Question 7.136: If the 45-m-long cable has a mass per unit length of 5 kg/m,...
If the 45-m-long cable has a mass per unit length of 5 kg/m, determine the equation of the catenary curve of the cable and the maximum tension developed in the cable.
As shown in Fig. a, the orgin of the x, y coordinate system is set at the lowest point of the cable. Here, w(s)=5(9.81) \mathrm{N} / \mathrm{m}=49.05 \mathrm{~N} / \mathrm{m}
\frac{d^2 y}{d x^2}=\frac{49.05}{F_H} \sqrt{1+\left(\frac{d y}{d x}\right)^2}
\text { Set } u=\frac{d y}{d x} \text {, then } \frac{d u}{d x}=\frac{d^2 y}{d x^2} \text {, then }\frac{d u}{\sqrt{1+u^2}}=\frac{49.05}{F_H} d x
Integrating,
\ln \left(u+\sqrt{1+u^2}\right)=\frac{49.05}{F_H} x+C_1
Applying the boundary condition u=\frac{d y}{d x}=0 \text { at } x=0 \text { results in } C_1=0 . Thus,
\ln \left(u+\sqrt{1+u^2}\right)=\frac{49.05}{F_H} x
u+\sqrt{1+u^2}=e^{\frac{49.05}{F_H} x}
\frac{d y}{d x}=u=\frac{e^{\frac{49.05}{F_H} x}-e^{-\frac{49.05}{F_H} x}}{2}
Since sinh x=\frac{e^x-e^{-x}}{2}, then
\frac{d y}{d x}=\sinh \frac{49.05}{F_H} x
Integrating,
y=\frac{F_H}{49.05} \cosh \left(\frac{49.05}{F_H} x\right)+C_2
Applying the boundary equation y = 0 at x = 0 results in C_2=-\frac{F_H}{49.05} . Thus,
y=\frac{F_H}{49.05}\left[\cosh \left(\frac{49.05}{F_H} x\right)-1\right] \mathrm{m}
If we write the force equation of equilibrium along the x and y axes by referring to the free-body diagram shown in Fig. b,
\overset{+}{\longrightarrow } \Sigma F_x=0 ; \quad T \cos \theta-F_H=0
+↑\Sigma F_y=0 ; \quad T \sin \theta-5(9.81) s=0
Eliminating T,
\frac{d y}{d x}=\tan \theta=\frac{49.05 s}{F_H}
Equating Eqs. (1) and (3) yields
\frac{49.05 s}{F_H}=\sinh \left(\frac{49.05}{F_H} x\right)
s=\frac{F_H}{49.05}=\sinh \left(\frac{49.05}{F_H}\right)
Thus, the length of the cable is
L=45=2\left\{\frac{F_H}{49.05} \sinh \left(\frac{49.05}{F_H}(20)\right)\right\}
Solving by trial and error,
F_H=1153.41 \mathrm{~N}
Substituting this result into Eq. (2),
y=23.5[\cosh 0.0425 x-1] \mathrm{m}
The maximum tension occurs at either points A or B where the cable makes the greatest angle with the horizontal. Here
\theta_{\max }=\tan ^{-1}\left(\left.\frac{d y}{d x}\right|_{\mathrm{x}=20 \mathrm{~m}}\right)=\tan ^{-1}\left\{\sinh \left(\frac{49.05}{F_H}(20)\right)\right\}=43.74^{\circ}
Thus,
T_{\max }=\frac{F_H}{\cos \theta_{\max }}=\frac{1153.41}{\cos 43.74^{\circ}}=1596.36 \mathrm{~N}=1.60 \mathrm{kN}