Q11-38E - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

The minimum diameter is 0.97 mm

Step 1: The given data

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

Step 2: Formula used

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

Where ${\mathit{F}}_{\mathbf{\perp }}$ 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.

$\mathit{A}\mathbf{=}\mathit{\pi }\frac{{\mathbf{d}}^{\mathbf{2}}}{\mathbf{4}}$……………..(2)

Step 3: Find the area

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

$A=\frac{F_{\perp }}{\phi }$

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

$F_{\perp }=350 N$

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

Therefore,

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

Step 4: Find the diameter

Using equation (2), we can write, that 

$d=\sqrt{\frac{4A}{\pi }}$

Substituting the values in the above expression, we get,

$\begin{array}{rcl}d& =& \sqrt{\frac{4\times \left(7.45\times {10}^{-7}{m}^2\right)}{3.14}}\\ & =& 0.97 m\end{array}$mm

Hence, the diameter of the wire is 0.97 mm.