Chapter 7.4, Problem 7.116P

(a)

To determine

The vertical distance $\left(a\right)$.

Answer to Problem 7.116P

The vertical distance $\left(a\right)$ is $\underset{\_}{a=4.05\text{m}}$.

Explanation of Solution

The free body diagram for the portion AC is depicted below:

Refer fig 1,

Write the equation of net force along vertical direction,

$\begin{array}{l}{A}_y-9w=0\\ A_y=9w\uparrow \end{array}$ (I)

Here, the forces are $A_y,w$.

Write the equation of momentum,

$T_{o}a-\left(9w\right)r_1=0$ (II)

Here, the tension is $T_o$, the vertical distance is $a$, the distance is $r_1$.

The free body diagram for the portion CB is depicted below:

Refer fig 2.

Write the equation of net force along vertical direction,

$\begin{array}{l}{B}_y-6w=0\\ B_y=6w\uparrow \end{array}$ (III)

Here, the forces are $B_y$.

The free body diagram for the entire cable is depicted below:

Refer fig 3.

Write the equation of momentum,

$15w{r}_2-B_y{r}_3-T_o{r}_4=0$ (IV)

Here, the distance is $r_2,r_3,r_4$.

Conclusion:

Substitute $7.5\text{m}$ for $r_2$, $6w$ for $B_y$, $15\text{m}$ for $r_3$, $2.25\text{m}$ for $r_4$ in Equation (IV).

$\begin{array}{l}\left[\left(15w\right)\left(7.5\text{m}\right)\right]-\left(6w\right)\left(15\text{m}\right)-T_o\left(2.25\text{m}\right)=0\\ T_o=10w\end{array}$

Substitute $10w$ for $T_o$, $4.5\text{m}$ for $r_1$ in Equation (II).

$\begin{array}{l}10wa-\left(9w\right)\left(4.5\text{m}\right)=0\\ a=4.05\text{m}\end{array}$

Thus, the vertical distance $\left(a\right)$ is $\underset{\_}{a=4.05\text{m}}$.

(b)

To determine

The length of the cable.

Answer to Problem 7.116P

The length of the cable is $\underset{\_}{L=16.41\text{m}}$.

Explanation of Solution

Write the equation of length of the cable

$L=s_{AC}+s_{CB}$ (V)

Here, the length of the cable is $L$, the length of the portion is $s_{AC}$, the length of the portion CB is $s_{CB}$.

Write the equation of the length of the portion AC,

$s_{AC}=x_A\left[1+\frac{2}{3}{\left(\frac{y_A}{x_A}\right)}^2-\frac{2}{5}{\left(\frac{y_A}{x_A}\right)}^4+\mathrm{....}\right]$ (VI)

Here, the coordinates are $x_A,y_A$.

Write the equation of the length of the portion CB,

$s_{CB}=x_B\left[1+\frac{2}{3}{\left(\frac{y_B}{x_B}\right)}^2-\frac{2}{5}{\left(\frac{y_B}{x_B}\right)}^4+\mathrm{....}\right]$ (VII)

Here, the coordinates are $x_B,y_B$.

Conclusion:

Substitute $9\text{m}$ for $x_A$, $4.05\text{m}$ for $y_A$ in Equation (VI).

$\begin{array}{c}{s}_{AC}=\left(9\text{m}\right)\left[1+\frac{2}{3}{\left(\frac{\left(4.05\text{m}\right)}{\left(9\text{m}\right)}\right)}^2-\frac{2}{5}{\left(\frac{\left(4.05\text{m}\right)}{\left(9\text{m}\right)}\right)}^4+\mathrm{....}\right]\\ =10.067\text{m}\end{array}$

Substitute $\left(6\text{m}\right)$ for $x_B$, $\left(4.05\text{m}-2.25\text{m}\right)$ for $y_B$ in Equation (VII).

$\begin{array}{c}{s}_{CB}=\left(6\text{m}\right)\left[1+\frac{2}{3}{\left(\frac{\left(4.05\text{m}-2.25\text{m}\right)}{\left(6\text{m}\right)}\right)}^2-\frac{2}{5}{\left(\frac{\left(4.05\text{m}-2.25\text{m}\right)}{\left(6\text{m}\right)}\right)}^4+\mathrm{....}\right]\\ =6.341\text{m}\end{array}$

Substitute, $10.067\text{m}$ for $s_{AC}$, $6.341\text{m}$ for $s_{CB}$ in equation (V)

$\begin{array}{c}L=\left(10.067\text{m}\right)+\left(6.341\text{m}\right)\\ =16.41\text{m}\end{array}$

Thus, the length of the cable is $\underset{\_}{L=16.41\text{m}}$.

(c)

To determine

The component of the reaction a A.

Answer to Problem 7.116P

The component of the reaction a A is $\underset{\_}{A_x=5890\text{N}\leftarrow \text{, }{A}_y=5300\text{N}\uparrow }$.

Explanation of Solution

Write the equation of vertical component of reaction A

$A_y=9w$ (VIII)

Write the equation of the horizontal component of reaction A

$\begin{array}{l}{A}_x=T_o\\ A_x=10w\end{array}$ (IX)

Conclusion:

Substitute $\left[\left(60\text{kg/m}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\right]$ for $w$ in Equation (VIII).

$\begin{array}{c}{A}_y=9\left[\left(60\text{kg/m}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\right]\\ =5297.4\text{N}\\ \approx \text{5300N, }\uparrow \end{array}$

Substitute $\left[\left(60\text{kg/m}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\right]$ for $w$ in Equation (IX).

$\begin{array}{c}{A}_x=\left(10\right)\left[\left(60\text{kg/m}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\right]\\ =5890\text{N,}\leftarrow \end{array}$

Thus, the component of the reaction a A is $\underset{\_}{A_x=5890\text{N}\leftarrow \text{, }{A}_y=5300\text{N}\uparrow }$.