Question

Two blocks of masses $1kg$ and $2kg$ are connected by a metal wire going over a smoother pulley as shown in figure. The breaking stress of the metal is $2\times {10}^{9}N/m^2$. What would be the minimum radius $\left(in\mu m\right)$ of the wire used if it is not to break? Take $g=10m/s^2$.

Answer 1

Draw a free body diagram of given problem


Calculate tension in the wire

Let tension in the wire be $T$,
$20-T=2a\dots \left(i\right)$
$T-20=a\dots \left(ii\right)$
Therefore, on solving both equations we get
$\frac{20-T}{T-10}=2$
$\Rightarrow 20-T=2T-20$
$T=\frac{40}{3}N$

Calculate area of cross section of the wire

Formula used: $\sigma =\frac{T}{A}$
As we know,
stress,$\sigma =\frac{T}{A}$
$\Rightarrow 2\times {10}^9=\frac{40/3}{A}$
$A=\frac{20}{3}\times {10}^{-9}{m}^2$

Calculate minimum radius of the wire

Let $r$ be the radius of the wire, and area will be equal to $\pi r^2$ Therefore,
$\pi r^2=20\times {10}^{-9}/3$
$r={\left(\frac{20\times {10}^{-9}}{3\times 3.14}\right)}^{\frac{1}{2}}$
$=0.46\times {10}^{-4}m=46\mu m$