Question

A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10¯² cm² 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.

Answer 1

Given: The length of steel wire is 1.0m, cross-sectional area is 0.50× 10 −2 cm 2 and the mass of 100 g is suspended from the midpoint of the wire.

By Pythagoras theorem,

AD=DB= AC 2 + CD 2 = l 2 + x 2 .

The increase in length can be calculated as,

ΔL=( AD+DB )−AB =2 l 2 + x 2 −2l =2l[ ( 1+ x 2 l 2 ) −1 ]

Using binomial theorem in above expression, we get

( 1+x ) n =1+nx.

2l[ ( 1+ x 2 l 2 ) -1 ]=2l x 2 2 l 2 = x 2 l

The longitudinal strain is given as,

Δl 2l = x 2 2 l 2

Applying the equilibrium condition in the above figure, we get

Tcosθ+Tcosθ=mg 2Tcosθ=mg T= mg 2 x x 2 + l 2 T= mgl 2x

The stress in the wire is given as,

Stress= T A = mgL 2xA

The Young’s modulus is given as,

Y= stress strain .

By substituting the values of stress and strain in Young’s modulus, we get

Y= mgl 2xA x 2 2 l 2 Y= mg l 3 A x 3 x=l [ mg YA ] 1 3

By substituting the values in the above equation, we get

x= 1 2 [ 0.1×9.8 2× 10 11 ×0.50× 10 −2 × 10 −4 ] 1 3 =1.074× 10 −2  m =1.074 cm

Thus, the depression at the midpoint is 1.074 cm.