Question

Example 23. A 60 kg motor rests on four cylindrical
rubber blocks. Each cylinder has a height of 3 cm and a
cross-sectional area of 15 cnt. The shear modulus for this
rubber is 2 x 106 Nm 2. If a sideways force of 300 N is
applied to the motor, how far will it move sideways ?​

Answer 1

Force applied on each block:F=3004=75 NDistance moved sideways:∆L=FLAη=75×0.0315×10-4×2×106=0.00075 m=0.075 cm

Hope this information will clear your doubts about Mechanical properties of solids.

Answer 2

The change in the length after a force is applied is 0.075 cm.

Explanation:

The share modulus of the body is given as

$\eta=\frac{Fl}{Ax}$

Here, F is the force applied on the cylinder, l is the initial length, A is the cross-sectional area, x is the change in length.

As four cylinder are there, so force applied on each cylinder is

F = 300 N/4 =75 N.

Given l = 3 cm = 0.03 m, $\eta=2\times10^6N/m^2, A=15cm^2=15\times10^-^4m^2$.

Substitute the given value, we get

$2\times10^6N/m^2=\frac{75N\times0.03m}{15\times10^-^4m^2\times x}$

$x=0.00075m$

x = 0.075 cm.

Thus, change in the length after a force is applied is 0.075 cm.

#Learn More: share modulus.

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