A cube of aluminium of each side 0.10 m is subjected to a tangential force of `10^(6)N`. If the top surface slides through `3.0xx10^(-4)` with respect to the bottom face then calculate (i) Shearing strain (ii) shearing stress (iii) modulus of rigidity (iv) elastic potential energy.

Explanation

Given `a=0.1m, a^(3)=V=(0.1)^(3)=10^(-3)m^(3)`
`F=10^(6)N,x=3.0xx10^(-4)m,a^(2)=(0.1)^(2)=10^(-2)m^(2)`
(i) Shearing strain `= theta =x/a =(3xx10^(-4))/0.1=3xx10^(3)` rad
(ii) Shearing stress `=F/A=(10^(6))/(10^(-2))=10^(8)Nm^(-2)`
(iii) Modulus of rigidity`eta=("shearing stress")/("shearing strain")=(10^(8))/(3xx10^(-3))=3.33xx10^(10)Nm^(-2)`
(iv) Elastic P.E. `=1/2xx"force"xx"extension"=1/2xx"stress"xx"strain"xx"volume"`
`=1/2x10^(8)xx3xx10^(-3)xx10^(-3)=1.5xx10^(2)J`