Question

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

Answer 1

Answer: 2

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 $3\hat{k}+4\hat{j}+3\hat{i}-4\hat{i}-4\hat{j}+{\vec{F}}_4=0$
$\therefore {\vec{F}}_4=\hat{i}-3\hat{k}$
$\therefore {\vec{F}}_4=\sqrt{1+9}=\sqrt{10}$
$\therefore n=2$