Four forces are acting at a point P in equili | vedclass

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(D) For the system to be in equilibrium,the net force in both the horizontal $(x)$ and vertical $(y)$ directions must be zero.Let the horizontal direc

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$3$

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(D) For the system to be in equilibrium,the net force in both the horizontal $(x)$ and vertical $(y)$ directions must be zero.Let the horizontal direction be the $x$-axis and the vertical direction be the $y$-axis.Resolving the forces into components:$1$. Force $1 	ext{ N}$ at $45^{\circ}$ to the $x$-axis: $x$-component $= 1 \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,$y$-component $= 1 \sin 45^{\circ} = \frac{1}{\sqrt{2}}$.$2$. Force $2 	ext{ N}$ at $135^{\circ}$ to the $x$-axis: $x$-component $= 2 \cos 135^{\circ} = -2 \sin 45^{\circ} = -\sqrt{2}$,$y$-component $= 2 \sin 135^{\circ} = 2 \cos 45^{\circ} = \sqrt{2}$.$3$. Force $F_{1}$ along the positive $x$-axis: $x$-component $= F_{1}$,$y$-component $= 0$.$4$. Force $F_{2}$ along the negative $y$-axis: $x$-component $= 0$,$y$-component $= -F_{2}$.Sum of forces in $x$-direction: $F_{1} + \frac{1}{\sqrt{2}} - \sqrt{2} = 0 \implies F_{1} = \sqrt{2} - \frac{1}{\sqrt{2}} = \frac{2-1}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.Sum of forces in $y$-direction: $\frac{1}{\sqrt{2}} + \sqrt{2} - F_{2} = 0 \implies F_{2} = \sqrt{2} + \frac{1}{\sqrt{2}} = \frac{2+1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.Therefore,the ratio $F_{1} : F_{2} = \frac{1}{\sqrt{2}} : \frac{3}{\sqrt{2}} = 1 : 3$.Thus,$x = 3$.

Four forces are acting at a point $P$ in equilibrium as shown in the figure. The ratio of force $F_{1}$ to $F_{2}$ is $1: x$,where $x = ....$

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