Three forces acting on a body are shown in th | vedclass

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(A) First,resolve all forces into their $x$ and $y$ components.For the $1 \ N$ force at $60^{\circ}$ with the $x$-axis: $F_{1x} = 1 \cos 60^{\circ} =

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$0.5 \ N$

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(A) First,resolve all forces into their $x$ and $y$ components.For the $1 \ N$ force at $60^{\circ}$ with the $x$-axis: $F_{1x} = 1 \cos 60^{\circ} = 0.5 \ N$,$F_{1y} = 1 \sin 60^{\circ} = \frac{\sqrt{3}}{2} \ N$.For the $4 \ N$ force at $30^{\circ}$ with the $y$-axis (or $120^{\circ}$ with the $x$-axis): $F_{4x} = -4 \sin 30^{\circ} = -2 \ N$,$F_{4y} = 4 \cos 30^{\circ} = 2\sqrt{3} \ N$.For the $2 \ N$ force at $30^{\circ}$ with the $y$-axis (or $-60^{\circ}$ with the $x$-axis): $F_{2x} = 2 \sin 30^{\circ} = 1 \ N$,$F_{2y} = -2 \cos 30^{\circ} = -\sqrt{3} \ N$.The net force in the $x$-direction is $F_x = F_{1x} + F_{4x} + F_{2x} = 0.5 - 2 + 1 = -0.5 \ N$.To make the resultant force act only along the $y$-direction,the net force in the $x$-direction must be zero.Therefore,we need an additional force $F_{add}$ such that $F_x + F_{add,x} = 0$,which means $F_{add,x} = 0.5 \ N$.The minimum magnitude of this additional force is $0.5 \ N$.

Three forces acting on a body are shown in the figure. To have the resultant force only along the $y$-direction,the magnitude of the minimum additional force needed is:

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