An object is in equilibrium under four concurrent forces in the directions shown in figure. Find the magnitudes of $F_{1}$ and $F_{2}$ .

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$F_{1} = \frac{4}{\sqrt{3}} N$ , $F_{2} = \frac{4}{\sqrt{3}} N$

$F_{1} = \frac{8}{\sqrt{3}} N,$ $F_{2} = \frac{20}{\sqrt{3}} N$

$F_{1} = \frac{4}{\sqrt{3}} N, F_{2} = \frac{50}{\sqrt{3}} N$

$F_{1} = \frac{4}{\sqrt{3}} N, F_{2} = \frac{20}{\sqrt{3}} N$

answer is D.

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Detailed Solution

The object is in equilibrium. Hence,

$Σ F_{x} = 0$

$∴ 8 + 4 \cos 60 ° - F_{2} \cos 30 ° = 0$

$8 + 2 - F_{2} \frac{\sqrt{3}}{2} = 0$

$F_{2} = \frac{20}{\sqrt{3}} N$

$Σ F_{y} = 0$

$∴ F_{1} + 4 \sin 60 ° - F_{2} \sin 30 ° = 0$

$F_{1} + \frac{4 \sqrt{3}}{2} - \frac{F_{2}}{2} = 0$

$F_{1} = \frac{F_{2}}{2} - 2 \sqrt{3} = \frac{10}{\sqrt{3}} - 2 \sqrt{3}$

$F_{1} = \frac{4}{\sqrt{3}} N$

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