91影视

Length of wire $L_0=4$m

Cross-sectional Area $A=0.05c{m}^2\left(\frac{{10}^{-4}{m}^2}{1c{m}^2}\right)=5脳{10}^{-6}{m}^2$

For steel, Young鈥檚 Modulus is $Y=20脳{10}^{10}Pa$

Proportional stress ${蠁}_P=1.6脳{10}^{-3}Y$

Breaking stress${蠁}_B=6.5脳{10}^{-3}Y$

Stress on wire =$\frac{{\mathbf{F}}_{\mathbf{鈯�}}}{\mathbf{A}}$

Where${\mathit{F}}_{\mathbf{鈯�}}$is the tensile force applied to an object, and A is the area over which force is exerted.

$\mathit{Y}\mathbf{=}\frac{{\mathbf{L}}_{\mathbf{o}}{\mathbf{F}}_{\mathbf{鈯�}}}{\mathbf{A}\mathbf{螖}\mathbf{L}}$

Where Y is young鈥檚 modulus, L₀ the length of wire, and$\mathit{螖}\mathit{L}$the change in length.

Here, weight$w=F_{鈯�}$

So $w=ProportionalStressonwire脳Area$

$\begin{array}{rcl}w& =& 1.6脳{10}^{-3}Y\left(5脳{10}^{-6}{m}^2\right)\\ & =& \left(1.6脳{10}^{-3}\right)\left(20脳{10}^{10}Pa\right)\left(5脳{10}^{-6}{m}^2\right)\\ & =& 1.60脳{10}^{3}N\end{array}$

Hence, the weight that can be hung without exceeding the proportional limit is $1.60脳{10}^{-3}$N.

$\begin{array}{rcl}螖L& =& \left(\frac{F_{鈯�}}{A}\right)\frac{L_0}{Y}\\ & =& {蠁}_P\frac{L_0}{Y}\\ & =& 1.6脳{10}^{-3}Y脳\frac{L_0}{Y}\\ & =& 1.6脳{10}^{-3}脳L_0\end{array}$

So

$\begin{array}{rcl}螖L& =& \left(1.6脳{10}^{-3}\right)\left(4m\right)\\ & =& 6.4脳{10}^{-3}m\end{array}$

Hence, the wire stretches by 6.4 mm.

Here, weight$w=F_{鈯�}$

So $w=ProportionalStressonwire脳Area$

$\begin{array}{rcl}w& =& 6.5脳{10}^{-3}Y\left(5脳{10}^{-6}{m}^2\right)\\ & =& \left(6.5脳{10}^{-3}\right)\left(20脳{10}^{10}Pa\right)\left(5脳{10}^{-6}{m}^2\right)\\ & =& 6.5脳{10}^{3}N\end{array}$

Hence, the maximum weight that can be hung is$6.5脳{10}^{3}N$