Question

Indian rubber cube of side 7 cm has one side fixed while a tangential force equal to the weight of 200 kilogram is applied to the opposite face find the shearing strain produced and distance through which the strained side moves modulus of rigidity for rubber is 2 into 10 to the power 7 dyne per CM square

Answer 1

Answer:

The shearing strain produced is 2 and distance through which the strained side moves is 14 cm

Explanation:

As we know that the force is applied tangentially

So here we will have

$F = mg$

$F = 200(9.81)$

$F = 1962 N$

now we will have

$\eta = \frac{stress}{strain}$

$strain = \frac{stress}{\eta}$

$stress = \frac{F}{A}$

$stress = \frac{1962}{0.07^2}$

$stress = 4 \times 10^6 N/m^2$

now we have

$strain = \frac{4 \times 10^6}{2 \times 10^6}$

$strain = 2$

now the distance that it will move due to force is given as

$\frac{x}{L} = strain$

$\frac{x}{7} = 2$

$x = 14 cm$

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Topic : Shear Strain

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