Precalculus (6th Edition) Blitzer

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

Chapter 6 - Test - Page 800: 21


The magnitude is $\text{323 pounds}$ and the direction of the resultant force is $\text{E3}\text{.4 }\!\!{}^\circ\!\!\text{ N}$.

Work Step by Step

Here, ${{\text{F}}_{\text{1}}}=\text{250 pounds, N60 }\!\!{}^\circ\!\!\text{ E and }{{\text{F}}_{\text{2}}}=\text{150 pounds, S45 }\!\!{}^\circ\!\!\text{ E}$ The standard equation used to denote the position of force $ F\left( r,\theta \right)$ is: $ F\left( r,\theta \right)=F\text{cos}\theta \mathbf{i}+\text{Fsin}\theta \mathbf{j}$ …… (1) ${{\text{F}}_{\text{1}}}$ makes a $90{}^\circ -60{}^\circ =30{}^\circ $ angle with $\text{F}$. Then, ${{\text{F}}_{\text{1}}}$ is denoted in equation (1) as ${{\text{F}}_{\text{1}}}=\text{250 pounds}$ $\begin{align} & {{\text{F}}_{1}}\left( 250,30{}^\circ \right)=250\text{cos}30\text{ }\!\!{}^\circ\!\!\text{ }\mathbf{i}+\text{250sin30}{}^\circ \mathbf{j} \\ & \text{=250}\left( \frac{\sqrt{3}}{2} \right)\mathbf{i}+250\left( \frac{1}{2} \right)\mathbf{j} \\ & \text{=216}\text{.5}\mathbf{i}+\text{125}\mathbf{j} \end{align}$ Now, ${{\text{F}}_{2}}$ makes a $360{}^\circ -45{}^\circ =315{}^\circ $ angle with $\text{F}$. Then ${{\text{F}}_{2}}$ is denoted in equation (1) as ${{\text{F}}_{2}}\text{=150 pounds}$ $\begin{align} & {{\text{F}}_{2}}\left( 150,315{}^\circ \right)=150\text{cos315 }\!\!{}^\circ\!\!\text{ }\mathbf{i}+\text{150sin315}{}^\circ \mathbf{j} \\ & \text{=150}\left( \frac{1}{\sqrt{2}} \right)\mathbf{i}+250\left( -\frac{1}{\sqrt{2}} \right)\mathbf{j} \\ & \text{=106}\mathbf{i}-\text{106}\mathbf{j} \end{align}$ Since the resultant force $ F\left( r,\theta \right)$ is given by, $\begin{align} & F\left( r,\theta \right)={{\text{F}}_{1}}\left( r,\theta \right)+{{\text{F}}_{2}}\left( r,\theta \right) \\ & \text{=}\left( \text{216}\text{.5}\mathbf{i}+\text{125}\mathbf{j} \right)+\left( \text{106}\mathbf{i}-\text{106}\mathbf{j} \right) \\ & \text{=322}\text{.5}\mathbf{i}+\text{19}\mathbf{j} \end{align}$ Then, $\begin{align} & \text{F}\left( r,\theta \right)=\sqrt{{{322.5}^{2}}+{{19}^{2}}} \\ & \approx \text{323}\,\text{pounds} \end{align}$ And, the angle is represented as below: $\begin{align} & \cos \theta =\frac{322.5}{323} \\ & \theta ={{\cos }^{-1}}\left( \frac{322.5}{323} \right) \\ & =3.4{}^\circ \end{align}$
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