Question

A).a circular steel wire 2.00m long must stretch no more than 0.25cm when a tensile force of 400N is applied to each end of the wire .What minimum diameter is required for the wire ?

Answer 1

Determine the strain:

$$\epsilon=\frac{\Delta l}{l}.$$

Determine the stress:

$$\sigma=E\epsilon=E\frac{\Delta l}{l}.$$

Determine the cross-sectional area:

$$A=\frac{F}{\sigma}=\frac{Fl}{E\Delta l}.$$

Determine the diameter:

$$d=2\sqrt{\frac{A}{\pi}}=2\sqrt{\frac{Fl}{E\Delta l\pi}}.$$

Substitute the values (take Young's modulus for steel E=200 GPa):

$$A=2\sqrt{\frac{400\cdot2}{200\cdot10^9\cdot0.0025\pi}}=1.42\cdot10^{-3}\text{ m}.$$