Inclined Surfaces and Resolving Forces

Many mechanics problems involve objects on slopes — a trolley on a ramp, a skier on a hill, a book on a tilted desk. On a slope, gravity is no longer aligned with a convenient axis, and the normal reaction is no longer simply equal to the weight. To analyse these situations you must resolve forces — break a vector into two perpendicular components — and then apply Newton's second law along each direction separately.

This lesson covers the resolving techniques used throughout OCR Module 3.2 (Forces in action) and is a skill that reappears in circular motion, projectile problems, electric and gravitational fields, and even the Young modulus practical.

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The Idea of Components

A single vector can always be written as the sum of two perpendicular vectors called its components. On a horizontal surface, gravity has components (0, −mg): simple. On a slope of angle θ to the horizontal, it is far more convenient to choose axes along and perpendicular to the slope — because the motion is along the slope.

Consider a block of mass m on a smooth slope at angle θ. The weight W = mg points straight down. Split it into:

• Component parallel to slope (pointing down the slope):

  W_|| = mg sin θ

• Component perpendicular to slope (pressing into the slope):

  W_⊥ = mg cos θ

Exam Tip: Many students muddle sin and cos. Remember: the angle is measured between the slope and the horizontal. Directly beneath the slope, the full weight acts. As the angle increases from 0° (flat) to 90° (vertical wall), the parallel component grows from 0 to mg (hence sin), while the perpendicular component shrinks from mg to 0 (hence cos).

A quick sanity check: at θ = 0°, W_|| = 0 (nothing slides) and W_⊥ = mg (full weight on the floor). At θ = 90°, W_|| = mg (free fall) and W_⊥ = 0 (no contact with "floor"). These limiting cases are always worth checking.

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Normal Reaction on a Slope

Because the block does not accelerate perpendicular to the slope (it stays on it), the forces perpendicular to the slope must balance:

N = mg cos θ

Notice how N is less than mg for any non-zero angle — the slope does not carry the full weight. This has a nice physical consequence: the harder you tilt, the less friction the slope can provide (friction ∝ N), which is why steep slopes are slippery.

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Acceleration Down a Smooth Slope

If the slope is smooth (frictionless), the only force along the slope is the parallel component of weight:

F_|| = mg sin θ

Applying Newton II:

ma = mg sin θ ⇒ a = g sin θ

So on a 30° slope, a = 9.81 × 0.500 = 4.91 m s⁻². As expected, 0 at θ = 0° and 9.81 at θ = 90°.

Worked Example — Smooth Ramp

A 2.5 kg trolley is released from rest on a smooth ramp of angle 25°. The ramp is 1.60 m long.

(a) Acceleration down the ramp:

• a = g sin 25° = 9.81 × 0.4226 = 4.15 m s⁻²

(b) Speed at the bottom (from SUVAT, v² = u² + 2as):

• v² = 0 + 2 × 4.15 × 1.60 = 13.3
• v = 3.64 m s⁻¹

(c) Time to slide down:

• v = u + at ⇒ t = v / a = 3.64 / 4.15 = 0.878 s

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Inclined Surface With Friction

If the slope is rough, friction acts up the slope (opposing tendency to slide). Let μ be the coefficient of friction (although OCR A-Level does not examine μ explicitly, the idea of a friction force F_f is important).

The friction force satisfies F_f ≤ μN, where at the point of slipping F_f = μN = μ mg cos θ.

Condition to be on the verge of slipping:

mg sin θ = μ mg cos θ tan θ = μ

This defines the angle of repose: the maximum angle at which the block remains stationary. For wood on wood, μ ≈ 0.5, so θ_repose ≈ 27°. This is why scree slopes in mountains have a characteristic angle — it is the angle of repose of the rock fragments.

Worked Example — Block on a Rough Slope

A 4.0 kg block sits on a slope at 30°. Friction force is 15 N up the slope. Find the acceleration.

• W_|| = 4.0 × 9.81 × sin 30° = 4.0 × 9.81 × 0.500 = 19.62 N down the slope
• Friction = 15 N up the slope
• Resultant force = 19.62 − 15 = 4.62 N down the slope
• a = F/m = 4.62 / 4.0 = 1.16 m s⁻² down the slope

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Worked Example — Object on a Slope With a Rope

A 10 kg crate is held stationary on a smooth slope of 20° by a rope parallel to the slope. Find (a) the tension in the rope and (b) the normal reaction.

Resolve weight:

• W_|| = 10 × 9.81 × sin 20° = 98.1 × 0.342 = 33.6 N
• W_⊥ = 10 × 9.81 × cos 20° = 98.1 × 0.940 = 92.2 N