Chapter 10, Problem 101P

(a)

To determine

The strain in the cable.

Answer to Problem 101P

The strain in the cable is $\underset{\_}{8\times {10}^{-4}}$.

Explanation of Solution

Consider the segment of the rope between the fixed end and the walker’s position. A right triangle can be constructed with the non-stretched segment and stretched segment of the cable as shown in Figure 1.

Write the expression for the strain in the cable.

$\begin{array}{c}\text{strain}=\frac{\Delta L}{L}\\ =\frac{L^{\prime }-L}{L}\end{array}$ (I)

Here, $\Delta L$ is the change in length, $L$ is the length of non-stretched segment of the cable, and is the stretched length of the segment of the cable.

Write the expression for $L$ from the right triangle in Figure 1.

$L=L^{\prime }\cos \theta$ (II)

Use equation (II) in (I).

$\begin{array}{c}\text{strain}=\frac{L^{\prime }-L^{\prime }\cos \theta }{L^{\prime }\cos \theta }\\ =\frac{1}{\cos \theta }-1\end{array}$ (III)

Conclusion:

Substitute $0.040\text{rad}$ for $\theta$ in equation (III) to find the strain.

$\begin{array}{c}\text{strain}=\frac{1}{\cos \left(0.040\text{rad}\right)}-1\\ =8\times {10}^{-4}\end{array}$

Therefore, the strain in the cable is $\underset{\_}{8\times {10}^{-4}}$.

(b)

To determine

The tension in the cable when the walker is at the middle of the rope.

Answer to Problem 101P

The tension in the cable when the walker is at the middle of the rope is $\underset{\_}{8.0\text{kN}}$.

Explanation of Solution

The free-body diagram of the walker is given in Figure 2.

Write the equilibrium conditions on the forces in $y$ direction.

${\displaystyle \sum F_y=0}$ (IV)

Apply the condition in equation (I) to the free-body diagram.

$2T\sin \theta -W=0$ (V)

Here, $T$ is the tension in the cable, and $W$ is the weight of the walker.

Solve equation (V) for $T$.

$T=\frac{W}{2\sin \theta }$ (VI)

Conclusion:

Substitute $640\text{N}$ for $W$, and $0.040\text{rad}$ for $\theta$ in equation (VI) to find $T$.

$\begin{array}{c}T=\frac{640\text{N}}{2\sin \left(0.040\text{rad}\right)}\\ =8002\text{N}\\ =8002\text{N}\times \frac{1\text{kN}}{1000\text{N}}\\ =8.0\text{kN}\end{array}$

Therefore, the tension in the cable when the walker is at the middle of the rope is $\underset{\_}{8.0\text{kN}}$.

(c)

To determine

The cross sectional area of the cable.

Answer to Problem 101P

The cross sectional area of the cable is $\underset{\_}{5\times {10}^{-5}{\text{m}}^2}$.

Explanation of Solution

Equation (VI) gives the tension in the cable.

$T=\frac{W}{2\sin \theta }$

Divide equation (VI) by the cross sectional area of the cable $A$.

$\frac{T}{A}=\frac{W}{2A\sin \theta }$ (VII)

Write the expression of Hooke’s law when force equal to the tension $T$ applied to the cable.

$\frac{T}{A}=Y\frac{\Delta L}{L}$ (VIII)

Here, $Y$ is the Young’s modulus (for steel $Y=200\times 10\text{Pa}$)

Use equation (I) in (VIII).

$\frac{T}{A}=Y\left(\text{strain}\right)$ (IX)

Equate the right hand sides of the equations (VII) and (IX) and solve for $A$.

$\begin{array}{c}\frac{W}{2A\sin \theta }=Y\left(\text{strain}\right)\\ A=\frac{W}{2Y\left(\text{strain}\right)\sin \theta }\end{array}$ (X)

Conclusion:

Substitute $640\text{N}$ for $W$, $200\times {10}^9\text{Pa}$ for $Y$, $8\times {10}^{-4}$ for strain, and $0.040\text{rad}$ for $\theta$ in equation (X) to find $A$.

$\begin{array}{c}A=\frac{640\text{N}}{2\left(200\times {10}^9\text{Pa}\right)\left(8\times {10}^{-4}\right)\sin \left(0.040\text{rad}\right)}\\ =\frac{640\text{N}}{2\left(200\times {10}^9\text{Pa}\right)\left(8\times {10}^{-4}\right)\sin \left(0.040\text{rad}\right)}\\ =5\times {10}^{-5}{\text{m}}^2\end{array}$

Therefore, the cross sectional area of the cable is $\underset{\_}{5\times {10}^{-5}{\text{m}}^2}$.

(d)

To determine

Whether the cable stretched beyond its elastic limit.

Answer to Problem 101P

The cable did not stretched beyond its elastic limit.

Explanation of Solution

Write the expression for the stress in terms of the tension in the cable.

$\text{stress}=\frac{T}{A}$ (XI)

Conclusion:

Substitute $8.0\text{kN}$ for $T$, and $5\times {10}^{-5}{\text{m}}^2$ for $A$ in equation (XI) to find the stress in the cable.

$\begin{array}{c}\text{stress}=\frac{8.0\text{kN}}{5\times {10}^{-5}{\text{m}}^2}\\ =\frac{8.0\text{kN}\times \frac{1000\text{N}}{1\text{kN}}}{5\times {10}^{-5}{\text{m}}^2}\\ =1.6\times {10}^8\text{Pa}<2.5\times {10}^8\text{Pa}\end{array}$

Therefore, the cable did not stretched beyond its elastic limit.