Question 7.132: A chain is suspended between points at the same elevation an...

x=\int \frac{d s}{\left\{1+\frac{1}{F_H^2} \int\left(w_0 d s\right)^2\right\}^{\frac{1}{2}}}

Performing the integration yields:

x=\frac{F_H}{0.5}\left\{\operatorname{sin~}^{-1}\left[\frac{1}{F_H}\left(0.5 s+C_1\right)\right]+C_2\right\}       (1)

From Eq. 7-14

\frac{d y}{d x}=\frac{1}{F_H} \int w_0 d s

\frac{d y}{d x}=\frac{1}{F_H}\left(0.5 s+C_1\right)

\frac{d y}{d x}=\tan \theta=\frac{0.5 s}{F_H}     (2)

Applying boundary conditions at x = 0; s = 0 to Eq. (1) and using the result C_1=0 yields C_2=0 . Hence

s=\frac{F_H}{0.5} \sinh \left(\frac{0.5}{F_H} x\right)          (3)

Substituting Eq. (3) into (2) yields:

\frac{d y}{d x}=\sinh \left(\frac{0.5 x}{F_H}\right)       (4)

Performing the integration

y=\frac{F_H}{0.5} \cosh \left(\frac{0.5}{F_H} x\right)+C_3        (5)

Applying boundary conditions at x=0 ; y=0 \text { yields } C_3=-\frac{F_H}{0.5} \text {. } Therefore y=\frac{F_H}{0.5}\left[\cosh \left(\frac{0.5}{F_H} x\right)-1\right]

By trial and error F_H=75.25  \mathrm{lb}

\text { At } x=30  \mathrm{ft} ; \quad \theta=\theta_{\max }. From Eq. (4)

\tan \theta_{\max }=\left.\frac{d y}{d x}\right|_{x=30  \mathrm{ft}}=\sinh \left(\frac{0.5(30)}{75.25}\right) \quad \theta_{\max }=11.346^{\circ}

T_{\max }=\frac{F_H}{\cos \theta_{\max }}=\frac{75.25}{\cos 11.346^{\circ}}=76.7  \mathrm{lb}