Question

A body acted upon by three forces $\overrightarrow{F_1},\overrightarrow{F_2}$ and $\overrightarrow{F_3}$ is under equilibrium. Given $\left|\overrightarrow{F_1}\right|=10N,\left|\overrightarrow{F_2}\right|=6N$

• A. $F_3=8N$
• B. $F_3=6N$
• C. $\theta =53$
• D. $\theta =90$

Answer 1

Answer: (A), (D)

The correct options are
A $F_3=8N$
D $\theta =90$

$\sum F_x=0\Rightarrow 6\sin \theta -10\sin 37\sin \theta =\frac{10}{6}\sin 37\theta =90\sum F_y=0\Rightarrow 6\cos \theta +10\cos 37=F_{3}6\cos 90+10\cos 37F_3=8N$