Free body diagram — resolve forces on a body on an inclined plane

Question

A block of mass 5 kg is placed on a smooth inclined plane making an angle of 30° with the horizontal. Draw the free body diagram and find the acceleration of the block along the incline. Take $g = 10$ m/s².

(NEET 2023, similar pattern)

Solution — Step by Step

Three forces act on the block:

Weight ( $mg$ ) acting vertically downward
Normal reaction ( $N$ ) acting perpendicular to the inclined surface
Friction — zero here, since the plane is smooth

For $m = 5$ kg: $mg = 5 \times 10 = 50$ N

This is the critical step. Instead of using horizontal and vertical axes, we choose:

x-axis: along the incline (positive direction = down the slope)
y-axis: perpendicular to the incline (positive direction = away from surface)

Why? Because the motion happens along the incline, so resolving along the incline directly gives us the net force causing acceleration.

The weight $mg$ acts vertically downward. We resolve it into two components:

Along the incline: $mg\sin\theta = 50 \times \sin 30° = 50 \times 0.5 = 25$ N (down the slope)
Perpendicular to incline: $mg\cos\theta = 50 \times \cos 30° = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3}$ N (into the surface)

Perpendicular to incline (y-axis): No acceleration in this direction (block doesn’t fly off or sink into the surface).

N - mg\cos\theta = 0

N = mg\cos\theta = 25\sqrt{3} \approx 43.3 \text{ N}

Along the incline (x-axis): The only force component is $mg\sin\theta$ (no friction).

ma = mg\sin\theta

a = g\sin\theta = 10 \times \sin 30° = 10 \times 0.5

\boxed{a = 5 \text{ m/s}^2 \text{ (down the incline)}}

Why This Works

The key insight is that gravity pulls the block straight down, but the incline constrains the motion to be along its surface. By resolving gravity into parallel and perpendicular components, we separate the “driving force” ( $mg\sin\theta$ , which causes sliding) from the “pressing force” ( $mg\cos\theta$ , which the normal reaction balances).

Notice that on a smooth incline, the acceleration $a = g\sin\theta$ depends only on the angle — not on the mass. A 1 kg block and a 100 kg block slide down at the same rate (just like free fall, where all masses fall equally).

Alternative Method — Using horizontal and vertical axes

You can also resolve forces along standard horizontal and vertical axes, but then both $N$ and $a$ appear in both equations, making the algebra messier. The incline-aligned axes give cleaner equations with fewer unknowns per equation.

For inclined plane problems, always choose axes along and perpendicular to the incline. This is not just a suggestion — it’s the approach that saves you from algebraic nightmares. The only exception is when the question specifically asks for horizontal or vertical components of force.

Common Mistake

The classic error: students mix up $\sin\theta$ and $\cos\theta$ . Here’s the rule — the component along the incline is $mg\sin\theta$ and the component perpendicular to the incline is $mg\cos\theta$ . Think of it this way: when $\theta = 0°$ (flat surface), the entire weight should be perpendicular ( $mg\cos 0° = mg$ ) and nothing along the surface ( $mg\sin 0° = 0$ ). This sanity check instantly tells you which is which.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using horizontal and vertical axes

Common Mistake

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