Four forces are acting at a point P in equilibrium as shown in figure. The ratio of force F₁ to F₂ is 1: x where x = _____________. Numerical answer type. Work it out, then open the answer.

Answer: 3. 2 N is at 45° left of vertical, 1 N is at 45° above +x, F₁ is along +x and F₂ is along -y. x: F₁ + 1cos 45° - 2cos 45° = 0 ⇒ F₁ = 1/√(2) y: 2sin 45° + 1sin 45° - F₂ = 0 ⇒ F₂ = 3/√(2) F₁: F₂ = 1: 3, so x = 3.

Four forces are acting at a point P in equilibrium as shown in… | JEE Main 2022 (25 Jul, Shift 1)

Q14JEE Main 2022 · 25 JUL · SHIFT 1Medium

Equilibrium, Third Law & Frames

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 $x =$ _____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

$2\text{ N}$ is at $45^\circ$ left of vertical, $1\text{ N}$ is at $45^\circ$ above $+x$ , $F_1$ is along $+x$ and $F_2$ is along $-y$ .
$x$ : $F_1 + 1\cos 45^\circ - 2\cos 45^\circ = 0 \Rightarrow F_1 = \frac{1}{\sqrt{2}}$
$y$ : $2\sin 45^\circ + 1\sin 45^\circ - F_2 = 0 \Rightarrow F_2 = \frac{3}{\sqrt{2}}$
$F_1 : F_2 = 1 : 3$ , so $x = 3$ .

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