Questions: Friction on Inclines and Horizontal Surfaces

Question 1

A 10 kg block rests on a 30° incline. Taking g = 9.8 m/s², what is the normal force exerted by the incline on the block?

• A. mg·cos(30°) ≈ 84.9 N — the component of weight perpendicular to the slope
• B. mg = 98 N — the full weight of the block
• C. mg·sin(30°) = 49 N — the component of weight along the slope
• D. mg·tan(30°) ≈ 56.6 N — the ratio of parallel to perpendicular weight components

Answer: (A)

On an incline, you must decompose the weight into components relative to the tilted surface. The component perpendicular to the slope is W·cos(θ) = mg·cos(30°) ≈ 84.9 N, and equilibrium in the direction normal to the incline requires N = W·cos(θ). The full weight (98 N) is only the normal force on a horizontal surface where the entire weight presses into the surface. On an incline, the surface tilts away from vertical, so it supports only a fraction of the weight. This is the foundational step in every incline problem: N ≠ mg.

Question 2

Two blocks — one 1 kg and one 10 kg — both rest on the same incline with the same coefficient of static friction μₛ = 0.5. The angle is slowly increased. What happens?

• A. The lighter block slides first because its weight provides less friction force
• B. The heavier block slides first because gravity pulls it more strongly down the slope
• C. Both blocks start sliding at exactly the same angle, regardless of mass
• D. The result depends on the surface area in contact with the incline

Answer: (C)

Both blocks slide at the same critical angle φₛ = arctan(μₛ). Here is why: the driving force is W·sin(θ) = mg·sin(θ); the maximum friction force is μₛN = μₛ·mg·cos(θ). The condition for sliding is mg·sin(θ) > μₛ·mg·cos(θ), and dividing both sides by mg gives tan(θ) > μₛ — the mass cancels completely. The critical angle depends only on μₛ, not on mass, weight, or surface area. Option A and B represent opposite versions of the same misconception: that mass tips the balance. It does not, because both the driving force and the maximum friction force scale with mass identically.

Question 3

On an incline at angle θ, the normal force on an object equals its full weight mg.

• T. True
• F. False

Answer: False

False. The normal force on an incline equals mg·cos(θ), which is always less than mg for any positive angle θ. This follows directly from decomposing the weight vector into components perpendicular and parallel to the incline surface. The incline surface only pushes back against the perpendicular component of gravity — it cannot exert a force along its own surface (in the absence of friction, an object would slide freely). As the incline angle increases toward 90°, cos(θ) approaches 0 and the normal force approaches zero: a vertical wall exerts no normal force on an object placed against it from below.

Question 4

The critical angle at which a block begins to slide on an incline depends only on the coefficient of static friction, not on the block's mass or weight.

• T. True
• F. False

Answer: True

True. The sliding condition is tan(θ) > μₛ, which contains no mass term. This elegant result comes from the fact that both the gravitational driving force (mg·sin θ) and the maximum friction resisting force (μₛ·mg·cos θ) are proportional to mg. Dividing the condition mg·sin θ > μₛ·mg·cos θ by mg yields tan θ > μₛ, independent of mass. This is why the angle of repose — the steepest angle at which loose granular material sits without sliding — is a material property (depending on μₛ between particles), not a property of how much material is piled up.

Question 5

Explain why the critical angle at which a block begins to slide on an incline does not depend on the block's mass, using the force balance on the incline.

Model answer: On the incline, the driving force down the slope is mg·sin(θ) and the maximum static friction resisting it is μₛN = μₛ·mg·cos(θ). Setting the driving force equal to maximum friction to find the critical angle: mg·sin(θ_c) = μₛ·mg·cos(θ_c). The mass m cancels from both sides, giving tan(θ_c) = μₛ, or θ_c = arctan(μₛ). Because both the driving force and the friction force are proportional to the same mg, the mass factor divides out. The critical angle is a purely geometric and material property.

This mass-independence has practical consequences: a light pebble and a heavy boulder on the same slope with the same surface texture will begin sliding at the same angle. It also means that on a given slope, if any block slides, all blocks of the same material will slide — you cannot make a block stay by making it heavier. The only way to increase the critical angle is to increase μₛ (change the surface) or reduce the incline angle θ. Engineers designing storage bins, conveyor belts, and retaining walls rely on this principle when calculating whether materials will flow or stay put.