Chapter 5, Problem 62A

To determine

The net force acting on the ring.

Answer to Problem 62A

The net force acting on the ring is ${\overrightarrow{F}}_{net}=\left(-125.9\widehat{i}+627.81\widehat{j}\right)$ .

Explanation of Solution

Formula used:

The net force is given as

  ${\overrightarrow{F}}_{net}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2$

Calculation:

  $\begin{array}{l}{\overrightarrow{F}}_{net}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2\\ {\overrightarrow{F}}_{net}=\left(F_1\cos 50\widehat{i}+F_{1}sin50\widehat{j}\right)+\left(F_2\cos 40\widehat{i}+F_{2}sin40\widehat{j}\right)\\ {\overrightarrow{F}}_{net}=\left(400\cos 50\widehat{i}+400sin50\widehat{j}\right)+\left(500\cos 40\widehat{i}+500sin40\widehat{j}\right)\\ {\overrightarrow{F}}_{net}=\left(-125.9\widehat{i}+627.81\widehat{j}\right)\end{array}$

The angle is

  $\begin{array}{c}\theta =180°-{\tan }^{-1}\left(\frac{125.9}{627.83}\right)\\ =180°-{\tan }^{-1}(0.20054)\\ =180°-11.34°\\ =168.66°\end{array}$

Conclusion:

The net force acting on the ring is ${\overrightarrow{F}}_{net}=\left(-125.9\widehat{i}+627.81\widehat{j}\right)$ .