Chapter 2, Problem 2.136RP

Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables. (Hint: The tension is the same in both portions of cable BAC.)

Fig. P2.136

To determine

The tension in each of the three cables $AD$, $AE$, and $BAC$.

Answer to Problem 2.136RP

The tension in each of the three cables are $\underset{\_}{T_{BAC}=75.4\text{lb}}$, $\underset{\_}{T_{AD}=26.4\text{lb}}$, and $\underset{\_}{T_{AE}=48.33\text{lb}}$.

Explanation of Solution

The free body diagram is sketched in the figure 1 below

From the free body diagram the tension along of cable $BAC$ is equal to the tension acting along cable $AC$ , and $AB$ which are the two parts of $BAC$ on the either sides of the ring.

Write the expression for the magnitude of tension acting along $AB$

${\text{T}}_{AB}=T_{BAC}{\text{λ}}_{AB}$ (I)

Here, $T_{BAC}$ is the vector form of  tension, ${\text{T}}_{AB}$ is the magnitude of tension, ${\text{λ}}_{AB}$ is representing the direction of tension

Write the expression for the direction of tension

${\lambda }_{AB}=\frac{\overrightarrow{AB}}{AB}$

Here, $\overrightarrow{AB}$ is the vector representation of cable, and $AB$ is the magnitude of the cable.

Thus equation 1 can be rewritten as

${\text{T}}_{AB}=T_{BAC}\frac{\overrightarrow{AB}}{AB}$ $\left(\text{I'}\right)$

Write the expression for the magnitude of tension acting along $AC$

$\begin{array}{c}{\text{T}}_{AC}=T_{BAC}{\text{λ}}_{AC}\\ =T_{BAC}\frac{\overrightarrow{AC}}{AC}\end{array}$ (II)

Here, $T_{BAC}$ is the vector form of tension, ${\text{T}}_{AC}$ is the magnitude of tension, ${\text{λ}}_{AC}$ is representing the direction of tension, $\overrightarrow{AC}$ is the vector representation of cable, and $AC$ is the magnitude of the cable

Write the expression for the magnitude of tension acting along $AD$

$\begin{array}{l}{\text{T}}_{AD}=T_{AD}{\text{λ}}_{AD}\\ =T_{AD}\frac{\overrightarrow{AD}}{AD}\end{array}$ (III)

Here, $T_{AD}$ is the vector form of tension, ${\text{T}}_{AD}$ is the magnitude of tension, ${\text{λ}}_{AD}$ is representing the direction of tension, $\overrightarrow{AD}$ is the vector representation of cable, and $AD$ is the magnitude of the cable

Similarly, the expression for the magnitude of tension acting along $AE$

$\begin{array}{l}{\text{T}}_{AE}=T_{AE}{\text{λ}}_{AE}\\ =T_{AE}\frac{\overrightarrow{AE}}{AE}\end{array}$ (IV)

Here, $T_{AE}$ is the vector form of tension, ${\text{T}}_{AE}$ is the magnitude of tension, ${\text{λ}}_{AE}$ is representing the direction of tension, $\overrightarrow{AE}$ is the vector representation of cable, and $AE$ is the magnitude of the cable

Under equilibrium, the total force will balances and the sum of all forces will be equal to zero.

${\displaystyle \sum F}=0$ (V)

Conclusion:

Write the vector representation of $AB$

$\overrightarrow{AB}=\left(-17.5\text{in}\text{.}\right)i+\left(60\text{in}\text{.}\right)j$

The magnitude of vector $AB$ is

$\begin{array}{c}AB=\sqrt{{\left(17.5\text{in}.\right)}^2+{\left(60\text{in}\right)}^2{}^{}}\\ =62.5\text{in}\text{.}\end{array}$

Write the vector representation of $AC$

$\overrightarrow{AC}=\left(\text{60in}\text{.}\right)i+\left(25\text{in}\text{.}\right)k$

The magnitude of vector $AC$ is

$\begin{array}{c}AC=\sqrt{{\left(60\text{in}.\right)}^2+{\left(25\text{in}\right)}^2{}^{}}\\ =65\text{in}\text{.}\end{array}$

Write the vector representation of $AD$

$\overrightarrow{AD}=\left(-80\text{in}\text{.}\right)i+\left(60\text{in}\text{.}\right)j$

The magnitude of vector $AD$ is

$\begin{array}{c}AD=\sqrt{{\left(80\text{in}.\right)}^2+{\left(60\text{in}\right)}^2{}^{}}\\ =100\text{in}\text{.}\end{array}$

Write the vector representation of $AE$

$\overrightarrow{AE}=\left(60\text{in}\text{.}\right)j+\left(-45\text{in}\text{.}\right)k$

The magnitude of vector $AE$ is

$\begin{array}{c}AD=\sqrt{{\left(60\text{in}.\right)}^2+{\left(-45\text{in}\right)}^2{}^{}}\\ =75\text{in}\text{.}\end{array}$

Substitute $\left(-17.5\text{in}\text{.}\right)i+\left(60\text{in}\text{.}\right)j$ for $\overrightarrow{AB}$, and $62.5\text{in}\text{.}$ for $AB$ in equation $\left(\text{I'}\right)$

$\begin{array}{c}{\text{T}}_{AB}=T_{BAC}\frac{\left(-17.5\text{in}\text{.}\right)i+\left(60\text{in}\text{.}\right)j}{62.5\text{in}\text{.}}\\ =T_{BAC}\left(\frac{-17.5}{62.5}i+\frac{60}{62.5}j\right)\end{array}$

Substitute $\left(\text{60in}\text{.}\right)i+\left(25\text{in}\text{.}\right)k$ for $\overrightarrow{AC}$, and $65\text{in}\text{.}$ for $AC$ in equation (II)

$\begin{array}{c}{\text{T}}_{AB}=T_{BAC}\frac{\left(\text{60in}\text{.}\right)i+\left(25\text{in}\text{.}\right)k}{65\text{in}\text{.}}\\ =T_{BAC}\left(\frac{60}{65}j+\frac{25}{65}k\right)\end{array}$

Substitute $\left(-80\text{in}\text{.}\right)i+\left(60\text{in}\text{.}\right)j$  for $\overrightarrow{AD}$, and $\text{100}\text{in}\text{.}$ for $AD$ in equation (III)

$\begin{array}{c}{\text{T}}_{AB}=T_{AD}\frac{\left(\text{80in}\text{.}\right)i+\left(60\text{in}\text{.}\right)j}{100\text{in}\text{.}}\\ =T_{AD}\left(\frac{80}{100}i+\frac{60}{100}j\right)\end{array}$

Substitute $\left(60\text{in}\text{.}\right)j+\left(-45\text{in}\text{.}\right)k$ for $\overrightarrow{AE}$, and $\text{75}\text{in}\text{.}$ for $AE$ in equation (IV)

$\begin{array}{c}{\text{T}}_{AE}=T_{AE}\frac{\left(\text{60in}\text{.}\right)j-\left(45\text{in}\text{.}\right)k}{75\text{in}\text{.}}\\ =T_{AE}\left(\frac{4}{5}i-\frac{3}{5}j\right)\end{array}$

Substitute $-200\text{lb}$ for $P$, and the values of ${\text{T}}_{AB}$, ${\text{T}}_{AC}$, ${\text{T}}_{AE}$, and ${\text{T}}_{AD}$ in equation (V). since ${\displaystyle \sum F=0}$ equate each coefficients of $i$, $j$, and $k$ to zero gives three sets of linear equations.

$\frac{-17.5}{62.5}{T}_{BAC}+\frac{4}{5}{T}_{AD}=0$ (VI)

$\left(\frac{61}{62.5}+\frac{62}{65}\right)T_{BAC}+\frac{3}{5}{T}_{AD}+\frac{4}{5}{T}_{AE}-200\text{lb=0}$ (VII)

$\frac{25}{65}{T}_{BAC}-\frac{3}{5}{T}_{AE}=0$ (VIII)

Rearrange equation (VI) to obtain $T_{AD}$

$T_{AD}=\left(\frac{17.5}{62.5}{T}_{BAC}\right)\times \frac{5}{4}$ (IX)

Rearrange equation (VIII) to obtain $T_{AE}$

$T_{AE}=\left(\frac{25}{65}{T}_{BAC}\right)\times \frac{5}{3}$ (X)

Substitute equation (IX) and (X) in equation (VII)

$\begin{array}{c}\left(\frac{61}{62.5}+\frac{62}{65}\right)T_{BAC}+\frac{3}{5}\left(\frac{17.5}{62.5}{T}_{BAC}\right)\times \frac{5}{4}+\frac{4}{5}\left(\frac{25}{65}{T}_{BAC}\right)\times \frac{5}{3}-200\text{lb=0}\\ T_{BAC}=75.4\text{lb}\end{array}$

Substitute $75.4\text{lb }$ for $T_{BAC}$ in equation (VI) to find $T_{AD}$

$\begin{array}{c}\frac{-17.5}{62.5}\left(75.4\text{lb}\right)+\frac{4}{5}{T}_{AD}=0\\ T_{AD}=\frac{17.5}{62.5}\left(75.4\text{lb}\right)\times \frac{5}{4}\\ =26.39\text{lb}\end{array}$

Substitute $75.4\text{lb }$ for $T_{BAC}$ in equation (VII) to find $T_{AE}$

$\begin{array}{c}\frac{25}{65}\left(75.4\text{lb}\right)-\frac{3}{5}{T}_{AE}=0\\ T_{AE}=\frac{25}{65}\left(75.4\text{lb}\right)\times \frac{5}{3}\\ =48.33\text{lb}\end{array}$

Therefore, the tension in each of the three cables are $\underset{\_}{T_{BAC}=75.4\text{lb}}$, $\underset{\_}{T_{AD}=26.4\text{lb}}$, and $\underset{\_}{T_{AE}=48.33\text{lb}}$.