A box of mass m is in equilibrium under the a | vedclass

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(C) All forces are resolved into two perpendicular axes ($X$ and $Y$) as shown in the figure.Since the block of mass $m$ is in equilibrium, the net fo

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$20$

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(C) All forces are resolved into two perpendicular axes ($X$ and $Y$) as shown in the figure.Since the block of mass $m$ is in equilibrium, the net force in both $x$ and $y$ directions must be zero.Resolving the forces in the $x$-direction:$|F_2| \cos(60^{\circ}) = |F_1| \cos(30^{\circ})$$|F_2| 	imes \frac{1}{2} = 10 	imes \frac{\sqrt{3}}{2} \quad (\because |F_1| = 10 \ N 	ext{ is given})$$|F_2| = 10\sqrt{3} \ N$Resolving the forces in the $y$-direction:$|F_3| = |F_1| \sin(30^{\circ}) + |F_2| \sin(60^{\circ})$$|F_3| = 10 	imes \frac{1}{2} + 10\sqrt{3} 	imes \frac{\sqrt{3}}{2}$$|F_3| = 5 + 15 = 20 \ N$Thus, the magnitude of $F_3$ is $20 \ N$.

$A$ box of mass $m$ is in equilibrium under the application of three forces as shown below. If the magnitude of $F_1$ is $10 \ N$, what is the magnitude of $F_3$ (in $N$)?

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