Chapter 7.4, Problem 94P

The cable supports the three loads shown. Determine the sags y_B and y_D of B and D. Take P₁ = 800 N, P₂ = 500 N.

Probs. 7–94/95

To determine

The sags $y_B$ and $y_D$ of points B and D.

Answer to Problem 94P

The sag $y_B$ at point B of the cable is $\underset{\_}{y_B=2.22\text{m}}$.

The sag $y_D$ at point D of the cable is $\underset{\_}{y_D=1.55\text{m}}$.

Explanation of Solution

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 member as tension where the force is pulling the member and as compression where the force is pushing the member.
• Consider the force indicating right side as positive and left side as negative in the horizontal components of forces.
• Consider the force indicating upward is taken as positive and downward as negative in the vertical components of forces.
• Consider clockwise moment as negative and anticlockwise moment as positive wherever applicable.
• Apply the Equation of equilibrium wherever applicable.

Given information:

• Take the vertical external loads $P_1=800\text{N}$ and $P_2=500\text{N}$.

Segment BCDE:

Show the free body diagram of segment BCDE as in Figure (1).

Using Figure (1),

Determine the value of $\sin \theta$:

$\sin \theta =\frac{\text{Opposite}\text{side}}{\text{Hypotenuse}}$ (I)

Determine the value of $\cos \theta$:

$\cos \theta =\frac{\text{Adjacent}\text{side}}{\text{Hypotenuse}}$ (II)

Moment about the point E:

Determine the force in cable segment AB and the sag $y_B$ by taking moment about point E.

$\begin{array}{l}{\displaystyle \sum M_E=0}\\ P_2\left(3\right)+P_1\left(9\right)+P_2\left(15\right)-F_{AB}\cos {\theta }_1\left(y_B+1\right)-F_{AB}\sin {\theta }_1\left(15\right)=0\end{array}$ (III)

Conclusion:

Substitute $y_B$ for the opposite side and $\sqrt{y_B^2+9}$ for hypotenuse in Equation (I).

$\sin {\theta }_1=\frac{y_B}{\sqrt{y_B^2+9}}$ (IV)

Substitute 3 m for the adjacent side and $\sqrt{y_B^2+9}$ for hypotenuse in Equation (II).

$\cos {\theta }_1=\frac{3}{\sqrt{y_B^2+9}}$ (V)

Substitute 500 N for $P_2$, 800 N for $P_1$, $\frac{y_B}{\sqrt{y_B^2+9}}$ for $\sin {\theta }_1$, and $\frac{3}{\sqrt{y_B^2+9}}$ for $\cos {\theta }_1$ Equation (III).

$\begin{array}{l}\left[\begin{array}{l}500\left(3\right)+800\left(9\right)+500\left(15\right)-F_{AB}\left(\frac{3}{\sqrt{y_B^2+9}}\right)\left(y_B+1\right)\\ -F_{AB}\left(\frac{y_B}{\sqrt{y_B^2+9}}\right)\left(15\right)\end{array}\right]=0\\ 1,500+7,200+7,500-F_{AB}\left(\frac{3y_B+3}{\sqrt{y_B^2+9}}\right)-F_{AB}\left(\frac{15y_B}{\sqrt{y_B^2+9}}\right)=0\\ 16,200-F_{AB}\left(\frac{18y_B+3}{\sqrt{y_B^2+9}}\right)=0\\ F_{AB}\left(\frac{18y_B+3}{\sqrt{y_B^2+9}}\right)=16,200\end{array}$ (VI)

Segment BC:

Show the free body diagram of segment BC as in Figure (2).

Using Figure (2),

Moment about the point C:

Determine the force in cable segment AB and the sag $y_B$ by taking moment about point C.

$\begin{array}{l}{\displaystyle \sum M_C=0}\\ P_2\left(6\right)+F_{AB}\cos {\theta }_1\left(4-y_B\right)-F_{AB}\sin {\theta }_1\left(6\right)=0\end{array}$ (VII)

Conclusion:

Substitute 500 N for $P_2$, $\frac{y_B}{\sqrt{y_B^2+9}}$ for $\sin {\theta }_1$, and $\frac{3}{\sqrt{y_B^2+9}}$ for $\cos {\theta }_1$ Equation (VII).

$\begin{array}{l}500\left(6\right)+F_{AB}\left(\frac{3}{\sqrt{y_B^2+9}}\right)\left(4-y_B\right)-F_{AB}\left(\frac{y_B}{\sqrt{y_B^2+9}}\right)\left(6\right)=0\\ 3,000+F_{AB}\left(\frac{12-3y_B}{\sqrt{y_B^2+9}}\right)-F_{AB}\left(\frac{6y_B}{\sqrt{y_B^2+9}}\right)=0\\ 3,000-F_{AB}\left(\frac{9y_B-12}{\sqrt{y_B^2+9}}\right)=0\\ F_{AB}\left(\frac{9y_B-12}{\sqrt{y_B^2+9}}\right)=3,000\end{array}$ (VIII)

Divide Equation (VI) with Equation (VIII).

$\begin{array}{l}\frac{F_{AB}\left(\frac{18y_B+3}{\sqrt{y_B^2+9}}\right)}{F_{AB}\left(\frac{9y_B-12}{\sqrt{y_B^2+9}}\right)}=\frac{16,200}{3,000}\\ \frac{18y_B+3}{9y_B-12}=\frac{16,200}{3,000}\end{array}$

$\begin{array}{l}54,000y_B+9,000=145,800y_B-194,400\\ 91,800y_B=203,400\\ y_B=2.22\text{m}\end{array}$

Substitute 2.22 m for $y_B$ in Equation (VIII).

$\begin{array}{l}{F}_{AB}\left(\frac{9\left(2.22\right)-12}{\sqrt{{2.22}^2+9}}\right)=3,000\\ F_{AB}\left(\frac{7.98}{3.732}\right)=3,000\\ F_{AB}=1,403.04\text{N}\end{array}$

Thus, the sag $y_B$ at point B of the cable is $\underset{\_}{y_B=2.22\text{m}}$.

Joint B:

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

Using Figure (3),

Determine the value of $\theta$:

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

Along the vertical direction:

Determine the force in cable segment BC by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{AB}\sin {\theta }_1-P_2-F_{BC}\sin {\theta }_2=0\end{array}$ (X)

Conclusion:

Substitute 2.22 m for the opposite side and $\sqrt{{\left(2.22\right)}^2+9}$ for hypotenuse in Equation (I).

$\begin{array}{l}\sin {\theta }_1=\frac{2.22}{\sqrt{{\left(2.22\right)}^2+9}}\\ \sin {\theta }_1=0.5948\\ {\theta }_1=36.5°\end{array}$

Substitute 1.78 m for the opposite side and 6 m for adjacent side in Equation (IX).

$\begin{array}{l}\tan {\theta }_2=\frac{\text{1}\text{.78}}{\text{6}}\\ {\theta }_2=16.52°\end{array}$

Substitute 1,403.04 N for $F_{AB}$, $36.50°$ for ${\theta }_1$, 500 N for $P_2$, and $16.52°$ for ${\theta }_2$ in Equation (X).

$\begin{array}{l}1,403.04\sin 36.50°-500-F_{BC}\sin 16.52°=0\\ 834.56-500-0.284F_{BC}=0\\ 0.284F_{BC}=334.56\\ F_{BC}=1,178.03\text{N}\end{array}$

Joint C:

Show the free-body diagram of joint C as in Figure (4).

Using Figure (4),

Along the vertical direction:

Determine sag $y_D$ at point D by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_{BC}\sin {\theta }_2+F_{CD}\sin {\theta }_3-P_1=0\end{array}$ (XI)

Along the horizontal direction:

Determine sag $y_D$ at point D by resolving the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -F_{BC}\cos {\theta }_2+F_{CD}\cos {\theta }_3=0\end{array}$ (XII)

Conclusion:

Substitute $\left(4-y_D\right)$ for the opposite side and $\sqrt{36+{\left(4-y_D\right)}^2}$ for hypotenuse in Equation (I).

$\sin {\theta }_3=\frac{\left(4-y_D\right)}{\sqrt{36+{\left(4-y_D\right)}^2}}$

Substitute 6 m for the adjacent side and $\sqrt{36+{\left(4-y_D\right)}^2}$ for hypotenuse in Equation (II).

$\cos {\theta }_3=\frac{6}{\sqrt{36+{\left(4-y_D\right)}^2}}$

Substitute 1,178.03 N for $F_{BC}$, $16.52°$ for ${\theta }_2$, $\frac{\left(4-y_D\right)}{\sqrt{36+{\left(4-y_D\right)}^2}}$ for $\sin {\theta }_3$, and 800 N for $P_1$ in Equation (XI).

$\begin{array}{l}1,178.03\sin 16.52°+F_{CD}\left(\frac{\left(4-y_D\right)}{\sqrt{36+{\left(4-y_D\right)}^2}}\right)-800=0\\ F_{CD}\left(\frac{\left(4-y_D\right)}{\sqrt{36+{\left(4-y_D\right)}^2}}\right)=465.04\end{array}$ (XIII)

Substitute 1,178.03 N for $F_{BC}$, $16.52°$ for ${\theta }_2$, and $\frac{6}{\sqrt{36+{\left(4-y_D\right)}^2}}$ for $\cos {\theta }_3$ in Equation (XII).

$\begin{array}{l}-1,178.03\cos 16.52°+F_{CD}\left(\frac{6}{\sqrt{36+{\left(4-y_D\right)}^2}}\right)=0\\ F_{CD}\left(\frac{6}{\sqrt{36+{\left(4-y_D\right)}^2}}\right)=1,129.40\end{array}$ (XIV)

Conclusion:

Divide Equation (XIII) with Equation (XIV).

$\begin{array}{l}\frac{F_{CD}\left(\frac{\left(4-y_D\right)}{\sqrt{36+{\left(4-y_D\right)}^2}}\right)}{F_{CD}\left(\frac{6}{\sqrt{36+{\left(4-y_D\right)}^2}}\right)}=\frac{465.04}{1,129.40}\\ \frac{\left(4-y_D\right)}{6}=\frac{465.04}{1,129.40}\end{array}$

$\begin{array}{l}4,517.6-1,129.40y_D=2,790.24\\ 1,129.40y_D=1,727.36\\ y_D=1.55\text{m}\end{array}$

Thus, sag $y_D$ at point D of the cable is $\underset{\_}{y_D=1.55\text{m}}$.