Question

Three forces $F_1,F_2$ and $F_3$ together keep a body in equilibrium. If $F_1=3N$ along the positive $x$-axis, $F_2=4N$ along the positive $y$-axis, then the third force $F_3$ is

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

Answer 1

Answer: (C)

Resultant, $R=\sqrt{x^2+y^2}$
$R=\sqrt{3^2+4^2}$
$R=\sqrt{25}=5N$ 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.