Chapter 7.4, Problem 7.99P

Knowing that d_c = 9 ft, determine (a) the distances d_B and d_D (b) the reaction at E.

Fig. P7.99 and P7.100

To determine

The distances $d_B$ and $d_D$.

Answer to Problem 7.99P

The distance $d_B$ is $5.20\text{ft}$. The distance $d_D$ is $12.60\text{ft}$.

Explanation of Solution

Refer Fig P7.99.

The figure 1 below shows the free body diagram of the portion ABC.

The total moment about the point C is zero.

Refer the free body diagram and write the equation for the moment about point C.

$9A_x-12A_y+\left(1\text{kip}\right)\left(6\text{ft}\right)=0$

Here $A_x$ is the horizontal reaction at point A, $A_y$ is the vertical reaction at point A.

Re-write the above equation to get an expression for $A_x$ .

$A_x=\frac{4}{3}{A}_y-\frac{2}{3}$ (I)

The figure 2 below shows the free body diagram of the entire cable.

The moment about point E is zero.

Refer the free body diagram of the entire cable and write the equation of the moment about point E.

$12A_x-30A_y+\left(1\text{kip}\right)\left(18\text{ft}\right)+\left(1\text{kip}\right)\left(24\text{ft}\right)+\left(2\text{kips}\right)\left(9\text{ft}\right)=0$

Simplify the above equation.

$12A_x-30A_y+60=0$ (II)

Since the system is in equilibrium the total vertical and horizontal components will be zero.

Refer figure 2 and write the equation for total horizontal force.

$-A_x+E_x=0$ (III)

Here $E_x$ is the horizontal reaction force at point E.

Refer figure 2 and write the equation for the total vertical force.

$A_y+E_y-1\text{kip}-1\text{kip}-2\text{kip=0}$ (IV)

The figure 4 below shows the free body diagram of the portion AB.

The moment about point B is zero.

Refer figure 4 and write the equation for the moment about point B.

$-A_y\left(6\text{ft}\right)+A_x{d}_B=0$ (V)

The figure 5 below shows the free body diagram of the portion DE.

Refer figure 5 and write the formula for the distance $h$.

$h=\left(9\text{ft}\right)\tan 3.8°$ (VI)

Here $\text{h}$ is the vertical distance between point D and E.

Refer figure 5 and write the formula for distance $d_D$.

$d_D=12\text{ft+h}$ (VII)

Conclusion:

Substitute equation (I) in equation (II).

$\begin{array}{c}12\left(\frac{4}{3}{A}_y-\frac{2}{3}\right)-30A_y+60=0\\ 16A_y-30A_y-8+60=0\\ 14A_y=52\\ A_y=3.7143\text{kips}\end{array}$

Substitute $3.7143\text{kips}$ for $A_y$ in equation (I) to get

$\begin{array}{c}{A}_x=\frac{4}{3}\left(3.7143\text{kips}\right)-\frac{2}{3}\\ =4.257\text{kips}\end{array}$

Substitute $4.257\text{kips}$ for $A_x$ in equation (III) to determine $E_x$.

$E_x=4.257\text{kips}$

Substitute $3.7143\text{kips}$ for $A_y$ in equation (IV) to determine $E_y$.

$\begin{array}{c}{E}_y=-3.7143\text{kips+}1\text{kip+}1\text{kip+}2\text{kip}\\ \text{=0}\text{.2857}\text{kips}\end{array}$

Substitute $4.257\text{kips}$ for $A_x$, $3.7143\text{kips}$ for $A_y$ in equation (V) to determine $d_D$.

$\begin{array}{c}{d}_B=\frac{3.7143\text{kips}\left(6\text{ft}\right)}{4.257\text{kips}}\\ =5.20\text{ft}\end{array}$

Calculate $h$ from equation (VI).

$\begin{array}{c}h=\left(9\text{ft}\right)\tan 3.8°\\ =0.599\text{ft}\end{array}$

Substitute $0.599\text{ft}$ for $h$ in equation (VII) to determine $d_D$.

$\begin{array}{c}{d}_D=12\text{ft+}0.599\text{ft}\\ =12.60\text{ft}\end{array}$

The distance $d_B$ is $5.20\text{ft}$. The distance $d_D$ is $12.60\text{ft}$.

To determine

The reaction at point E.

Answer to Problem 7.99P

The reaction at point E is $4.30\text{kips}$ making an angle $3.81°$ with the horizontal.

Explanation of Solution

Refer Fig P7.99.

The figure 1 below shows the free body diagram of the portion ABC.

The total moment about the point C is zero.

Refer the free body diagram and write the equation for the moment about point C.

$9A_x-12A_y+\left(1\text{kip}\right)\left(6\text{ft}\right)=0$

Here $A_x$ is the horizontal reaction at point A, $A_y$ is the vertical reaction at point A.

Re-write the above equation to get an expression for $A_x$ .

$A_x=\frac{4}{3}{A}_y-\frac{2}{3}$ (I)

The figure 2 below shows the free body diagram of the entire cable.

The moment about point E is zero.

Refer the free body diagram of the entire cable and write the equation of the moment about point E.

$12A_x-30A_y+\left(1\text{kip}\right)\left(18\text{ft}\right)+\left(1\text{kip}\right)\left(24\text{ft}\right)+\left(2\text{kips}\right)\left(9\text{ft}\right)=0$

Simplify the above equation.

$12A_x-30A_y+60=0$ (II)

Since the system is in equilibrium the total vertical and horizontal components will be zero.

Refer figure 2 and write the equation for total horizontal force.

$-A_x+E_x=0$ (III)

Here $E_x$ is the horizontal reaction force at point E.

Refer figure 2 and write the equation for the total vertical force.

$A_y+E_y-1\text{kip}-1\text{kip}-2\text{kip=0}$ (IV)

Write the formula for the magnitude of the reaction at point E.

$E=\sqrt{E_x^2+E_y^2}$ (V)

Here $E$ is the magnitude of the reaction at point E.

Write the formula for the angle made by the reaction at point E with horizontal.

$\theta ={\tan }^{-1}\left(\frac{E_y}{E_x}\right)$ (VI)

Here $\theta$ is the angle made by the reaction at point E with horizontal.

Conclusion:

Substitute equation (I) in equation (II).

$\begin{array}{c}12\left(\frac{4}{3}{A}_y-\frac{2}{3}\right)-30A_y+60=0\\ 16A_y-30A_y-8+60=0\\ 14A_y=52\\ A_y=3.7143\text{kips}\end{array}$

Substitute $3.7143\text{kips}$ for $A_y$ in equation (I) to get

$\begin{array}{c}{A}_x=\frac{4}{3}\left(3.7143\text{kips}\right)-\frac{2}{3}\\ =4.257\text{kips}\end{array}$

Substitute $4.257\text{kips}$ for $A_x$ in equation (III) to determine $E_x$.

$E_x=4.257\text{kips}$

Substitute $3.7143\text{kips}$ for $A_y$ in equation (IV) to determine $E_y$.

$\begin{array}{c}{E}_y=-3.7143\text{kips+}1\text{kip+}1\text{kip+}2\text{kip}\\ \text{=0}\text{.2857}\text{kips}\end{array}$

Substitute $\text{0}\text{.2857}\text{kips}$ for $E_y$, $4.257\text{kips}$ for $E_y$ in equation (V) to determine $E$.

$\begin{array}{c}E=\sqrt{{\left(4.257\text{kips}\right)}^2+{\left(\text{0}\text{.2857}\text{kips}\right)}^2}\\ =4.30\text{kips}\end{array}$

Substitute $\text{0}\text{.2857}\text{kips}$ for $E_y$, $4.257\text{kips}$ for $E_y$ in equation (VI) to determine $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\text{0}\text{.2857}\text{kips}}{4.257\text{kips}}\right)\\ =3.81°\end{array}$

Thus the reaction at point E is $4.30\text{kips}$ making an angle $3.81°$ with the horizontal.