Chapter 7.4, Problem 7.125P

To determine

The expression for the curve made by the cable, maximum and minimum values of the tension in the cable.

Answer to Problem 7.125P

The curve represented by the cable is $\underset{\_}{\left(\frac{w_0{L}^2}{T_0{\pi }^2}\right)\left(1-\cos \frac{\pi x}{L}\right)}$, the minimum tension on the cable is $\underset{\_}{\frac{w_0{L}^2}{T_0{\pi }^2}}$ and maximum tension on the cable is $\underset{\_}{\left(\frac{w_{0}L}{\pi }\right)\sqrt{1+{\left(\frac{L}{\pi h}\right)}^2}}$.

Explanation of Solution

The figure 1 below shows the cable and which makes the curve due the load w.

Write the expression for the load distributed.

$w\left(x\right)=w_0\cos \frac{\pi x}{L}$

Refer the problem 77268-7.4-7.124P and writhe the differential equation for the curve.

Write the expression for the differential equation for the curve.

$\frac{d^{2}y}{d{x}^2}=\frac{w\left(x\right)}{T_0}$

Substitute $w_0\cos \frac{\pi x}{L}$ for $w\left(x\right)$ in the above equation to rewrite.

$\frac{d^{2}y}{d{x}^2}=\frac{w_0\cos \frac{\pi x}{L}}{T_0}$

Integrate both side of the above equation and apply the condition ${\frac{dy}{dx}|}_{x=0}=0$.

$\begin{array}{c}\frac{dy}{dx}={\displaystyle \int \frac{w_0\cos \frac{\pi x}{L}}{T_0}dx}\\ =\frac{w_{0}L}{T_0\pi }\sin \frac{\pi x}{L}\end{array}$

Integrate the above equation and apply the condition ${y|}_{x=0}=0$.

$\begin{array}{c}y={\displaystyle \int \left(\frac{w_{0}L}{T_0\pi }\right)\sin \frac{\pi x}{L}dx}\\ =\left(\frac{w_0{L}^2}{T_0{\pi }^2}\right)\left(-\cos \frac{\pi x}{L}\right)+C\end{array}$

Here C is the integration constant.

Apply the condition ${y|}_{x=0}=0$ to calculate C.

$\begin{array}{l}0=\left(\frac{w_0{L}^2}{T_0{\pi }^2}\right)\left(-\cos 0\right)+C\\ C=\frac{w_0{L}^2}{T_0{\pi }^2}\end{array}$

Substitute $\frac{w_0{L}^2}{T_0{\pi }^2}$ for C in the equation for y to rewrite.

$\begin{array}{c}y=\left(\frac{w_0{L}^2}{T_0{\pi }^2}\right)\left(-\cos \frac{\pi x}{L}\right)+\frac{w_0{L}^2}{T_0{\pi }^2}\\ =\left(\frac{w_0{L}^2}{T_0{\pi }^2}\right)\left(1-\cos \frac{\pi x}{L}\right)\end{array}$

At $x=\frac{L}{2}$ the tension on the cable might have minimum value.

Substitute $x=\frac{L}{2}$ in the above equation for y to calculate the minimum tension.

$\begin{array}{c}y=\left(\frac{w_0{L}^2}{T_0{\pi }^2}\right)\left(1-\cos \frac{\pi }{L}\left(\frac{L}{2}\right)\right)\\ =\left(\frac{w_0{L}^2}{T_0{\pi }^2}\right)\left(1-\cos \frac{\pi }{2}\right)\\ =\frac{w_0{L}^2}{T_0{\pi }^2}\end{array}$

The value of y at $x=\frac{L}{2}$ is same as the sag h.

Rewrite the above equation in terms of h.

$\begin{array}{l}h=\frac{w_0{L}^2}{T_0{\pi }^2}\\ T_0=\frac{w_0{L}^2}{h{\pi }^2}\end{array}$

Here $T_0$ is the minimum tension on the cable.

Therefore the minimum tension on the cable is $\frac{w_0{L}^2}{T_0{\pi }^2}$.

To find the maximum tension on the cable the slope of the above equation for y at $x=\frac{L}{2}$ which is equal to the ratio $\frac{T_{By}}{T_0}$

Here $T_{By}$ is the y-component of the tension at A.

$\begin{array}{c}\frac{dy}{dx}=\frac{T_{By}}{T_0}\\ =\frac{d}{dx}\left(\left(\frac{w_0{L}^2}{T_0{\pi }^2}\right)\left(1-\cos \frac{\pi x}{L}\right)\right)\\ =\left(\frac{w_0{L}^2}{T_0{\pi }^2}\right)\left(\frac{\pi }{L}\right)\sin \frac{\pi x}{L}\\ \frac{T_{By}}{T_0}=\left(\frac{w_{0}L}{T_0\pi }\right)\sin \frac{\pi x}{L}\end{array}$

Substitute $x=\frac{L}{2}$ in the above equation to calculate $T_{By}$.

$\begin{array}{c}\frac{T_{By}}{T_0}=\left(\frac{w_{0}L}{T_0\pi }\right)\sin \frac{\pi }{L}\left(\frac{L}{2}\right)\\ =\left(\frac{w_{0}L}{T_0\pi }\right)\sin \frac{\pi }{2}\\ =\frac{w_{0}L}{T_0\pi }\end{array}$

Rewrite the above equation in terms of $T_{By}$.

$\begin{array}{c}{T}_{By}=\left(\frac{w_{0}L}{T_0\pi }\right)T_0\\ =\frac{w_{0}L}{\pi }\end{array}$

Write the expression to calculate the maximum tension on the cable.

$T_{\max }=\sqrt{T_{By}^2+T_0^2}$

Here $T_{\max }$ is the maximum tension in the cable.

Substitute $\frac{w_{0}L}{\pi }$ for $T_{By}$ and $\frac{w_0{L}^2}{h{\pi }^2}$ for $T_0$ in the above equation to calculate $T_{\max }$.

$\begin{array}{c}{T}_{\max }=\sqrt{{\left(\frac{w_{0}L}{\pi }\right)}^2+{\left(\frac{w_0{L}^2}{h{\pi }^2}\right)}^2}\\ =\left(\frac{w_{0}L}{\pi }\right)\sqrt{1+{\left(\frac{L}{\pi h}\right)}^2}\end{array}$

Conclusion:

Thus, the curve represented by the cable is $\underset{\_}{\left(\frac{w_0{L}^2}{T_0{\pi }^2}\right)\left(1-\cos \frac{\pi x}{L}\right)}$, the minimum tension on the cable is $\underset{\_}{\frac{w_0{L}^2}{T_0{\pi }^2}}$ and maximum tension on the cable is $\underset{\_}{\left(\frac{w_{0}L}{\pi }\right)\sqrt{1+{\left(\frac{L}{\pi h}\right)}^2}}$.