Chapter 7.4, Problem 94P

To determine

The tension in each segment of the cable and the cable’s total length.

Answer to Problem 94P

• The tension in the cable BD is $\underset{\_}{T_{BD}=78.19\text{lb}}$.
• The tension in the cable AC is $\underset{\_}{T_{AC}=74.75\text{lb}}$.
• The tension in the cable CD is $\underset{\_}{T_{CD}=43.72\text{lb}}$.
• The total length of the chain is $\underset{\_}{l=15.75\text{ft}}$.

Explanation of Solution

Given information:

• Set $P=80\text{lb}$.

Assumption:

• The self-weight of the cable is ignored.
• The cable is perfectly flexible and inextensible.
• The tensile force acting in the cable is always tangent to the cable at points along its length.
• Being inextensible, the cable has a constant length both before and after the load is applied.
• The cable or a segment of the cable can be treated as a rigid body.
• Method of joints is used to determine the force in the cable.
• Consider the state of the member as tension where the force is pulling the member as compression where the force is pushing the member.
• Consider the force indicating the right side as positive and left side as negative in horizontal components of forces.
• Consider the force indicating upward is taken as positive and downward as negative in vertical components of forces.
• Consider clockwise moment as negative and anti-clock wise moment as positive wherever applicable.
• Apply Equation of equilibrium wherever applicable.

Segment ACD:

Show the free-b0dy diagram of the segment ACD as in Figure (1).

Using Figure (1),

Determine the value of $\theta$ using the relation:

$\tan \theta =\frac{\text{Opposite}\text{side}}{\text{Adjacent}\text{side}}$ (I)

Moment about point A:

Determine the tension in the BD by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ T_{BD}\cos {\theta }_1\left(3\right)+T_{BD}\sin {\theta }_1\left(7\right)-50\left(7\right)-P\left(3\right)=0\end{array}$ (II)

Along the vertical direction:

Determine the vertical reaction at point A by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ T_{BD}\sin {\theta }_1-50-P+A_y=0\end{array}$ (III)

Along the horizontal direction:

Determine the horizontal reaction at point A by resolving the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ T_{BD}\cos {\theta }_1-A_x=0\end{array}$ (IV)

Conclusion:

Substitute 5 ft for opposite side and 3 ft for adjacent side in Equation (I).

$\begin{array}{l}\tan {\theta }_1=\frac{\text{5}}{\text{3}}\\ {\theta }_1=59.04°\end{array}$

Substitute $59.04°$ for ${\theta }_1$ and 80 lb for P in Equation (II).

$\begin{array}{l}{T}_{BD}\cos 59.04°\left(3\right)+T_{BD}\sin 59.04°\left(7\right)-350-80\left(3\right)=0\\ 1.543T_{BD}+6.003T_{BD}-350-240=0\\ 7.546T_{BD}=590\\ T_{BD}=78.19\text{lb}\end{array}$

Substitute 78.10 lb for $T_{BD}$, $59.04°$ for ${\theta }_1$, and 80 lb for P in Equation (III).

$\begin{array}{l}78.10\sin 59.04°-50-80+A_y=0\\ A_y=63.03\text{lb}\end{array}$

Substitute 78.10 lb for $T_{BD}$ and $59.04°$ for ${\theta }_1$ in Equation (IV).

$\begin{array}{l}78.10\cos 59.04°-A_x=0\\ A_x=40.18\text{lb}\end{array}$

Thus, the tension in the cable BD is $\underset{\_}{T_{BD}=78.19\text{lb}}$.

Joint A:

Show the free-body diagram of the joint A as in Figure (2).

Using Figure (2),

Along the vertical direction:

Determine the tension in the cable AC and the angle ${\theta }_2$ by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-T_{AC}\sin {\theta }_2=0\end{array}$ (V)

Along the horizontal direction:

Determine the tension in the cable AC and the angle ${\theta }_2$ by resolving the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ T_{AC}\cos {\theta }_2-A_x=0\end{array}$ (VI)

Conclusion:

Substitute 63.03 lb for $A_y$ in Equation (V).

$\begin{array}{l}63.03-T_{AC}\sin {\theta }_2=0\\ T_{AC}\sin {\theta }_2=63.03\end{array}$ (VII)

Substitute 40.18 lb for $A_x$ in Equation (VI).

$\begin{array}{l}{T}_{AC}\cos {\theta }_2-40.18=0\\ T_{AC}\cos {\theta }_2=40.18\end{array}$ (VIII)

Divide Equation (VII) by Equation (VIII).

$\begin{array}{l}\frac{T_{AC}\sin {\theta }_2}{T_{AC}\cos {\theta }_2}=\frac{63.03}{40.18}\\ \tan {\theta }_2=\frac{63.03}{40.18}\\ {\theta }_2=57.48°\end{array}$

Substitute $57.48°$ for ${\theta }_2$ in Equation (VII).

$\begin{array}{l}{T}_{AC}\sin 57.48°=63.03\\ T_{AC}=74.75\text{lb}\end{array}$

Thus, the tension in the cable AC is $\underset{\_}{T_{AC}=74.75\text{lb}}$.

Joint D:

Show the free-body diagram of the joint D as in Figure (3).

Using Figure (3),

Along the vertical direction:

Determine the tension in the cable CD and the angle ${\theta }_3$ by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ T_{BD}\sin {\theta }_1-T_{CD}\sin {\theta }_3-50=0\end{array}$ (IX)

Along the horizontal direction:

Determine the tension in the cable CD and the angle ${\theta }_3$ by resolving the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ T_{BD}\cos {\theta }_1-T_{CD}\cos {\theta }_3=0\end{array}$ (X)

Conclusion:

Substitute 78.19 lb for $T_{BD}$ and $59.04°$ for ${\theta }_1$ in Equation (IX).

$\begin{array}{l}78.19\sin 59.04°-T_{CD}\sin {\theta }_3-50=0\\ T_{CD}\sin {\theta }_3=17.05\end{array}$ (XI)

Substitute 78.19 lb for $T_{BD}$ and $59.04°$ for ${\theta }_1$ in Equation (X).

$\begin{array}{l}78.19\cos 59.04-T_{CD}\cos {\theta }_3=0\\ T_{CD}\cos {\theta }_3=40.22\end{array}$ (XII)

Divide Equation (XI) by Equation (XII).

$\begin{array}{l}\frac{T_{CD}\sin {\theta }_3}{T_{CD}\cos {\theta }_3}=\frac{17.05}{40.22}\\ \tan {\theta }_3=\frac{17.05}{40.22}\\ {\theta }_3=22.97°\end{array}$

Substitute $22.97°$ for ${\theta }_3$ in Equation (XI).

$\begin{array}{l}{T}_{CD}\sin 22.97°=17.05\\ T_{CD}=43.72\text{lb}\end{array}$

Thus, the tension in the cable CD is $\underset{\_}{T_{CD}=43.72\text{lb}}$.

Entire segment:

Show the free-body diagram of the entire segment with the length of each cable as in Figure (4).

Using Figure (4),

Determine the entire length of the cable:

$\begin{array}{c}l=\frac{5}{\sin 59.04°}+\frac{4}{\cos 22.97°}+\frac{3}{\cos 57.48°}\\ =5.83+4.34+5.58\\ =15.75\text{ft}\end{array}$

Thus, the total length of the chain is $\underset{\_}{l=15.75\text{ft}}$.