Question 7.131: The uniform beam weighs 500 lb and is held in the horizontal...

T=\frac{250}{\sin 30^{\circ}}=500  \mathrm{lb}

F_H=500 \cos 30^{\circ}=433.0  \mathrm{lb}

From Example 7 – 13

\frac{d y}{d x}=\frac{1}{F_H}\left(w_0 s+C_1\right)

\therefore C_1=433.0(0.577)=250

x=\frac{F_H}{w_0}\left\{\sinh ^{-1}\left[\frac{1}{F_H}\left(w_0 s+C_1\right)\right]+C_2\right\}

=\frac{433.0}{5}\left\{\sinh ^{-1}\left[\frac{1}{433.0}(5 s+250)\right]+C_2\right\}

s=0 \text { at } x=0, \quad C_2=-0.5493

Thus,

x=86.6\left\{\sinh ^{-1}\left[\frac{1}{433.0}(5 s+250)\right]-0.5493\right\}

When x = 15 ft.

s = 18.2 ft