian}$

$\text{Shear strain}\left(\theta \right)=\frac{x}{L}=0.3$
$\Rightarrow x=0.3\times 5\times {10}^{-2}=1.5\times {10}^{-2}\text{m}=1.5\text{cm}$