Question #258445

A wire of length 0.50 m is fixed horizontally between two supports separated by 0.50 m. When a

mass of 8 kg hangs from the middle of the wire, the mid-point sags by 1 cm. The diameter of the

wire is 2.8 mm. Calculate the Young’s modulus of the wire.


Expert's answer

Find the change in length. First, find the stretched length:


Ls=2(0.5/2)2+0.012=0.5004 m.L_s=2\sqrt{(0.5/2)^2+0.01^2}=0.5004\text{ m}.

Find the tension in the wire:


2Tsin⁡arctan⁡LsagL/2=mg, T=981 N.2T\sin\arctan\frac{L_\text{sag}}{L/2}=mg,\\\space\\ T=981\text{ N}.

Strain:


ϵ=Ls−LL=σE, E=σLLs−L, σ=T/A, E=TLA(Ls−L), E=199 GPa.\epsilon=\frac{L_s-L}{L}=\frac\sigma E,\\\space\\ E=\frac{\sigma L}{L_s-L},\\\space\\ \sigma=T/A,\\\space\\ E=\frac{TL}{A(L_s-L)},\\\space\\ E=199\text{ GPa}.


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