A brass wire is to withstand a tensile force of 350 N without breaking. What minimum diameter must the wire have?

Given that a brass wire is to withstand a tensile force of 350 N without breaking.

Breaking stress of wire = $\frac{{\mathbf{F}}_{\mathbf{\perp }}}{\mathbf{A}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$……………..(1)

Where${\mathbf{F}}_{\mathbf{\perp }}$is the tensile force applied to an object. A is the area over which force is exerted.

Let d be the diameter of the wire.

$\mathbf{A}\mathbf{=}\mathbf{\tau \tau }\frac{{\mathbf{d}}^{\mathbf{2}}}{\mathbf{4}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$……………..(2)

Using equations (1) and (2), we can write,

$\mathrm{A}=\frac{{\mathrm{F}}_{\perp }}{\mathrm{\phi }}$

Where ${\mathrm{F}}_{\perp }$ is the tensile force applied to an object, A is the area over which force is exerted, and$\phi$ is breaking stress of brass

${\mathrm{F}}_{\perp }=350\mathrm{N}$

Breaking stress of brass is $\mathrm{\phi }=4.7\times {10}^8\mathrm{Pa}$

Therefore,

$\begin{array}{r}A=\frac{350\mathrm{N}}{4.7\times {10}^8\mathrm{Pa}}\\ =7.45\times {10}^{-7}{\mathrm{m}}^2\end{array}$

Using equation (2), we can write, that

$\mathrm{d}=\sqrt{\frac{4\mathrm{A}}{\mathrm{\tau \tau }}}$

Substituting the values in the above expression, we get,

$\mathrm{d}=\sqrt{\frac{4\times \left(7.45\times {10}^{-7}{\mathrm{m}}^2\right)}{3.14}}$mm

= 0.97 m

Hence, the diameter of the wire is 0.97 mm.