Two opposite forces`F_(1) = 120 N` and `F_(2) = 80 N` act on an elastic plank of modulus of elasticity `Y= 2 x 10^(11) N//m^(2)` and length `l=1m` placed over a smooth horizontal surface. The cross-sectional area of the plank is `S = 0.5 m^(2)`. The change in length of the plank is `x xx 10^(-11)m`. Find the value of `x`.

Answer

1

Explanation

Consider an element of thickness `dx`. Change in the length of the element is `dl=T/S (ds)/Y` and `T=F_(1)=(F_(1)-F_(2))x/l`

`int_(0)^(/_\l)dl=int_(0)^(l)(F_(1)-((F_(1)-F_(2))x)/l dx)/(SY)`
`/_\l=((F_(1)+F_(2)))/(2SY)l=(200xx1)/(2xx0.5xx2xx10^(11))`
`=1xx10^(-9)m`
`x=1`