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(B) The object is in equilibrium,so the net force in both $x$ and $y$ directions must be zero: $\Sigma F_x = 0$ and $\Sigma F_y = 0$.From the figure,r

Vedclass · Accepted Answer

$\frac{4}{\sqrt{3}} 	ext{ N}$ and $\frac{20}{\sqrt{3}} 	ext{ N}$

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(B) The object is in equilibrium,so the net force in both $x$ and $y$ directions must be zero: $\Sigma F_x = 0$ and $\Sigma F_y = 0$.From the figure,resolving the forces into components:For $\Sigma F_x = 0$:$8 + 4 \cos(60^{\circ}) - F_2 \cos(30^{\circ}) = 0$$8 + 4(0.5) - F_2(\frac{\sqrt{3}}{2}) = 0$$8 + 2 = F_2(\frac{\sqrt{3}}{2})$$10 = F_2(\frac{\sqrt{3}}{2}) \Rightarrow F_2 = \frac{20}{\sqrt{3}} 	ext{ N}$.For $\Sigma F_y = 0$:$F_1 + 4 \sin(60^{\circ}) - F_2 \sin(30^{\circ}) = 0$$F_1 + 4(\frac{\sqrt{3}}{2}) - (\frac{20}{\sqrt{3}})(\frac{1}{2}) = 0$$F_1 + 2\sqrt{3} - \frac{10}{\sqrt{3}} = 0$$F_1 = \frac{10}{\sqrt{3}} - 2\sqrt{3} = \frac{10 - 2(3)}{\sqrt{3}} = \frac{4}{\sqrt{3}} 	ext{ N}$.Thus,$F_1 = \frac{4}{\sqrt{3}} 	ext{ N}$ and $F_2 = \frac{20}{\sqrt{3}} 	ext{ N}$.

An object is in equilibrium when four concurrent forces,acting in the same plane,are in the directions shown in the figure. Find the magnitudes of $F_1$ and $F_2$.

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