Chapter 10, Problem 41RE

To determine

To find: The tension in each of the supporting cables handling a load of 300 pounds.

Answer to Problem 41RE

The tension ineach of the supporting cables are $649.75N$ , $468.75N$ and $468.75N$ respectively.

Explanation of Solution

Given information:

A load of $300$ pounds is supported by three cables, as shown in the figure below.

  

Figure(1)

Calculation:

Three upward forces $OA,OB$ and $OC$ are acting on the load. The forces are in the form vector.

The vector $OA$ is,

  $\overrightarrow{OA}=\langle 0,10,10\rangle$

The vector $OB$ is,

  $\overrightarrow{OB}=\langle -4,-6,10\rangle$

The vector $OC$ is,

  $\overrightarrow{OC}=\langle 4,-6,10\rangle$

One downward force is also acting on the load that is the weight 300 pounds. Let $w$ be the force and its vector form is:

  $w=\langle 0,0,-300\rangle$

Theunit vector in direction of $OA$ is $\frac{\overrightarrow{OA}}{\Vert \overrightarrow{OA}\Vert }$ . Let $a$ be the force in direction of $OA$ can be calculated as:

  $\begin{array}{c}a=\Vert a\Vert \frac{\overrightarrow{OA}}{\Vert \overrightarrow{OA}\Vert }\\ =\Vert a\Vert \frac{\langle 0,10,10\rangle }{\sqrt{0+}}\\ =\Vert a\Vert \frac{\langle 0,10,10\rangle }{10\sqrt{2}}\\ =\Vert a\Vert \langle 0,0.707,0.707\rangle \end{array}$

The unit vector in direction of $OB$ is $\frac{\overrightarrow{OB}}{\Vert \overrightarrow{OB}\Vert }$ . Let $b$ be the force in direction of $OB$ can be calculated as:

  $\begin{array}{c}b=\Vert b\Vert \frac{\overrightarrow{OB}}{\Vert \overrightarrow{OB}\Vert }\\ =\Vert b\Vert \frac{\langle -4,-6,10\rangle }{\sqrt{16+36+100}}\\ =\Vert b\Vert \frac{\langle -4,-6,10\rangle }{12.33}\\ =\Vert b\Vert \langle -0.32,-0.49,0.81\rangle \end{array}$

The unit vector in direction of $OC$ is $\frac{\overrightarrow{OC}}{\Vert \overrightarrow{OC}\Vert }$ . Let $c$ be the force in direction of $OC$ can be calculated as:

  $\begin{array}{c}c=\Vert c\Vert \frac{\overrightarrow{OC}}{\Vert \overrightarrow{OC}\Vert }\\ ==\Vert c\Vert \frac{\langle 4,-6,10\rangle }{\sqrt{16+36+100}}\\ =\Vert c\Vert \frac{\langle 4,-6,10\rangle }{12.33}\\ =\Vert c\Vert \langle 0.32,-0.49,0.81\rangle \end{array}$

At equilibrium, the resultant action on load is zero.Therefore

  $a+b+c-w=0$

The equations formedare:

  $\begin{array}{c}0.707\Vert a\Vert -0.81\Vert b\Vert =0\\ 0.707\Vert a\Vert -0.81\Vert b\Vert -0.81\Vert c\Vert =0\\ 0.707\Vert a\Vert -0.81\Vert b\Vert -0.81\Vert c\Vert =-300\end{array}$

Solve the above equations.

  $\begin{array}{l}\Vert b\Vert =\Vert c\Vert =468.75N\\ \Vert a\Vert =649.75N\end{array}$

Therefore,the tension ineach of the supporting cables are $649.75N$ , $468.75N$ and $468.75N$ respectively.