Two parallel and opposite forces each 5000 N are applied tangentially to the upper and lower faces of a cubical metal block of side 25 cm. the angle of shear is (The shear modulus of the metal is 80 Gpa)

Explanation

Here, `F=500 kgf = 500 xx 10 N` ,
`L=25 xx 10^(-2) m` ,
`A=(25 xx 10^(-2))^(2) = 6.25 xx 10^(-2) m^(-2)`,
`G=8 xx 10^(10) Nm^(-2)`
Angle of shear or shear strain.
`theta=(F)/(AG) = (500 xx 10)/((6.25 xx 10^(-2)) xx (8 xx 10^(10))`
`=10^(-6)rad`
Now `theta =(Delta L)/(L)`
or `Delta L=L theta = (25 xx 10^(-2)) xx 10^(-6) rad`
`=25 xx 10^(-8)m`.