Question

A 50 kg motor rests on four cylindrical rubber blocks. Each block has a height of 4 cm and a cross-sectional area of 16 $c{m}^2$. The shear modulus of rubber is $2\times {10}^{6}N/m^2$ A sideways force of 500 N is applied to the motor. The distance that the motor moves sideways is

• A. 0.156 cm
• B. 1.56 cm
• C. 0.312 cm
• D. 0.204 cm

Answer 1

Answer: (A)

0.156 cm

$\begin{array}{c}\text{Given,}\\ \text{weight of motorboat}=50kg\\ \text{cross sectional Area of each rubber}=16{cm}^2\\ \text{shear modulus of rubber}=2\times {10}^{6}N/m^2\\ \text{Sheared force}=500N\\ \text{height of rubber}=4cm\end{array}$

$\begin{array}{c}\text{Now,}\\ \text{we know}\\ \text{Shear modulus =}\end{array}$

$\frac{\text{Sheare stress}}{\text{shear strain}}$

$\begin{array}{cc}2\times {10}^6& =\frac{F_{1}L}{A\Delta x}\\ \left(\Delta x\right)& =\frac{500\times 4\times {10}^{-2}}{4\times 16\times {10}^{-4}\times 2\times {10}^6}\\ \Delta x& =\frac{5}{32}\times {10}^{-2}m\\ \Delta x& =0.156cm\end{array}$