A mild steel wire of length 1.0 m and cross sectional area `0.50 xx 10 ^(-2) cm^(2)` is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

Explanation

Let AB a mild steel wire of length = 1 m and its cross sectional area `A =0.50 xx 10 ^(-2) cm ^(2)`
Mass suspended `m = 100 g = 0.1 kg and C ` is mid-point of wire AB
`therefore AC = CB =L = 1/2 and CD =x`

`therefore AD = sqrt (AC ^(2) + CD ^(2)) and DB = sqrt (L ^(2) + x ^(2))= sqrt(L ^(2) + x ^(2))`
Increas in length `= AD +DB -2L`
`Delta L = sqrt( L ^(2) + x ^(2)) + sqrt (L^(2) + x ^(2)) - 2L`
`=2 sqrt ( L ^(2) + x ^(2)) - 2L`
`=2L [ (1 + (x ^(2))/( L ^(2))) -1 ]`
`= 2L [(1 + (x ^(2))/( 2L ^(2)) + ...)-1]`
Only two terms taken from bionomial theorem.s expansion
`=2L [1+ (x ^(2))/( 2 L ^(2)) -1 ]`
`= 2L xx (x ^(2))/( 2 L ^(2))`
`Delta L = (x ^(2))/(L) " "...(1)`
Longitudinal strain `= (Delta L)/(2L) = (x ^(2))/(2L^(2)) " "...(2)`
If T be the tension in the wire, then in equilibrium
`2T cos theta = mg `
`therefore T= (mg )/( 2 cos theta )`
`= ( mg )/( 2x (x )/( sqrt ( x ^(2) + L ^(2))))`
`[because "From figure, " cos theta (CD)/(AD) = (x)/( sqrt (x^(2) +L^(2)))]= (mg sqrt (x ^(2) + L ^(2)))/( 2x)`
`If x lt lt L implies x` neglected `T = (mgL)/(2x)" "...(3)`
`therefore "Stress" sigma = (T)/(A) = (mgL)/(2 x A) " "...(4)`
`[because` Equation (2)]
`therefore` Young modulus `Y = ("Stress")/("Strain")`
`= (mgL)/(2xA) xx (2 L ^(2))/(x ^(2))`
`Y = ( mg L ^(3))/( A x ^(3))`
`therefore x ^(3) = (mg L ^(3))/( AY)`
`therefore x = L [ (mg)/(AY) ] ^(1/3)`
`therefore x = 1/2 [ ( 0.1 xx 9.8 )/( 0.5 xx 10 ^(-6) xx 2 xx 10 ^(11) )] ^(1/3)`
`therefore x = 1/2[(0.98)/(10^(5))]^(1/3)`
`therefore x 1/2 [ 9.8 xx 10 ^(-6) ] ^(1/3)`
`log x = 1/2 [ (1)/(3) log 9.8 xx 10 ^(-6) ]`
`= 1/2 {(1)/(3) [0.9912 +bar6.0000]}`
`=1/2 { (1)/(3) [ bar (6).9912]}=1/2 [bar (2).3304]`
By antilog,
`x = 2.14 xx10 ^(-2)m`
`thereforx=1. 07xx10 ^(-3)m`
`thereforex ~~ 1 cm = 0.01 m`