Question

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

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

Answer 1

Answer: (C)

$7N$ making an angle $\theta =ta{n}^{-1}\left(\frac{3}{4}\right)$ with negative y-axis

$F_3=\sqrt{3^2+4^2}$

$=5N$

(Refer to Image)

From figure,

$\theta ={\tan }^{-1}\left(\frac{3}{4}\right)$ with negative $y$-axis