A steel of area of cross-section `A` and length `2L` is calmped family between two rigid supports separated by a distance `'2L'`. A body is hung from the middle point sags by a distance `x`. Calculate the mass of the body body and the angle made by the string with the horizontal. (Young's modulus of Steel `= Y`)

Explanation

Since `theta` is small, `sin theta approx tan theta = (x)/(L)`
`Y = (F)/(A) (L)/(e), F = (YA)/(L) [(L^(2) + x^(2))^(1//2) - L]`
`= (YA)/(L) [L (1 + (x^(2))/(2L^(2)))-L] = (YA)/(L) [L + (x^(2))/(2L) - L]`
`F = (Yax^(2))/(2L^(2)).....(1)` From the fig.
`2F sin theta = Mg, 2F theta = Mg`, from (1)
`2((YA x^(2))/(2L^(2)))theta = Mg rArr (YAx^(2))/(L^(2)) xx (x)/(L) = Mg rArr M = (YA x^(3))/(L^(3) g)`
`(x^(3))/(L^(3)) = (Mg)/(YA) rArr (x)/(L) = ((Mg)/(YA))^(1//3)`, Since, `Tan theta = (x)/(L)`
`Tan theta = ((Mg)/(YA))^(1//3) rArr theta = Tan^(-1) ((MG)/(YA))^(1//3)`