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.

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Accepted Answer

As shown in Fig. a, the orgin of the x, y coordinate system is set at the lowest point of the cable. Here, [latex]w(s)=5(9.81) \mathrm{N} / \mathrm{m}=49.05 \mathrm{~N} / \mathrm{m} [/latex]

[latex] \frac{d^2 y}{d x^2}=\frac{49.05}{F_H} \sqrt{1+\left(\frac{d y}{d x} ight)^2}[/latex]

[latex] ext { Set } u=\frac{d y}{d x} ext {, then } \frac{d u}{d x}=\frac{d^2 y}{d x^2} ext {, then }[/latex]

[latex] \frac{d u}{\sqrt{1+u^2}}=\frac{49.05}{F_H} d x[/latex]

Integrating,

[latex] \ln \left(u+\sqrt{1+u^2} ight)=\frac{49.05}{F_H} x+C_1[/latex]

Applying the boundary condition [latex]u=\frac{d y}{d x}=0 ext { at } x=0 ext { results in } C_1=0 [/latex]. Thus,

[latex]\ln \left(u+\sqrt{1+u^2} ight)=\frac{49.05}{F_H} x [/latex]

[latex]u+\sqrt{1+u^2}=e^{\frac{49.05}{F_H} x} [/latex]

[latex]\frac{d y}{d x}=u=\frac{e^{\frac{49.05}{F_H} x}-e^{-\frac{49.05}{F_H} x}}{2} [/latex]

Since sinh [latex] x=\frac{e^x-e^{-x}}{2}[/latex], then

[latex] \frac{d y}{d x}=\sinh \frac{49.05}{F_H} x[/latex]

Integrating,

[latex]y=\frac{F_H}{49.05} \cosh \left(\frac{49.05}{F_H} x ight)+C_2 [/latex]

Applying the boundary equation y = 0 at x = 0 results in [latex]C_2=-\frac{F_H}{49.05} [/latex]. Thus,

[latex] y=\frac{F_H}{49.05}\left[\cosh \left(\frac{49.05}{F_H} x ight)-1 ight] \mathrm{m}[/latex]

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,

[latex] \overset{+}{\longrightarrow } \Sigma F_x=0 ; \quad T \cos heta-F_H=0[/latex]

+↑[latex]\Sigma F_y=0 ; \quad T \sin heta-5(9.81) s=0 [/latex]

Eliminating T,

[latex] \frac{d y}{d x}= an heta=\frac{49.05 s}{F_H}[/latex]

Equating Eqs. (1) and (3) yields

[latex] \frac{49.05 s}{F_H}=\sinh \left(\frac{49.05}{F_H} x ight)[/latex]

[latex] s=\frac{F_H}{49.05}=\sinh \left(\frac{49.05}{F_H} ight)[/latex]

Thus, the length of the cable is

[latex]L=45=2\left\{\frac{F_H}{49.05} \sinh \left(\frac{49.05}{F_H}(20) ight) ight\} [/latex]

Solving by trial and error,

[latex]F_H=1153.41 \mathrm{~N} [/latex]

Substituting this result into Eq. (2),

[latex] y=23.5[\cosh 0.0425 x-1] \mathrm{m}[/latex]

The maximum tension occurs at either points A or B where the cable makes the greatest angle with the horizontal. Here

[latex] heta_{\max }= an ^{-1}\left(\left.\frac{d y}{d x} ight|_{\mathrm{x}=20 \mathrm{~m}} ight)= an ^{-1}\left\{\sinh \left(\frac{49.05}{F_H}(20) ight) ight\}=43.74^{\circ}[/latex]

Thus,

[latex]T_{\max }=\frac{F_H}{\cos heta_{\max }}=\frac{1153.41}{\cos 43.74^{\circ}}=1596.36 \mathrm{~N}=1.60 \mathrm{kN} [/latex]

Question 7.136: If the 45-m-long cable has a mass per unit length of 5 kg/m,...

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