Q. Three forces $F_{1},\ F_{2}$ and $F_{3}$ together keep a body in equilibrium. If $F_{1} = 3\ N$ along the positive $x$-axis, $F_{2} = 4N$ along the positive $y$-axis, then the third force $F_{3}$ is
A) $5\ N$ making an angle $\theta = \tan^{- 1}\left( \frac{3}{4} \right)$ with negative $y$-axis
B) $5\ N$ making an angle $\theta = \tan^{- 1}\left( \frac{4}{3} \right)$ with negative $y$ axis
C) $7\ N$ making an angle $\theta = \tan^{- 1}\left( \frac{3}{4} \right)$ with negative $y$ axis
D) $7\ N$ making an angle $\theta = \tan^{- 1}\left( \frac{4}{3} \right)$ with negative $y$ axis
Explanation
The correct answer is Option A: $5\ N$ making an angle $\theta = \tan^{- 1}\left( \frac{3}{4} \right)$ with negative $y$-axis.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.