Consider an Indian rubber cube having modulus of rigidity of `2xx10^(7)` dyne `cm^(-2)` and of side 8 cm. If one side of the rubber is fixed, while a tangential force equal to the weight of 300 kg is applied to the opposite face, then find out the shearing strain produced and distance through which the strain side moves .

Explanation

Given, modulus of rigidity, `eta=2xx10^(7)` dyne `cm^(-2)`

Side of the cube, l=8 cm
Area, `A=l^(2)=64 cm^(2)`
Force or load, F=300 kgf`=300xx1000xx981` dyne
Modulus of rigidity, `eta=(F)/(Atheta)impliestheta=(F)/(Aeta)`
Shearing strain, `theta=(300xx1000x981)/(64xx2xx10^(7))=0.23` rad
As, `eta=(F)/(A)(l)/(Deltal)implies(Deltal)/(l)=theta`
Distance through which the strain side moves
`impliesDeltal="l"theta=7xx0.23,Delta l=1.84` cm