Chapter 9, Problem 75QAP

To determine

(a)

The stretched length of hair.

Answer to Problem 75QAP

The stretched length of hair is $6.8mm$.

Explanation of Solution

Given:

Original length of hair $L=20.0cm=20.0\times {10}^{-2}m$

Total active weight of object $W=250\times {10}^{-3}kg\times 9.8m/s^2=2.45N$

Radius of hair $r=75\mu m=75\times {10}^{-6}m$

Area of cross-section of hair $A=\pi r^2=\pi \times {\left(75\times {10}^{-6}m\right)}^2=1.8\times {10}^{-8}{m}^2$

Young's Modulus of hair $Y=4\times {10}^{9}N/m^2$

Let $\Delta L$ be the stretched length of hair

Formula used:

Young's Modulus $Y=\frac{longitudinalstress}{longitudinalstrain}=\frac{W/A}{\Delta L/L}$

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  $\begin{array}{l}Y=\frac{W/A}{\Delta L/L}\\ or,\Delta L=\frac{W/A}{Y/L}\\ or,\Delta L=\frac{2.45N}{\left(1.8\times {10}^{-8}{m}^2\right)}\times \frac{\left(20.0\times {10}^{-2}m\right)}{\left(4\times {10}^{9}N/m^2\right)}\\ or,\Delta L=6.8\times {10}^{-3}m\\ or,\Delta L=6.8mm\end{array}$

Hence, the stretched length of hair is $6.8mm$.

Conclusion:

Thus, the stretched length of hair is $6.8mm$.

To determine

(b)

The stretched length of aluminum wire.

Answer to Problem 75QAP

The stretched length of aluminum wire is $0.39mm$.

Explanation of Solution

Given:

The stretched length of hair $\Delta L_{hair}=6.8mm$.

Young's Modulus of hair $Y_{hair}=4\times {10}^{9}N/m^2$

Young's Modulus of aluminum wire $Y_{\text{al}}=70\times {10}^{9}N/m^2$

Let $\Delta L_{al}$ be the stretched length of aluminum wire

Formula used:

Young's Modulus $Y\propto \frac{1}{\Delta L}$

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  $\begin{array}{l}Y\propto \frac{1}{\Delta L}\\ or,\frac{\left(\Delta L_{al}\right)}{\left(\Delta L_{hair}\right)}=\frac{Y_{hair}}{Y_{al}}\\ or,\left(\Delta L_{al}\right)=\frac{Y_{hair}}{Y_{al}}\times \left(\Delta L_{hair}\right)\\ or,\left(\Delta L_{al}\right)=\frac{\left(4\times {10}^{9}N/m^2\right)}{\left(70\times {10}^{9}N/m^2\right)}\times \left(6.8\times {10}^{-3}m\right)\\ or,\left(\Delta L_{al}\right)=0.39\times {10}^{-3}m\\ or,\left(\Delta L_{al}\right)=0.39mm\end{array}$

Hence, the stretched length of aluminum wire is $0.39mm$.

Conclusion:

Thus, the stretched length of aluminum wire is $0.39mm$.

To determine

(c)

The spring constant of hair.

Answer to Problem 75QAP

The spring constant of hair is $3.6\times {10}^{2}N/m$.

Explanation of Solution

Given:

The stretched length of hair $\Delta L_{hair}=6.8mm$.

Total active weight of object $W=250\times {10}^{-3}kg\times 9.8m/s^2=2.45N$

Let 'k' be the spring constant

Formula used:

Apply Hooke's Law,

Weight of object $W=-k\Delta L_{hair}$

Here, all alphabets are in their usual meanings.

Calculation:

Apply Hooke's Law,

Weight of object $W=-k\Delta L_{hair}$

  $\begin{array}{l}or,k=\frac{W}{\Delta L_{hair}}\\ or,k=\frac{2.45N}{6.8\times {10}^{-3}m}\\ or,k=3.6\times {10}^{2}N/m\end{array}$

Hence, the spring constant of hair is $3.6\times {10}^{2}N/m$.

Conclusion:

Thus, the spring constant of hair is $3.6\times {10}^{2}N/m$.

To determine

(d)

To compare the spring constant of hair with the spring constant in laboratory.

Answer to Problem 75QAP

The spring constant of hair is about $250$ time smaller than the laboratory spring constant.

Explanation of Solution

Given:

The spring constant of hair $=3.6\times {10}^{2}N/m$.

The spring constant wire in laboratory (say) $=6.53\times {10}^{4}N/m$.

Formula used:

Ratio of the spring constant $=\frac{\text{Spring}\text{constant}\text{in}\text{lab}}{\text{Spring}\text{constant}\text{of hair}}$

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  $\begin{array}{l}\text{Ratio of the spring constant}=\frac{\text{Spring}\text{constant}\text{in}\text{lab}}{\text{Spring}\text{constant}\text{of hair}}\\ \text{Ratio of the spring constant}=\frac{6.53\times {10}^{4}N/m}{2.6\times {10}^{2}N/m}=250\end{array}$

Hence, the spring constant of hair is about $250$ time smaller than the laboratory spring constant.

Conclusion:

Thus, the spring constant of hair is about $250$ time smaller than the laboratory spring constant.