Question

Four forces are acting on a particle. One force of magnitude $3 \,N$ is directed upward, another is directed $37^{\circ}$ East of North having magnitude $5\, N$, third is directed in South-West direction is of magnitude $4 \sqrt{2} N$ and fourth force is $\sqrt{5 n} N$. If the particle is in equilibrium. The value on $n$ is ___

Answer 1

Here, $\vec{F}_{1}=3 \hat{k}$
$\vec{F}_{2}=5 \cos 37^{\circ}+5 \sin 37^{\circ} \hat{i}=4 \hat{i}+3 \hat{j} $
$\vec{F}_{3} =-4 \sqrt{2} \cos 45^{\circ} \hat{i}-4 \sqrt{2} \sin 45^{\circ} \hat{j}$
$=-4 \hat{i}-4 \hat{j}$
For equilibrium of the particle
$ \vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}+\vec{F}_{4}=0 $
or $ 3 \hat{k}+4 \hat{j}+3 \hat{i}-4 \hat{i}-4 \hat{j}+\vec{F}_{4}=0 $
$\therefore \vec{F}_{4}=\hat{i}-3 \hat{k} $
$\therefore \vec{F}_{4}=\sqrt{1+9}=\sqrt{10}$
$\therefore n=2$