Question

A 60 kg motor rests on four cylindrical rubber blocks. Each block has a height of 3 cm and a cross-sectional area of 15 $c{m}^2$ The shear modulus of rubber is $2\times {10}^{6}N/m^2$ A sideways force of 300 N is applied to the motor The distance that the motor moves sideways is

• A. 0.156 m
• B. 1.56 m
• C. 0.075 cm
• D. 0.204 m

Answer 1

The correct option is C 0.075 cm

$\begin{array}{c}\text{Tangantial force on each block,}f=\frac{300}{4}=\text{75}N\\ l=3cm\text{,}A=\mathrm{l5}{\mathrm{cm}}^2=15\times {10}^{-4}{m}^2\\ l=3\times {10}^{-2}m\end{array}$

$\begin{array}{c}\eta =2\times {10}^{6}N/m^2\\ \eta =\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\frac{\Delta l}{l}}=\frac{F}{A}\cdot \frac{l}{\Delta l}\\ \therefore \Delta l=\frac{Fl}{A^{}\eta }=\frac{75\times 3\times {10}^{-2}}{15\times {10}^{-4}\times 2\times {10}^6}\end{array}$

$\begin{array}{c}\therefore \Delta l=7.5\times {10}^{-4}m\\ [\Delta l=0.075cm]\\ \text{So, none of the given option is correct.}\end{array}$