Chapter 9, Problem 72QAP

To determine

(a)

The diameter of steel cable.

Answer to Problem 72QAP

The diameter of steel cable is $1.12cm$.

Explanation of Solution

Given:

Let the radius of steel cable be $r$

Total active weight of diesel engine $W_{\max }=4.0\times {10}^{3}kg\times 9.8m/s^2=3.92\times {10}^{4}N$

Ultimate breaking strength of steel $4.00\times {10}^{8}N/m^2$

Formula used:

  $\text{Ultimate breaking strength of steel }=\frac{W_{\max }}{\pi r^2}$

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  $\begin{array}{l}\text{Ultimate breaking strength of steel }=\frac{W_{\max }}{\pi r^2}\\ or,r=\sqrt{\frac{W_{\max }}{\pi \times \text{Ultimate breaking strength of steel}}}\\ or,r=\sqrt{\frac{3.92\times {10}^{4}N}{\pi \times \left(4.00\times {10}^{8}N/m^2\right)}}\\ or,r=0.56\times {10}^{-2}m\\ or,r=0.56cm\end{array}$

Hence, the required diameter $d=2r=2\times 0.56cm=1.12cm$

Conclusion:

Thus, the diameter of steel cable is $1.12cm$.

To determine

(b)

The stretched length of cable.

Answer to Problem 72QAP

The stretched length of cable is $2.0cm$.

Explanation of Solution

Given:

Original length of cable $L=10.0m$

Ultimate breaking strength of steel $4.00\times {10}^{8}N/m^2$

Young's Modulus of steel $Y=2.0\times {10}^{11}N/m^2$

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

Formula used:

Young's Modulus $Y=\frac{Ultimatestress}{Breakingstrain}=\frac{Ultimatestress}{\Delta L/L}$

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  $\begin{array}{l}Y=\frac{Ultimatestress}{\Delta L/L}\\ or,\Delta L=\frac{\left(4.00\times {10}^{8}N/m^2\right)}{\left(2.0\times {10}^{11}N/m^2\right)}\times \left(10.0m\right)\\ or,\Delta L=2.0\times {10}^{-2}m\\ or,\Delta L=2.0cm\end{array}$

Hence, the stretched length of cable is $2.0cm$.

Conclusion:

Thus, the stretched length of cable is $2.0cm$.