Q11-40E - University Physics with Modern Physics

Question

A 4.0-m-long steel wire has a cross-sectional area of 0.050 cm². Its proportional limit has a value of 0.0016 times its Young’s Modulus. Its breaking stress has a value of 0.0065 times its Young’s modulus. The wire is fastened at its upper end and hangs vertically. (a) How great a weight can be hung from the wire without exceeding the proportional limit? (b) How much will the wire stretch under this load? (c) What is the maximum weight that the wire can support?

Step-by-Step Solution

Answer

1. The weight that can be hung without exceeding the proportional limit is $1.60\times {10}^{-3}$N
2. The wire stretches by 6.4 mm
3. The maximum weight wire can support is $6.5\times {10}^3$N

Step 1: The given data

Length of wire $L_0=4$m

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

For steel, Young’s Modulus is $Y=20\times {10}^{10} Pa$

Proportional stress ${\phi }_P=1.6\times {10}^{-3} Y$

Breaking stress ${\phi }_B=6.5\times {10}^{-3} Y$

Step 2: Formula used

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

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

$\mathit{Y}\mathbf{=}\frac{{\mathbf{L}}_{\mathbf{o}}{\mathbf{F}}_{\mathbf{\perp }}}{\mathbf{A}\mathbf{\Delta }\mathbf{L}}$

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

Step 3: Find the weight that can be hung within the proportional limit (a)

Here, weight $w=F_{\perp }$

So $w=ProportionalStressonwire\times Area$

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

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

Step 4: Calculate how much the wire will stretch (b)

$\begin{array}{rcl}\Delta L& =& \left(\frac{F_{\perp }}{A}\right)\frac{L_0}{Y}\\ & =& {\phi }_P\frac{L_0}{Y}\\ & =& 1.6\times {10}^{-3}Y\times \frac{L_0}{Y}\\ & =& 1.6\times {10}^{-3}\times L_0\end{array}$

So

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

Hence, the wire stretches by 6.4 mm.

Step 5: Find the maximum weight that can be hanged (c)

Here, weight $w=F_{\perp }$

So $w=ProportionalStressonwire\times Area$

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

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