Question

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, $F_{2}=4\, N$ 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 the negative y-axis
• B. $7\, N$ making an angle $\theta=\tan ^{-1}\left(\frac{3}{4}\right)$ with the negative y-axis
• C. $5\,N $ making an angle $ \theta \tan^{-1}\left( \frac{4}{3} \right) $ with the negative y-axis
• D. $7\,N $ making an angle $ \theta =\tan^{-1}\left( \frac{4}{3} \right) $ with the negative y-axis

Answer 1

Answer: (C)

Resultant, $R=\sqrt{x^{2}+y^{2}}$
$R=\sqrt{3^{2}+4^{2}}$
$R=\sqrt{25}=5\, N$ and $\cos \theta=\frac{x}{R}=\frac{3}{5}$
$\sin \theta=\frac{y}{R}=\frac{4}{5}$
then $\tan \theta=\frac{y}{x}=\frac{4}{3}$
$\theta=\tan ^{-1}\left(\frac{4}{3}\right)$ with negative $y$-axis.