Chapter 6, Problem 16CT

To determine

The direction andmagnitude of the resultant forcesif forces with magnitudes of 250 pounds and 130 pounds act on an object at anglesof ${45}^0$ and $-{60}^0$ respectively, with the positive -axis.

Answer to Problem 16CT

The magnitude of resultant force is $250.16lb$ and angle with x-axis is ${14.87}^0$

Explanation of Solution

Given info:

  $\begin{array}{l}\Vert F_1\Vert =250lb\\ {\theta }_1={45}^0\\ {\Vert F\Vert }_2=130lb\\ {\theta }_2=-{60}^0\end{array}$

Formula used:

  $\overrightarrow{F}=\Vert F\Vert \cos \theta \widehat{i}+\Vert F\Vert \sin \theta \widehat{j}$

Calculation:

We have

  $\begin{array}{l}\overrightarrow{F}=\Vert F\Vert \cos \theta \widehat{i}+\Vert F\Vert \sin \theta \widehat{j}\\ \Vert F_1\Vert =250lb\\ {\theta }_1={45}^0\\ {\Vert F\Vert }_2=130lb\\ {\theta }_2=-{60}^0\\ {\overrightarrow{F}}_1=250\times \cos {45}^0\widehat{i}+250\sin {45}^0\widehat{j}=176.78\widehat{i}+176.78\widehat{j}\\ {\overrightarrow{F}}_2=130\times \cos \left(-{60}^0\right)\widehat{i}+130\sin \left(-{60}^0\right)\widehat{j}=65\widehat{i}-112.58\widehat{j}\\ {\overrightarrow{F}}_1+{\overrightarrow{F}}_2=241.78\widehat{i}+64.2\widehat{j}\\ \Vert {\overrightarrow{F}}_1+{\overrightarrow{F}}_2\Vert =\sqrt{{241.78}^2+{64.2}^2}=250.16lb\\ \theta ={\tan }^{-1}\left(\frac{64.2}{241.78}\right)={14.87}^0\end{array}$

The magnitude of resultant force is $250.16lb$ and angle with x-axis is ${14.87}^0$

Conclusion:

Thus,themagnitude of resultant force is $250.16lb$ and angle with x-axis is ${14.87}^0$