Chapter 6, Problem 6.83P

The cable carrying three 400-lb loads has a sag at C of $h_C=14\text{ft}\text{.}$ Calculate the force in each segment of the cable.

To determine

The force in each segment of the cable.

Answer to Problem 6.83P

  $F_{AB}=832.34lb$, $F_{BC}=555.7lb$, $F_{CD}=494.81lb$ and $F_{DE}=540.844lb$.

Explanation of Solution

Given Information:

Load for each segment are $400lb$.

Concept used:

Moment of force in different points.

Calculation:

  

For support reaction moment about point $E$ ,

  ${\displaystyle \sum M_E=0}$

Reaction force along $y$ -direction at $A$ ,

  $R_A\times 40-400\times 32-400\times 24-400\times 12=0$

Therefore,

  $R_A=680lb$ in upward direction.

Reaction force along $y$ -direction at point $E$ ,

  $R_E\times 40-400\times 28-400\times 16-400\times 8=0$

Therefore,

  $R_E=520lb$ in upward direction.

Moments about point $C$ ,

  ${\displaystyle \sum M_C=0}$

Reaction force along $x$ -direction at point $A$ ,

  $\begin{array}{c}-H_A\times 16+R_A\times 16-400\times 8=0\\ 16H_A=-400\times 8+16\times 680\end{array}$

Therefore,

  $H_A=480lb$ acts towards left

Similarly, resultant force along $x$ -direction at point $E$ ,

  $H_E=480lb$ acts towards right

Now, let us consider,

  ${\displaystyle \sum M_B=0}$

Therefore,

  $-H_A\times h_B+R_A\times 8=0$

Substituting the values and solving we get,

  $\begin{array}{c}-480\times h_B+680\times 8=0\\ h_B=\frac{680\times 8}{480}\\ h_B=11.33ft\end{array}$

Consider joint $A$ :

  

From the above figure,

  $\begin{array}{c}\tan \theta =\frac{11.33}{8}\\ \theta ={\tan }^{-1}\left(\frac{11.33}{8}\right)\\ \theta ={54}^{\text{o}}{46}^{/}\end{array}$

Now,

  ${\displaystyle \sum H=0}$

Therefore,

  $\begin{array}{c}{H}_A=F_{AB}\times \cos {54}^{\text{o}}{46}^{/}\\ F_{AB}=\frac{480}{\cos {54}^{\text{o}}{46}^{/}}\\ =832.34lb\end{array}$

Consider joint $B$ :

  

From the above figure,

  $\begin{array}{c}\tan {\theta }_1=\frac{\left(16-11.33\right)}{8}\\ {\theta }_1={30}^{\text{o}}{15}^{/}\end{array}$

Now,

  ${\displaystyle \sum H=0}$

  $F_{AB}\times \cos {54}^{\text{o}}{46}^{/}=F_{BC}\times \cos {50}^{\text{o}}{15}^{/}$

Solving we get,

  $F_{BC}=555.7lb$

Now, considering moment about point $D$.

  ${\displaystyle \sum M_D=0}$

Therefore, reaction force,

  $\begin{array}{c}{H}_E\times h_D-R_E\times 12=0\\ h_D=\frac{R_E\times 12}{H_E}\\ =\frac{520\times 12}{480}\\ =13ft\end{array}$

Considering joint $C$ :

  

From the above figure,

  $\begin{array}{c}\tan {\theta }_2=\frac{\left(16-13\right)}{12}\\ {\theta }_2={14}^{\text{o}}{2}^{/}\end{array}$

Now,

  $\begin{array}{c}{\displaystyle \sum H=0}\\ F_{BC}\times \cos {\theta }_1=F_{CD}\times \cos {\theta }_2\\ 555.71\times \cos {30}^{\text{o}}{15}^{/}=F_{CD}\times \cos {14}^{\text{o}}{2}^{/}\end{array}$

Solving we get,

  $F_{CD}=494.81lb$

Consider joint $D$ :

  ${\displaystyle \sum M_E}=0$

Therefore,

  $\begin{array}{c}-H_g\times 24+R_E\times 24-400\times 12=0\\ -12H_g=400\times 12-520\times 24\\ H_g=\frac{400\times 12-520\times 24}{-12}\\ =640\end{array}$

Now, considering moment about point $E$ ,

Therefore, Reaction force,

  $\begin{array}{c}{H}_g\times h_E-R_E\times 12=0\\ h_E=\frac{12R_E}{H_g}\\ =\frac{12\times 520}{640}\\ =9.75\text{ft}\end{array}$

  

From the above figure,

  $\begin{array}{c}\tan {\theta }_3=\frac{16-9.75}{12}\\ {\theta }_3={\tan }^{-1}\left(\frac{16-9.75}{12}\right)\\ ={27.51}^{\text{o}}\end{array}$

Now,

  ${\displaystyle \sum H=0}$

Therefore,

  $\begin{array}{c}{F}_{CD}\times \cos {\theta }_2=F_{DE}\times \cos {\theta }_3\\ 494.81\times \cos {14.20}^{\text{o}}=F_{DE}\times \cos {27.51}^{\text{o}}\\ F_{DE}=\frac{494.81\times \cos {14.20}^{\text{o}}}{\cos {27.51}^{\text{o}}}\\ F_{DE}=540.844\text{lb}\end{array}$

Conclusion:

The force in each segment of the cable are $F_{AB}=832.34lb$, $F_{BC}=555.7lb$, $F_{CD}=494.81lb$ and $F_{DE}=540.844lb$.