(a) A steel wire of mass u per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform an lateral strains `lt lt` longitudinal strains find the extension in the length of the wire. The density of steel is 7860 `kgm ^(-3)` and Young's modulus `=2 xx 10 ^(11) Nm ^(-2)`
(b) If the yield strength of steel is `2.5 xx 10 ^(8) Nm ^(-2),`
what is the maximum weight that can be hung at the lower end of the wire ?

Explanation

(a) Consider an element dx at a distance x from the load `(x =0) . If T (x) and T ( x + dx)` are tensions on the two cross sections a distance dx apart then,

`therefore T(x+dx) -T(x) = gdm = mu g dx `
(where dm = mass of wire dx and `mu =` is the mass per unit length `= (dm )/(dx))`
`therefore dT = mu g dx [ because T (x + dx) - T (x) = dT]`
Integration on both side,
`therefore T(x) = mu g x + C`
but at `x =0,` tension `T(0)=0 +C`
`therefore Mg = C` where M is suspended mass
`therefore T(x) = mu gx + Mg " "...(1)`
If element dx, increases by length dr then strain `= (dr)/(dx)`
Young modulus `Y= ((T (x))/( A))/( (dr)/( dx)) (dr)/(dx) = (T (x))/( YA)`
Integrating on both the sides,
`int _(0) ^(r) (dr)/(dx) = (1)/(YA) int T`
`therefore r =(1)/(YA) int _(0) ^(1) (mu g x + Mg ) dx`
`= (1)/(YA) [ ( mu g x ^(2))/( 2 ) + Mgx ] _(0 ) ^(L) `
`= (1)/(YA) [ (mu g L ^(2))/( 2) + MgL]`
Now `mu =(m)/(L) implies mu L = m`( let)
`= (1)/(YA) [ (mgL )/(2) + MgL]`
but m = area ` xx` length `xx` density
`= AL rho`
`r = (1)/(YA) [ ( A L rho xx gL )/(2) + MgL ]`
`= (1)/(YA) [ (A ro g L ^(2))/( 2 ) + MgL ] `
`=(1)/( 2 xx 10 ^(11) xx pi (0.1 xx 10 ^(-2)) ^(2))`
`[ ((pi xx (0.1) xx 10 ^(-2)) xx 7860 xx 10 xx (10) ^(2))/( 2)+ 25 xx 10 xx 10 ]`
`therefore r = 15.92 xx10 ^(-7)`
`((3.14 xx 786 xx 10 ^(-6) xx 10 ^(3))/( 2 ) + 2500)`
`= 15.72 xx 10 ^(-7) (1.234+ 2500)`
`=15.72 xx 10 ^(-2) xx 2501.234`
`= 3.98196xx 10 ^(-3)`
`therefore r ~~4 xx 10 ^(-3) m`
(b) The maximum tension would be at `x = L`
`T = mu g L + Mg [ because `Putting `x = L ` in eq. (1) ]
`= mg + Mg [ because mu = (m)/( L ) implies m=mu L]`
`T= (m +M) g" "...(2)`
and maximum tension force
= stress `xx` area
`=250 xx 10 ^(6) xx pi xx 10 ^(-6)`
`= 250 pi N " "...(3)`
`(m + M) g = 250 pi N`
here, `m lt lt M , m `is neglected compare to M
`Mg = 250 pi`
`therefore M = (250 pi)/(g) = ( 250 xx 3.14 )/(10)`
` therefore M =78.5 kg`