Resolving forces — component method for 2D and 3D equilibrium problems

Question

Three forces act on a particle: $\vec{F}_1 = 10$ N at $30°$ from the x-axis, $\vec{F}_2 = 8$ N along the negative x-axis, and $\vec{F}_3 = 6$ N along the positive y-axis. Find the net force magnitude and direction. Is the particle in equilibrium?

(JEE Main & CBSE 11 pattern)

Solution — Step by Step

This is the core technique: replace every angled force with its horizontal and vertical parts.

$F_{1x} = 10\cos 30° = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66$ N

$F_{1y} = 10\sin 30° = 10 \times \frac{1}{2} = 5$ N

$F_{2x} = -8$ N (negative x-direction), $F_{2y} = 0$

$F_{3x} = 0$ , $F_{3y} = 6$ N

\Sigma F_x = 5\sqrt{3} + (-8) + 0 = 5\sqrt{3} - 8 \approx 0.66 \text{ N}

\Sigma F_y = 5 + 0 + 6 = 11 \text{ N}

F_{\text{net}} = \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2} = \sqrt{(0.66)^2 + (11)^2} \approx \sqrt{0.44 + 121} \approx \mathbf{11.02 \text{ N}}

Direction: $\alpha = \tan^{-1}\left(\dfrac{\Sigma F_y}{\Sigma F_x}\right) = \tan^{-1}\left(\dfrac{11}{0.66}\right) \approx \mathbf{86.6°}$ from x-axis.

For equilibrium, we need $\Sigma F_x = 0$ AND $\Sigma F_y = 0$ . Since both are non-zero, the particle is not in equilibrium — it will accelerate in the direction of the net force.

Why This Works

Forces are vectors. You cannot simply add their magnitudes — $10 + 8 + 6 \neq$ net force. By resolving into perpendicular components, we convert vector addition into simple scalar addition along each axis. Then Pythagoras gives us the magnitude of the resultant.

This method works for any number of forces at any angles. It scales to 3D by adding a z-component.

graph TD
    A["Multiple forces at various angles"] --> B["Resolve each into x, y components"]
    B --> C["Sum all x-components: ΣFx"]
    B --> D["Sum all y-components: ΣFy"]
    C --> E["F_net = √(ΣFx² + ΣFy²)"]
    D --> E
    E --> F{"ΣFx = 0 AND ΣFy = 0?"}
    F -->|"Yes"| G["Equilibrium"]
    F -->|"No"| H["Acceleration = F_net / m"]

Alternative Method — Triangle or Polygon Law

For 2-3 forces, you can draw them head-to-tail and use the triangle law. The closing side gives the resultant. But this is graphical — it’s less accurate and slower for numerical answers. The component method is always preferred in exams.

For equilibrium problems in JEE: if three forces are in equilibrium, use Lami’s theorem: $\dfrac{F_1}{\sin\alpha_1} = \dfrac{F_2}{\sin\alpha_2} = \dfrac{F_3}{\sin\alpha_3}$ , where $\alpha_i$ is the angle opposite to force $F_i$ . This is faster than resolving components when angles between forces are given directly.

Common Mistake

When resolving forces on an inclined plane, students often resolve along horizontal and vertical instead of along and perpendicular to the incline. On an incline at angle $\theta$ : resolve $mg$ into $mg\sin\theta$ (along the plane) and $mg\cos\theta$ (normal to the plane). Using the wrong axes makes the equations much harder and introduces unnecessary coupling between the two directions.

Question

Solution — Step by Step

Why This Works

Alternative Method — Triangle or Polygon Law

Common Mistake

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