Four forces are acting on a particle is in equilibrium. One force of magnitude $3 N$ is directed upward, another is directed $37^{0}$ 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 $\sqrt{5 n} N .$
$(\cos 37^{0} = \frac{4}{5}, \sin 37^{0} = \frac{3}{5})$

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answer is 2.

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Detailed Solution

$F_{1} = 3 \hat{k} F_{2} = 5 \sin 37^{0} \hat{i} + 5 \cos 37^{0} \hat{j} = 3 \hat{i} + 4 \hat{j} F_{3} = - 4 \sqrt{2} \cos 45^{0} \hat{i} - 4 \sqrt{2} \sin 45^{0} \hat{j} = - 4 \hat{i} - 4 \hat{j} F_{4} F_{n e t} = 0 \Rightarrow F_{1} + F_{2} + F_{3} + F_{4} = 0 \Rightarrow 3 \hat{k} + 3 \hat{i} + 4 \hat{j} - 4 \hat{i} - 4 \hat{j} + F_{4} = 0 - \hat{i} + 3 \hat{k} + F_{4} = 0 F_{4} = \hat{i} - 3 \hat{k} | F_{4} | = \sqrt{1^{2} + 3^{2}} = \sqrt{10} n = 2$