Chapter 10, Problem 23P

(a)

To determine

The diameter and tensile stress in the wire.

Answer to Problem 23P

The diameter and tensile stress in the wire is respectively $1.3\text{mm}$ and $8.4\times {10}^7\text{N}/{\text{m}}^{\text{2}}$.

Explanation of Solution

The length of the wire is $3.0\text{m}$, the change in length (elongation) of the wire is $2.1\text{mm}$, the young’s modulus of the copper wire is $120\times {10}^9\text{Pa}$ and the force or weight at the end is $120\text{N}$.

Write the ex

Write the expression to calculate the power output of the pump.

$P=\frac{mgy}{t}$

Here, P is the power, m is the mass of the water, g is the acceleration due to gravity, y is the depth of the well.

Write the expression for the force and young’s modulus of the wire.

$\frac{F}{A}=Y\frac{\Delta L}{L}$ (I)

Here, F is the force at the end, A is the cross-sectional area of the wire, Y is the young’s modulus, L is the length of the wire and $\Delta L$ is the change in length.

Write the expression to calculate the cross-sectional area of the wire.

$A=\pi {\left(\frac{d}{2}\right)}^2$

Here, d is the diameter of the wire.

Use the above expression in (I) to rewrite.

$\frac{F}{\pi {\left(\frac{d}{2}\right)}^2}=Y\frac{\Delta L}{L}$

Rewrite the above equation in terms of d.

$d=2\sqrt{\frac{F}{Y\pi }\left(\frac{L}{\Delta L}\right)}$

Substitute $3.0\text{m}$ for L, $2.1\text{mm}$ for $\Delta L$, $120\times {10}^9\text{Pa}$ for Y and $120\text{N}$ for F in the above equation to calculate d.

$\begin{array}{c}d=2\sqrt{\left(\frac{120\text{N}}{120\times {10}^9\text{Pa}\left(\pi \right)}\right)\left(\frac{3.0\text{m}}{2.1\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\right)}\\ d=0.0013\text{m}\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\\ =1.3\text{mm}\end{array}$

Write the expression for the tensile stress.

$\frac{F}{A}=Y\frac{\Delta L}{L}$

Here, $\frac{F}{A}$ is the tensile stress.

Substitute $3.0\text{m}$ for L, $2.1\text{mm}$ for $\Delta L$, $120\times {10}^9\text{Pa}$ for Y in the above equation to calculate $\frac{F}{A}$.

$\begin{array}{c}\frac{F}{A}=\left(120\times {10}^9\text{Pa}\right)\frac{2.1\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}{3.0\text{m}}\\ =8.4\times {10}^7\text{N}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the diameter and tensile stress in the wire is respectively $1.3\text{mm}$ and $8.4\times {10}^7\text{N}/{\text{m}}^{\text{2}}$.

(b)

To determine

The maximum weight in order to hang the wire.

Answer to Problem 23P

The maximum weight in order to hang the wire is $531\text{N}$.

Explanation of Solution

The tensile strength is $400\text{MPa}$.

Write the expression for the tensile stress.

$\frac{W}{A}=F^{\prime }$

Here, $F^{\prime }$ is the tensile strength and W is the maximum weight.

Rewrite the above equation using $A=\pi {\left(\frac{d}{2}\right)}^2$ in terms of W.

$\begin{array}{c}\frac{W}{\pi {\left(\frac{d}{2}\right)}^2}=F^{\prime }\\ W=\frac{\pi F{d}^2}{4}\end{array}$

Substitute $400\text{MPa}$ for $F^{\prime }$ and $1.3\text{mm}$ for d in the above equation to calculate W.

$\begin{array}{c}W=\frac{\pi \left(400\text{MPa}\right)\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right){\left(1.3\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}^2}{4}\\ =531\text{N}\end{array}$

Conclusion:

Therefore, the maximum weight in order to hang the wire is $531\text{N}$.