Question

Three forces of magnitudes \(F_1\), \(F_2\) and \(F_3\) act on a body located at the origin as shown in the figure. The condition that gives zero net force is

• A. \(F_2=-(2+\sqrt{3})F_1,\qquad F_3=\dfrac{-4}{\sqrt{6}-\sqrt{2}}F_1\)
• B. \(F_2=-(2-\sqrt{3})F_1,\qquad F_3=\dfrac{-4}{\sqrt{6}+\sqrt{2}}F_1\)
• C. \(F_2=-(2+\sqrt{3})F_1,\qquad F_3=\dfrac{-2}{\sqrt{6}+\sqrt{2}}F_1\)
• D. \(F_2=-(2+\sqrt{2})F_1,\qquad F_3=\dfrac{-2}{\sqrt{6}-\sqrt{2}}F_1\)

Hint:

For equilibrium of concurrent forces, \[ \sum F_x=0 \] and \[ \sum F_y=0. \] Always resolve each force into horizontal and vertical components before applying equilibrium conditions.

Answer 1

The Correct Option is A

Step 1: Resolve each force into x and y components.
F₁ = (F₁√3/2, F₁/2), F₂ = (–F₂/2, F₂√3/2), F₃ = (F₃/√2, –F₃/√2).

Step 2: Set ΣF_x = 0 and ΣF_y = 0.
x: (√3/2)F₁ – (1/2)F₂ + F₃/√2 = 0 → √3 F₁ – F₂ + √2 F₃ = 0. y: (1/2)F₁ + (√3/2)F₂ – F₃/√2 = 0 → F₁ + √3 F₂ – √2 F₃ = 0.

Step 3: Solve the linear system for F₂ and F₃ in terms of F₁.
From first: F₂ = √3 F₁ + √2 F₃. Substitute into second: F₁ + √3(√3F₁+√2F₃) – √2F₃ = 0 → 4F₁ + (√6–√2)F₃ = 0 → F₃ = –4F₁/(√6–√2). Then F₂ = √3F₁ + √2(–4F₁/(√6–√2)) = –(2+√3)F₁.

Step 4: Final Answer:
F₂ = –(2+√3)F₁, F₃ = –4F₁/(√6–√2).