Four forces are acting at a point $$P$$ in equilibrium as shown in figure. The ratio of force $$F_1$$ to $$F_2$$ is $$1:x$$ where $$x =$$ ______.

We need to determine the value of $$x$$ given that four concurrent forces acting at point $$P$$ are in a state of translational equilibrium.

1. Resolve the Forces into Components

For the particle to be in equilibrium, the net force along both the horizontal ($$x$$-axis) and vertical ($$y$$-axis) directions must equal zero. Let's resolve the angled forces ($$2\text{ N}$$ and $$1\text{ N}$$) into their components using $$\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}}$$:

• $$2\text{ N}$$ Force (Top-Left):
  • Horizontal component: $$2 \cos 45^\circ = \frac{2}{\sqrt{2}} = \sqrt{2}\text{ N}$$ (pointing in the negative $$x$$-direction)
  • Vertical component: $$2 \sin 45^\circ = \frac{2}{\sqrt{2}} = \sqrt{2}\text{ N}$$ (pointing in the positive $$y$$-direction)
• $$1\text{ N}$$ Force (Top-Right):
  • Horizontal component: $$1 \cos 45^\circ = \frac{1}{\sqrt{2}}\text{ N}$$ (pointing in the positive $$x$$-direction)
  • Vertical component: $$1 \sin 45^\circ = \frac{1}{\sqrt{2}}\text{ N}$$ (pointing in the positive $$y$$-direction)

2. Set Up the Equilibrium Equations

• Along the horizontal ($$x$$) direction ($$\Sigma F_x = 0$$):

  $$F_1 + 1 \cos 45^\circ - 2 \cos 45^\circ = 0$$

  $$F_1 + \frac{1}{\sqrt{2}} - \frac{2}{\sqrt{2}} = 0$$

  $$F_1 - \frac{1}{\sqrt{2}} = 0 \implies F_1 = \frac{1}{\sqrt{2}}\text{ N}$$

• Along the vertical ($$y$$) direction ($$\Sigma F_y = 0$$):

  $$2 \sin 45^\circ + 1 \sin 45^\circ - F_2 = 0$$

  $$\frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} - F_2 = 0$$

  $$F_2 = \frac{3}{\sqrt{2}}\text{ N}$$

3. Find the Ratio ($$F_1 : F_2$$)

Taking the ratio of the calculated forces:

$$\frac{F_1}{F_2} = \frac{\frac{1}{\sqrt{2}}}{\frac{3}{\sqrt{2}}} = \frac{1}{3}$$

Comparing this result with the given ratio format ($$1 : x$$)

$$\frac{1}{x} = \frac{1}{3} \implies x = 3$$

Conclusion

The value of $$x$$ is 3.