What is R on a rough inclined plane of angle α?

R = mg cos α (for a block with no other perpendicular forces). R is never equal to mg on a slope.

What is F_MAX on an inclined plane?

F_MAX = μR = μ mg cos α. Same formula as horizontal, but R = mg cos α on a slope.

Which direction does friction act on an inclined plane?

Parallel to the plane, opposite to the direction the object is moving or about to move.

A block slides down a rough slope. Which way does friction act?

Up the slope. Friction opposes motion, so it acts opposite to the direction of motion (down the slope).

Component of weight down a slope of angle α

mg sin α. This is the component parallel to the plane that tends to pull the block down.

Why is R not equal to mg on an inclined plane?

The normal reaction is perpendicular to the plane. Only the perpendicular component of weight (mg cos α) is balanced by R. The parallel component (mg sin α) causes motion.

Block of 2 kg on rough 30° slope, μ = 0.3, g = 10. Find F_MAX.

R = 2×10×cos 30° = 17.3 N. F_MAX = 0.3×17.3 = 5.2 N.

Will the block above slide down? (mg sin 30° = 10 N)

Yes. Component down slope = 10 N > F_MAX = 5.2 N. Block accelerates. F = 5.2 N up the slope.

When is a block in limiting equilibrium on a slope?

When it is on the point of moving. F = μR. The component of weight down the slope equals F_MAX (if about to slide down).

Coefficient of Friction & Inclined Planes | Pearson Edexcel International AS Maths

Q: Step 1 for friction on inclined plane

Resolve weight (and any other forces) into components parallel and perpendicular to the plane.

Dynamics & Statics of a Particle

Coefficient of Friction & Inclined Planes

Summary

On inclined planes, friction problems use the same principles as horizontal surfaces but resolve parallel and perpendicular to the plane. R is never simply mg—it equals mg cos α. Resolve weight, find R, calculate F_MAX = μR, then apply F = ma in the parallel direction.

1. Key Difference from Horizontal Surfaces

On an inclined plane, the important directions are parallel and perpendicular to the plane (not horizontal and vertical).

Critical: On an inclined plane, (R \neq mg)! The normal reaction is never simply equal to the weight.

(R) acts perpendicular to and away from the plane
(R) is such that the total perpendicular force is zero

2. Step-by-Step Procedure

Step 1. Resolve weight (and any other forces) into components parallel and perpendicular to the plane.

Parallel (down slope): (mg\sin\alpha)
Perpendicular: (mg\cos\alpha)

Step 2. Calculate the normal reaction (R).

Perpendicular equilibrium: (R = mg\cos\alpha) (for a block on a slope with no other perpendicular forces)

Step 3. Calculate (F_{MAX} = \mu R) and find the resultant of all forces parallel to the plane.

Step 4. Use (F = ma) in the parallel direction to find acceleration.

3. Friction on Inclined Planes

Friction (F) acts parallel to the plane, opposite to the direction of motion or potential motion.

If resultant of other parallel forces ≤ F_MAX: object remains stationary, friction balances
If resultant > F_MAX: object accelerates, (F = \mu R) opposing motion
When moving: (F = \mu R) always

4. Force Diagram – Rough Inclined Plane

5. Worked Example

Block: mass 1 kg, rough plane at 20° to horizontal, μ = 0.2. Released from rest.

Perpendicular: (R = mg\cos 20° = 1 \times 9.8 \times 0.94 = 9.21) N

F_MAX = (\mu R = 0.2 \times 9.21 = 1.84) N

Parallel (down slope): Component of weight = (mg\sin 20° = 1 \times 9.8 \times 0.342 = 3.35) N

3.35 > 1.84 → block accelerates. (F = 1.84) N up the slope.

F = ma: (3.35 - 1.84 = 1 \times a) → (a = 1.51) m s⁻² down the slope

6. Important Reminders

7. Smooth vs Rough