Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.6 - Vectors - Exercise Set - Page 783: 73


The resultant force is $2038\text{ kg}$ and the angle is $162.8{}^\circ $.

Work Step by Step

Magnitude of the first force is $1610\text{ kg}$. The direction of the first forceis $\text{N35 }\!\!{}^\circ\!\!\text{ E}$. Magnitude of the second forceis $\text{1250 kg}$. The direction of the second forceis $\text{S55 }\!\!{}^\circ\!\!\text{ W}$. The vector component along the x-axis or horizontal is i and the vector component along the y-axis or the vertical direction is j. Let ${{F}_{1}}$ and ${{F}_{2}}$ denote the first and second force acting on the object, respectively. Therefore, the equation becomes $\begin{align} & {{F}_{1}}=1610\cos {{125}^{0}}\mathbf{i}+1610\sin {{125}^{0}}\mathbf{j} \\ & =-923.46\mathbf{i}+1318.83\mathbf{j} \\ & {{F}_{2}}=1250\cos {{215}^{0}}\mathbf{i}+1250\sin {{215}^{0}}\mathbf{j} \\ & =-1023.94\mathbf{i}-716.97\mathbf{j} \end{align}$ And, $\begin{align} & F=\left( -923.46-1023.94 \right)\mathbf{i}+\left( 1318.83-716.97 \right)\mathbf{j} \\ & =-1947.40\mathbf{i}+601.86\mathbf{j} \end{align}$ So, the magnitude of force vector is as shown below $\begin{align} & \left| \left| F \right| \right|=\sqrt{{{\left( -1947.40 \right)}^{2}}+{{\left( 601.86 \right)}^{2}}} \\ & \approx 2038.28 \end{align}$ Also, the angle is $\cos \theta =\frac{-1947.40}{2038.28}$ Therefore, $\theta =162.8{}^\circ $
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.