A force vec{F_1} of magnitude 4 N acts on an | vedclass

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(A) The components of force $\vec{F_1}$ are: $F_{1x} = 4 \cos 30^{\circ} = 4 	imes \frac{\sqrt{3}}{2} = 2\sqrt{3} \ N$ and $F_{1y} = 4 \sin 30^{\circ

Vedclass · Accepted Answer

$6.9$

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(A) The components of force $\vec{F_1}$ are: $F_{1x} = 4 \cos 30^{\circ} = 4 	imes \frac{\sqrt{3}}{2} = 2\sqrt{3} \ N$ and $F_{1y} = 4 \sin 30^{\circ} = 4 	imes \frac{1}{2} = 2 \ N$.The components of force $\vec{F_2}$ are: $F_{2x} = 0 \ N$ and $F_{2y} = 4 \ N$.The net force components are: $F_x = F_{1x} + F_{2x} = 2\sqrt{3} \ N$ and $F_y = F_{1y} + F_{2y} = 2 + 4 = 6 \ N$.The magnitude of the resultant force is $F_R = \sqrt{F_x^2 + F_y^2} = \sqrt{(2\sqrt{3})^2 + 6^2} = \sqrt{12 + 36} = \sqrt{48} \approx 6.928 \ N$.Using Newton's second law,the acceleration $a = \frac{F_R}{m} = \frac{6.928}{1} \approx 6.9 \ m \ s^{-2}$.

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