Chapter 6.3, Problem 81E

To determine

To find:The direction and magnitude of the resultant of these forces.

Answer to Problem 81E

Thedirection and the magnitude of resultant vector are $71.33°\text{and 228}\text{.49}\text{pounds}$ , respectively.

Explanation of Solution

Given information:

Three forces with magnitudes of 75 pounds, 100 pounds, and 125 pounds act on an object at angles of $30°,45°\text{and 120}°$ , respectively, with the positive x -axis.

Calculations:

From the given information, the force with magnitude of 75 pounds is

  $\begin{array}{l}{F}_1=75\langle \cos 30°,\sin 30°\rangle \\ =75\langle \frac{\sqrt{3}}{2},\frac{1}{2}\rangle \\ =\langle 75\times \frac{\sqrt{3}}{2},75\times \frac{1}{2}\rangle \\ =\langle 37.5\sqrt{3},37.5\rangle \end{array}$

And the force with magnitude of 100 pounds is

  $\begin{array}{l}{F}_2=100\langle \cos (45°),\sin (45°)\rangle \\ =100\langle \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\rangle \\ =\langle 100\times \frac{\sqrt{2}}{2},100\times \frac{\sqrt{2}}{2}\rangle \\ =\langle 50\sqrt{2},50\sqrt{2}\rangle \end{array}$

And the force with magnitude of 125 pounds is

  $\begin{array}{l}{F}_3=125\langle \cos (120°),\sin (120°)\rangle \\ =125\langle -\frac{1}{2},\frac{\sqrt{3}}{2}\rangle \\ =\langle 125\times -\frac{1}{2},125\times \frac{\sqrt{3}}{2}\rangle \\ =\langle -62.5,62.5\sqrt{3}\rangle \end{array}$

Thus, the resultant force is

  $F_1+F_2+F_3=\langle 37.5\sqrt{3}+50\sqrt{2}-62.5,37.5+50\sqrt{2}+62.5\sqrt{3}\rangle$ .

The direction of the resultant force is

  $\theta ={\tan }^{-1}\left(\frac{37.5+50\sqrt{2}+62.5\sqrt{3}}{37.5\sqrt{3}+50\sqrt{2}-62.5}\right)=71.33°$ .

The magnitude of the resultant vector is

  $\begin{array}{l}\Vert F_1+F_2+F_3\Vert =\sqrt{{\left(37.5\sqrt{3}+50\sqrt{2}-62.5\right)}^2+{\left(37.5+50\sqrt{2}+62.5\sqrt{3}\right)}^2}\\ =228.49\end{array}$ .

Conclusion:

Therefore, the direction and the magnitude of resultant vector are $71.33°\text{and 228}\text{.49}\text{pounds}$ , respectively.