Chapter 9, Problem 26CT

To determine

The minimum tension each cable must be able to endure, where a $1200$ pound chandelier is to be suspended over a large ballroom; the chandelier will be hung on two cables of equal length whose ends will be attached to the ceiling, $16$ feet apart. The chandelier will be free-hanging so that the ends of the cable will make equal angles with the ceiling. The top of the chandelier is to be $16$ feet from the ceiling.

Answer to Problem 26CT

Solution:

The cable must be able to endure a tension of approximately $670.82$ lbs.

Explanation of Solution

Given information:

A $1200$ pound chandelier is to be suspended over a large ballroom; the chandelier will be hung on two cables of equal length whose ends will be attached to the ceiling, $16$ feet apart. The chandelier will be free-hanging so that the ends of the cable will make equal angles with the ceiling. The top of the chandelier is to be $16$ feet from the ceiling.

Draw a force diagram by using the force vectors $\text{F1}$ and $\text{F2}$ .

The tensions in the cable are the magnitudes $\Vert \text{F1}\Vert$ and $\Vert \text{F2}\Vert$ of the force vectors $\text{F1}$ and $\text{F2}$ .

The magnitude of the force vector $\text{F3}$ equals $1200$ pounds, the weight of the box.

Here $\alpha$ is the angle between the vector $\text{F}$ and the positive $x-$ axis.

  

Using the right angle triangle in the sketch it is observed that,

  $\tan \alpha =\frac{16}{8}$

  $\Rightarrow \tan \alpha =2$

  $\Rightarrow \alpha ={\tan }^{-1}2$

  $\Rightarrow \alpha \approx {63.43494882}^{\text{o}}$

  $\Rightarrow \alpha \approx {63.435}^{\text{o}}$

Now write each force vector in terms of the unit vectors $i$ and $j$ , using $\text{F}=\Vert \text{F}\Vert \left(\cos \alpha i+\sin \alpha j\right)$ .

Here, $180-\alpha$ is the angle between the vector $\text{F1}$ and the positive $x-$ axis.

Hence, the force vectors $\text{F1}$ can be written as

  $\text{F1}=\Vert \text{F1}\Vert \left(\cos {116.565}^{\text{o}}i+\sin {116.565}^{\text{o}}j\right)$

  $\Rightarrow \text{F1}=\Vert \text{F1}\Vert \left(-0.44721i+0.89443j\right)$

Here, $\alpha$ is the angle between the vector $\text{F2}$ and the positive $x-$ axis.

Hence, the force vectors $\text{F2}$ can be written as

  $\text{F2}=\Vert \text{F2}\Vert \left(\cos {63.435}^{\text{o}}i+\sin {63.435}^{\text{o}}j\right)$

  $\Rightarrow \text{F2}=\Vert \text{F2}\Vert \left(0.44721i+0.89443j\right)$

The angle between vector $\text{F3}$ and positive $x-$ axis is ${270}^{\text{o}}$ .

Hence, the force vectors $\text{F3}$ can be written as

  $\text{F3}=\text{1200}\left(\cos {270}^{\text{o}}i+\sin {270}^{\text{o}}j\right)$

  $\Rightarrow \text{F3}=\text{1200}\left(\left(0\right)i+\left(-1\right)j\right)$

  $\text{F3}=-1200j$

Since the system is in equilibrium,

  $\Rightarrow \text{F1}+\text{F2}+\text{F3}\text{=}\text{0i}\text{+}\text{0j}$

  $\Rightarrow \Vert \text{F1}\Vert \left(-0.44721i+0.89443j\right)+\Vert \text{F2}\Vert \left(0.44721i+0.89443j\right)-\text{1200}j\text{=}0i+0j$

  $\Rightarrow \left(\Vert \text{F1}\Vert -0.44721+\Vert \text{F2}\Vert 0.44721\right)i+\left(\Vert \text{F1}\Vert 0.89443+\Vert \text{F2}\Vert 0.89443-1200\right)j\text{=}0i+0j$

By comparing components,

  $\Rightarrow \Vert \text{F1}\Vert -0.44721+\Vert \text{F2}\Vert 0.44721=0$ and $\Vert \text{F1}\Vert 0.89443+\Vert \text{F2}\Vert 0.89443-1200=0$

By solving $\Vert \text{F1}\Vert -0.44721+\Vert \text{F2}\Vert 0.44721=0$ implies that $\Vert \text{F1}\Vert 0.44721=\Vert \text{F2}\Vert 0.44721$ that is $\Vert \text{F1}\Vert =\Vert \text{F2}\Vert$ .

Consider $\Vert \text{F1}\Vert$ and $\Vert \text{F2}\Vert$ has common value $\Vert \text{F}\Vert$

Substitute $\Vert \text{F1}\Vert =\Vert \text{F2}\Vert =\Vert \text{F}\Vert$ in $\Vert \text{F1}\Vert 0.89443+\Vert \text{F2}\Vert 0.89443-1200=0$ ,

  $\Rightarrow \Vert \text{F}\Vert 0.89443+\Vert \text{F}\Vert 0.89443-1200=0$

  $\Rightarrow 2\Vert \text{F}\Vert 0.89443=1200$

  $\Rightarrow \Vert \text{F}\Vert 0.89443=600$

  $\Rightarrow \Vert \text{F}\Vert =\frac{600}{0.89443}$

  $\Rightarrow \Vert \text{F}\Vert =670.8182865$

  $\Rightarrow \Vert \text{F}\Vert \approx 670.82$

Therefore, the cable must be able to endure a tension of approximately $670.82$ lbs.