Chapter 5, Problem 5.1P

To determine

The tension in each segment of the cable and the vertical distance between points $A$ and $D$.

Answer to Problem 5.1P

The tension in segment $AB$ is $15.21\text{kN}$.

The tension in segment $BC$ is $8.49\text{kN}$.

The tension in segment $CD$ is $10.82\text{kN}$.

The vertical distance between points $A$ and $D$ that is $y_{\text{D}}$ .is $4.33\text{m}$.

Explanation of Solution

Concept Used:

Write the expression for the net moment about end $A$ of the system.

   ${\displaystyle \sum M_{\text{A}}}=0$   ......... (I)

Here ${\displaystyle \sum M_{\text{A}}}$ is the sum of all the moments taken about the end $A$.

Calculations:

The free body diagram for the system is shown below.

  

  Figure-(1)

Here, the unknown vertical reactions are $A_y$ and $D_y$ at points $A$ and $D$ respectively and the unknown horizontal reactions are $A_x$ and $D_x$ at points $A$ and $D$ respectively.

The slope of cable $CD$ is known, therefore resolve the tension in cable $CD$, that is, $T_{\text{CD}}$ in perpendicular and horizontal directions and determine the moment about point $A$.

Write the moment about $A$ from Equation (I).

   $\begin{array}{c}-D_{\text{x}}\times y_{\text{D}}+D_{\text{y}}\times \left(9\text{m}\right)-\left(3\text{KN}\right)\times \left(5\text{m}\right)-\left(20\text{KN}\right)\times \left(2\text{m}\right)=0\\ -\left(D_{\text{x}}\times y_{\text{D}}\text{kN}\right)+\left(9D_{\text{y}}\text{kN}\right)-\left(15\text{kN-m}\right)-\left(40\text{kN-m}\right)=0\\ -\left(D_{\text{x}}\times y_{\text{D}}\text{kN}\right)+\left(9D_{\text{y}}\text{kN}\right)-\left(55\text{kN-m}\right)=0\end{array}$   ......... (II)

Consider $\Delta BCE$.

Given $\begin{array}{c}BE=CE\\ =\text{3}\text{m}\end{array}$

Given that $\angle BEC$ is equal to $90°$

Therefore $\Delta \text{BCE}$ is a right angled isosceles triangle, therefore,

   $\begin{array}{c}\angle BCE=\angle CBE\\ =45°\end{array}$

Consider $\Delta DCF$.

Apply Pythagoras theorem in $\Delta DCF$.

   ${\left(DC\right)}^2={\left(DF\right)}^2+{\left(CF\right)}^2$   ......... (III)

Here, the distance between points $D$ and $C$ is $DC$, the distance between points $D$ and $F$ is $DF$ and the distance between points $C$ and $F$ is $CF$.

Substitute $6\text{m}$ for $DF$ and $4\text{m}$ for $CF$ in Equation (III).

   $\begin{array}{c}{\left(DC\right)}^2={\left(6\text{m}\right)}^2+{\left(4\text{m}\right)}^2\\ =\left(36{\text{m}}^2\right)+\left(16{\text{m}}^2\right)\\ =52{\text{m}}^2\end{array}$

Take square root on both sides.

   $\begin{array}{c}DC=\sqrt{\left(52{\text{m}}^2\right)}\\ =7.21\text{m}\end{array}$

Calculate $\angle DCF$.

   $\angle DCF={\text{sin}}^{-1}\left(\frac{DF}{CF}\right)$   ......... (IV)

Substitute $6\text{m}$ for $DF$ and $7.21\text{m}$ for $CF$ in Equation (IV).

   $\begin{array}{c}\angle DCF={\text{sin}}^{-1}\left(\frac{6\text{m}}{7.21\text{m}}\right)\\ ={\text{sin}}^{-1}\left(0.832\right)\\ =56.32°\end{array}$

Calculate $\angle DCE$.

   $\angle DCE=\angle DCF+\angle FCE$   ......... (V)

Substitute $56.32°$ for $\angle DCF$ and $90°$ for $\angle FCE$ in Equation (V).

   $\begin{array}{c}\angle DCE=56.32°+90°\\ =146.32°\end{array}$

Calculate $\angle DCB$.

   $\angle DCB+\angle DCE+\angle BCE=360°$   ......... (VI)

Substitute $146.32°$ for $\angle DCE$ and $45°$ for $\angle BCE$ in Equation (VI).

   $\begin{array}{c}\angle DCB+146.32°+45°=360°\\ \angle DCB=360°-146.32°-45°\\ =168.68°\end{array}$

The free body diagram at point C is shown below.

  

  Figure-(2)

Apply Sine Law at point $C$.

   $\begin{array}{l}\left(\frac{3\text{kN}}{\sin \left(168.68°\right)}\right)=\left(\frac{T_{BC}}{\sin \left(146.32°\right)}\right)=\left(\frac{T_{CD}}{\sin \left(45°\right)}\right)\\ \left(\frac{3\text{kN}}{0.196}\right)=\left(\frac{T_{BC}}{0.555}\right)=\left(\frac{T_{CD}}{0.707}\right)\end{array}$   ......... (VII)

Here, the tension in cable $BC$ is $T_{\text{BC}}$ and the tension in cable $CD$ is $T_{\text{CD}}$.

Determine $T_{\text{CD}}$ and $T_{\text{BC}}$ by taking two equations at a time from Equation (VII).

   $\begin{array}{c}\left(\frac{3\text{kN}}{0.196}\right)=\left(\frac{T_{\text{CD}}}{0.707}\right)\\ \left(15.306\text{kN}\right)=\left(\frac{\left(T_{\text{CD}}\right)}{0.707}\right)\\ \left(\left(15.306\text{kN}\right)\times 0.707\right)=\left(T_{\text{CD}}\right)\\ T_{\text{CD}}=10.821\text{kN}\end{array}$

Thus the tension in segment $CD$ is equal to $10.82\text{kN}$.

   $\begin{array}{c}\left(\frac{3\text{kN}}{0.196}\right)=\left(\frac{T_{\text{BC}}}{0.555}\right)\\ \left(15.306\text{kN}\right)=\left(\frac{\left(T_{\text{BC}}\right)}{0.555}\right)\\ \left(\left(15.306\text{kN}\right)\times 0.555\right)=\left(T_{\text{BC}}\right)\\ T_{\text{BC}}=8.49\text{kN}\end{array}$

Thus the tension in segment $BC$ is equal to $8.49\text{kN}$.

Analyze point $D$.

  

  Figure-(3)

Apply Angle Sum Property in $\Delta DCF$.

   $\angle DCF\text{+}\angle CDF\text{+}\angle DFC\text{=180}°$   ......... (VIII)

Substitute $56.32°$ for $\angle DCF$ and $90°$ for $\angle DFC$ in Equation (VIII).

   $\begin{array}{l}56.32°+\angle CDF+90°=180°\\ \angle CDF=180°-90°-56.32°\\ \angle CDF=33.68°\end{array}$

Calculate $D_y$.

   $D_{\text{y}}=T_{\text{CD}}\times \cos \left(\angle \text{CDF}\right)$   ......... (XI)

Substitute $10.82\text{kN}$ for $T_{\text{CD}}$ and $33.68°$ for $\angle CDF$ in Equation (XI).

   $\begin{array}{c}{D}_{\text{y}}=\left(10.82\text{kN}\right)\times \cos \left(33.68°\right)\\ =\left(10.82\text{kN}\right)\times 0.83\\ =8.98\text{kN}\end{array}$

Calculate $D_x$.

   $D_{\text{x}}=T_{\text{CD}}\times \sin \left(\angle \text{CDF}\right)$

Substitute $10.82\text{kN}$ for $T_{\text{CD}}$ and $33.68°$ for $\angle CDF$.

   $\begin{array}{c}{D}_{\text{x}}=\left(10.82\text{kN}\right)\times \sin \left(33.68°\right)\\ =\left(10.82\text{kN}\right)\times 0.554\\ =5.99\text{kN}\end{array}$

Calculate $y_{\text{D}}$ from equation (II).

Substitute $8.98\text{kN}$ for $D_y$ and $5.99\text{kN}$ for $D_x$ in equation (II).

   $\begin{array}{l}-\left(5.99\text{kN}\right)\times \left(y_{\text{D}}\right)+\left(\text{9}\text{m}\right)\times \left(8.98\text{kN}\right)-\left(55\text{kN-m}\right)=0\\ -\left(5.99\times y_{\text{D}}\text{kN}\right)+\left(80.82\text{kN-m}\right)-\left(55\text{kN-m}\right)=0\\ -\left(5.99\times y_{\text{D}}\text{kN}\right)+\left(25.82\text{kN-m}\right)=0\\ y_{\text{D}}=4.33\text{m}\end{array}$

Consider $\text{Δ}ABG$.

  

  Figure-(4)

Calculate the length $AG$.

   $AG=\left(9\text{m}\right)-y_{\text{D}}$   ......... (XII)

Substitute $4.33\text{m}$ for $y_{\text{D}}$ in Equation (XII).

   $\begin{array}{c}AG=\left(9\text{m}\right)-\left(4.33\text{m}\right)\\ =4.67\text{m}\end{array}$

Calculate $\angle ABG$.

   $\angle ABG={\tan }^{-1}\left(\frac{AG}{BG}\right)$   ......... (XIII)

Substitute $4.67\text{m}$ for $AG$ and $2\text{m}$ for $BG$ in Equation (XIII).

   $\begin{array}{c}\angle ABG={\tan }^{-1}\left(\frac{4.67\text{m}}{2\text{m}}\right)\\ ={\tan }^{-1}\left(2.335\right)\\ =66.81°\end{array}$

Analyze point $B$.

The free body diagram for point $B$ is shown below.

  

  Figure-(5)

   $\angle ABH\text{=}\angle ABG+90°$   ......... (XIV)

Substitute $66.81°$ for $\angle ABG$ in Equation (XIV).

   $\begin{array}{c}\angle ABH=66.81°+90°\\ =156.81°\end{array}$

Calculate $\angle CBH$.

   $\angle CBH\text{=}\angle CBE\text{+}\angle EBH$   ......... (XV)

Substitute $45°$ for $\angle CBE$ and $90°$ for $\angle EBH$ in Equation (XV).

   $\begin{array}{c}\angle CBH\text{=}45°+90°\\ \text{=}135°\end{array}$

Calculate $\angle ABC$.

   $\angle ABC+\angle ABH+\angle CBH\text{=360}°$   ......... (XVI)

Substitute $156.81°$ for $\angle ABH$ and $135°$ for $\angle CBH$ in Equation (XVI).

   $\begin{array}{c}\angle ABC+156.81°+135°=360°\\ \angle ABC=360°-156.81°-135°\\ =68.19°\end{array}$

Apply Sine Law at point $B$.

   $\begin{array}{l}\left(\frac{20\text{kN}}{\sin \left(68.19°\right)}\right)=\left(\frac{T_{\text{BC}}}{\sin \left(156.81°\right)}\right)=\left(\frac{T_{\text{AB}}}{\sin \left(135°\right)}\right)\\ \left(\frac{20\text{kN}}{0.93}\right)=\left(\frac{T_{\text{BC}}}{0.39}\right)=\left(\frac{T_{\text{AB}}}{0.707}\right)\\ \left(21.51\text{kN}\right)=\left(\frac{T_{\text{BC}}}{0.39}\right)=\left(\frac{T_{\text{AB}}}{0.707}\right)\end{array}$   ......... (XVII)

Here, the tension in cable $BC$ is $T_{\text{BC}}$ and the tension in cable $AB$ is $T_{\text{AB}}$.

Determine $T_{\text{CD}}$ and $T_{\text{BC}}$ by taking two equations at a time from Equation (XVII).

   $\begin{array}{l}21.51\text{kN}=\left(\frac{T_{\text{AB}}}{0.707}\right)\\ T_{\text{AB}}=\left(21.51\text{kN}\right)\times 0.707\\ T_{\text{AB}}=15.21\text{kN}\end{array}$

Thus the tension in segment $AB$ is $15.21\text{kN}$.

Conclusion:

The tension in segment $AB$ is $15.21\text{kN}$.

The tension in segment $BC$ is $8.49\text{kN}$.

The tension in segment $CD$ is $10.82\text{kN}$.

The vertical distance between points $A$ and $D$ that is $y_{\text{D}}$ .is $4.33\text{m}$.