Question #215344

A steel wire of length 2m and diameter 1 mm is stretched horizontally between two rigid supports spaced 2 m apart. A mass 250 g is suspended from the mid-point of the wire. Calculate the small sag of the wire at the midpoint.

 

Young’s modulus for steel = 2×1011 N m –2. 

 

Assume that   where x<<1. The figure below might aid you in your analysis.

 

                  


1
Expert's answer
2021-07-12T12:13:40-0400
T=mg2sinθ=mg2x0.52+x2T=0.25πϵEd2x=l(mg2πEd2)13T=\frac{mg}{2\sin \theta}=\frac{mg}{2x}\sqrt{0.5^2+x^2}\\T=0.25\pi \epsilon Ed^2\\x=l\left(\frac{mg}{2\pi Ed^2}\right)^{\frac{1}{3}}

x=2(0.259.82π(21011)(0.001)2)13=0.025 mx=2\left(\frac{0.25\cdot9.8}{2\pi (2\cdot10^{11})(0.001)^2}\right)^{\frac{1}{3}}=0.025\ m


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