Chapter 11, Problem 45RE

To determine

To calculate: The tension in each supporting cables which is supported by a load of $300$ pounds.

Answer to Problem 45RE

  $159.1$ Pounds, $115.6$ Pounds, $115.6$ Pounds.

Explanation of Solution

Given information: A load of $300$ pounds is supported by three cables $OA,OB,OC$

Formula Used: By using the force vector which is given by $force=magnitude\times direction$

  $u=$ $\Vert u\Vert \frac{\overrightarrow{{\displaystyle OA}}}{\Vert \overrightarrow{OA}\Vert }$ , $v=$ $\Vert v\Vert \frac{\overrightarrow{OB}}{\Vert \overrightarrow{OA}\Vert }$ , $z=$ $\Vert z\Vert \frac{\overrightarrow{OC}}{\Vert \overrightarrow{OC}\Vert }$

Calculation: The points are:

  $A\left(0,10,10\right)$ , $B\left(-4,-6,10\right)$ , $C\left(4,-6,10\right)$

Downward force $w=\langle 0,0,-300\rangle$

Force vectors corresponding to ropes:

  $u=\Vert u\Vert \frac{O\overrightarrow{A}}{\Vert O\overrightarrow{A}\Vert }=\frac{\langle 0,10,10\rangle }{\sqrt{0^2+{10}^2+{10}^2}}=\frac{\langle 0,10,10\rangle }{10\sqrt{2}}$

So,

  $u=\Vert u\Vert \langle 0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\rangle$

And

  $\begin{array}{c}v=\Vert v\Vert \frac{\overrightarrow{OB}}{\Vert \overrightarrow{OB}\Vert }\\ =\frac{\langle -4,-6,10\rangle }{\sqrt{-4^2+{\left(-6\right)}^2+{\left(-10\right)}^2}}\\ =\frac{\langle -4,-6,10\rangle }{2\sqrt{38}}\\ v=\Vert v\Vert \langle \frac{-2}{\sqrt{38}},\frac{-3}{\sqrt{38}},\frac{5}{\sqrt{38}}\rangle \end{array}$

Further

  $\begin{array}{c}z=\Vert z\Vert \frac{\overrightarrow{OB}}{\Vert \overrightarrow{OB}\Vert }\\ =\frac{\langle 4,-6,10\rangle }{\sqrt{-4^2+{\left(-6\right)}^2+{\left(-10\right)}^2}}\\ =\frac{\langle 4,-6,10\rangle }{2\sqrt{38}}\\ z=\Vert z\Vert \langle \frac{2}{\sqrt{38}},\frac{-3}{\sqrt{38}},\frac{5}{\sqrt{38}}\rangle \end{array}$

For the system in equilibrium

  $u+v+z+w=0$ Or $u+v+z=-w$

This gives the system of equations:

  $\begin{array}{l}\frac{-2}{\sqrt{38}}\Vert v\Vert +\frac{2}{\sqrt{38}}\Vert z\Vert =0\\ \frac{1}{\sqrt{2}}\Vert u\Vert -\frac{3}{\sqrt{38}}\Vert v\Vert -\frac{3}{\sqrt{38}}\Vert z\Vert =0\\ \frac{1}{\sqrt{2}}\Vert u\Vert +\frac{5}{\sqrt{38}}\Vert v\Vert +\frac{5}{\sqrt{38}}\Vert z\Vert =300\end{array}$

By solving, $\Vert v\Vert =\Vert z\Vert$

Subtracting last two equations, equation become

  $\begin{array}{l}-\frac{8}{\sqrt{38}}\Vert v\Vert -\frac{8}{\sqrt{38}}\Vert z\Vert =-300\\ -\frac{8}{\sqrt{38}}\Vert z\Vert -\frac{8}{\sqrt{38}}\Vert z\Vert =-300\\ \frac{16}{\sqrt{38}}\Vert z\Vert =300\\ \Vert z\Vert =115.6\\ \Vert v\Vert =115.6\end{array}$

Putting above two values in any one of the equation

  $\Vert u\Vert =159.1$