Q11-39E - University Physics with Modern Physics

Question

In a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of 90.8 N is applied perpendicular to each end. If the diameter of the wire is 1.84 mm, what is the breaking stress of the alloy?

Step-by-Step Solution

Answer

Breaking stress of alloy is $3.41\times {10}^7$ Pa.

Step 1: The given data

Given that in a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of 90.8N is applied perpendicular to each end.

The diameter of the wire, $d=1.84 mm\left(\frac{1 m}{1000 mm}\right)=1.84\times {10}^{-3} m$

Tensile force applied perpendicularly, $F_{\perp }=90.8 N$

Step 2: Formula used

Breaking stress of wire =$\frac{F_{\perp }}{A}$

Where $F_{\perp }$ 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.

$A=\pi \frac{d^2}{4}$

Step 3: Find the breaking stress

$\begin{array}{rcl}Area& =& 3.14\times \frac{{\left(1.84\times {10}^{-3}m\right)}^2}{4}\\ & =& 2.66\times {10}^{-6} m^2\end{array}$

 $\begin{array}{rcl}Breakingstress& =& \frac{90.8 N}{2.66\times {10}^{-6} m^2}\\ & =& 3.41\times {10}^7 Pa \end{array}$Pa

Hence, Breaking stress of alloy is $3.41\times {10}^7 Pa$