The elastic limit of a metal is frac{400}{pi} | vedclass

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(A) The stress $\sigma$ is defined as the force $F$ divided by the cross-sectional area $A$. The elastic limit is the maximum stress the material can

Vedclass · Accepted Answer

$2.2$

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(A) The stress $\sigma$ is defined as the force $F$ divided by the cross-sectional area $A$. The elastic limit is the maximum stress the material can withstand.Given: Elastic limit $\sigma = \frac{400}{\pi} 	ext{ MPa} = \frac{400}{\pi} 	imes 10^6 	ext{ Pa}$, Force $F = 484 	ext{ N}$.The area of the cross-section of a rod with diameter $d$ is $A = \pi \frac{d^2}{4}$.Using the formula $\sigma = \frac{F}{A}$, we have:$\frac{400}{\pi} 	imes 10^6 = \frac{484}{\pi \frac{d^2}{4}}$$\frac{400}{\pi} 	imes 10^6 = \frac{484 	imes 4}{\pi d^2}$Canceling $\pi$ from both sides:$400 	imes 10^6 = \frac{1936}{d^2}$$d^2 = \frac{1936}{400 	imes 10^6} = 4.84 	imes 10^{-6} 	ext{ m}^2$$d = \sqrt{4.84 	imes 10^{-6}} = 2.2 	imes 10^{-3} 	ext{ m} = 2.2 	ext{ mm}$.

The elastic limit of a metal is $\frac{400}{\pi} \text{ MPa}$. If a rod of this metal is to support a $484 \text{ N}$ load without exceeding its elastic limit, the minimum diameter of the rod is: (in $\text{ mm}$)

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