Chapter 8.1, Problem 38E

To determine

Find the magnitude and direction of resultant.

Answer to Problem 38E

  $72.11Nand119.74°$

Explanation of Solution

Given information:

Three workers are pulling on ropes attached to the tree as shown,

Calculation:

Consider force vectors, $\stackrel{\rightharpoonup }{F_1}=60N,\stackrel{\rightharpoonup }{F_2}=35N,\stackrel{\rightharpoonup }{F_3}=40N$

  $Anglebetween\stackrel{\rightharpoonup }{F_2}and\stackrel{\rightharpoonup }{F_3}is60°$

The magnitude of resultant $\stackrel{\rightharpoonup }{F_1}and,\stackrel{\rightharpoonup }{F_2}$

  $\begin{array}{l}\left|\stackrel{\rightharpoonup }{F_1}+\stackrel{\rightharpoonup }{F_2}\right|=\sqrt{{\left|\stackrel{\rightharpoonup }{F_1}\right|}^2+{\left|\stackrel{\rightharpoonup }{F_2}\right|}^2}\\ \left|\stackrel{\rightharpoonup }{F_1}+\stackrel{\rightharpoonup }{F_2}\right|=\sqrt{{60}^2+{35}^2}\\ \left|\stackrel{\rightharpoonup }{F_1}+\stackrel{\rightharpoonup }{F_2}\right|=69.46N\end{array}$

The direction between resultant $\stackrel{\rightharpoonup }{F_1}and,\stackrel{\rightharpoonup }{F_2}$ and horizontal,

  $\begin{array}{l}\tan \theta =\frac{60}{35}\\ \theta ={\tan }^{-1}\left(\frac{60}{35}\right)\\ \theta =59.74°\end{array}$

The $Anglebetween\stackrel{\rightharpoonup }{F_2}and\stackrel{\rightharpoonup }{F_3}is60°$ now, $Anglebetween\stackrel{\rightharpoonup }{F_1}and\stackrel{\rightharpoonup }{F_3}is$ ,

  $59.74°+60°=119.74°$

The horizontal component of $60N$ is,

  $\begin{array}{l}=60\cos 0°\\ =60N\end{array}$

The vertical component of $60N$ is,

  $\begin{array}{l}=60\sin 0°\\ =0N\end{array}$

The horizontal component of $40N$ is,

  $\begin{array}{l}=40\cos 119.74°\\ =-19.78N\end{array}$

The vertical component of $40N$ is,

  $\begin{array}{l}=40\sin 119.74°\\ =34.73N\end{array}$

Now calculate total horizontal and vertical force,

  $60+\left(-19.78\right)=40.22and0+34.73=34.73N$

Now the magnitude of resultant $\stackrel{\rightharpoonup }{F_1}and\stackrel{\rightharpoonup }{F_3}$ ,

  $\begin{array}{l}\left|\stackrel{\rightharpoonup }{F_1}+\stackrel{\rightharpoonup }{F_3}\right|=\sqrt{{\left|\stackrel{\rightharpoonup }{F_1}\right|}^2+{\left|\stackrel{\rightharpoonup }{F_3}\right|}^2}\\ \left|\stackrel{\rightharpoonup }{F_1}+\stackrel{\rightharpoonup }{F_3}\right|=\sqrt{{60}^2+{40}^2}\\ \left|\stackrel{\rightharpoonup }{F_1}+\stackrel{\rightharpoonup }{F_3}\right|=72.11N\end{array}$

Direction of resultant force is,

  $59.74°+60°=119.74°$

Hence, the magnitude and direction of resultant force is $72.11Nand119.74°$ .