A mass $2 \sqrt{3} \mathrm{~kg}$ is acted upon by two forces which are inclined to each other at $60^{\circ}$ and each of magnitude $1 \mathrm{~N}$. The acceleration of that mass in SI system is $\left[\sin 30^{\circ}=\cos 60^{\circ}=0.5\right]$

Question

Aicharya · Accepted Answer

$F_{n e t}=\sqrt{F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos 	heta}$ $=\sqrt{1+1+2 \cos 60}$ $=\sqrt{3}$ Now, $a=\mathrm{F} / \mathrm{m}=\sqrt{3} / 2 \sqrt 3=0.5 \mathrm{~m} / \mathrm{s}^{2}$

A mass $2 \sqrt{3} \mathrm{~kg}$ is acted upon by two forces which are inclined to each other at…

Solution