Question

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 $x=$ _____

Answer 1

Correct Answer: 3

Since the forces are in equilibrium, the net force in both horizontal and vertical directions should be zero. Consider horizontal forces:

\( F_1 \) acts to the right and \( 1 \, \text{N} \cos(45^\circ) \) and \( 2 \, \text{N} \cos(45^\circ) \) act to the left.

Equating the horizontal components: 
\( F_1 = 1 \cos(45^\circ) + 2 \cos(45^\circ) = 3 \cdot \frac{\sqrt{2}}{2} \)
\( F_1 = \frac{3\sqrt{2}}{2} \)

For vertical forces:

\( F_2 \) acts downward while \( 1 \, \text{N} \sin(45^\circ) \) and \( 2 \, \text{N} \sin(45^\circ) \) act upward.

Equating the vertical components: 
\( F_2 = 1 \sin(45^\circ) + 2 \sin(45^\circ) = 3 \cdot \frac{\sqrt{2}}{2} \)
\( F_2 = \frac{3\sqrt{2}}{2} \)

The ratio \( \frac{F_1}{F_2} = \frac{\frac{3\sqrt{2}}{2}}{\frac{3\sqrt{2}}{2}} = 1 \)

Thus, the value of \( x \) is \( \boxed{3} \), confirming it fits in the range [3,3].