Friction and Inclined Planes

Friction and Inclined Planes

Friction is the force that resists the relative sliding motion of two surfaces in contact. Understanding friction is essential for analysing motion on rough surfaces, particularly on inclined planes.

The Friction Model

The frictional force F between two surfaces in contact satisfies:

F ≤ μR

where:

• μ (mu) is the coefficient of friction — a dimensionless constant depending on the surfaces
• R is the normal reaction (the force pressing the surfaces together)
• The inequality means friction can take any value from zero up to the maximum μR

At limiting equilibrium (the point when the object is on the verge of sliding), friction reaches its maximum value:

F = μR (limiting friction)

When the object is sliding, the friction force equals μR and is called kinetic (dynamic) friction (at A-Level, usually assumed equal to the static maximum).

Direction of Friction

Friction always acts in the opposite direction to the motion (or to the direction in which the object would move if friction were absent). It acts along the surface (tangent to the contact).

Friction on a Horizontal Surface

Worked Example: A 10 kg box is on a rough horizontal floor (μ = 0.4). A horizontal force P is applied. Find the minimum P to start the box moving.

The normal reaction R = mg = 10(9.8) = 98 N.

Limiting friction F = μR = 0.4 × 98 = 39.2 N.

The box begins to move when P > F, so the minimum force is P = 39.2 N.

If P = 50 N (greater than limiting friction), the box accelerates:

F = ma: 50 - 39.2 = 10a, so a = 10.8/10 = 1.08 m/s²

Friction on an Inclined Plane

On a plane inclined at angle θ, resolve forces parallel and perpendicular to the slope:

Perpendicular: R = mgcosθ (no acceleration perpendicular to slope)

Parallel: The component of weight down the slope is mgsinθ. Friction acts up the slope if the object slides or tends to slide down.

Case 1: Object on the verge of sliding down

At limiting equilibrium: mgsinθ = μR = μmgcosθ

Dividing: tanθ = μ (at the critical angle)

The angle at which sliding is about to begin is called the angle of friction: θ_c = arctan(μ).

Worked Example: A block sits on a rough slope. The coefficient of friction is 0.5. Find the maximum angle of the slope before the block slides.

tanθ = μ = 0.5

θ = arctan(0.5) = 26.6°

Case 2: Object sliding down a rough slope

If the slope angle exceeds the critical angle, the block accelerates down:

ma = mgsinθ - μmgcosθ

a = g(sinθ - μcosθ)

Worked Example: A 4 kg block slides down a rough slope inclined at 35° with μ = 0.3. Find the acceleration.

a = 9.8(sin35° - 0.3cos35°)

= 9.8(0.5736 - 0.3 × 0.8192)

= 9.8(0.5736 - 0.2458)

= 9.8 × 0.3278

= 3.21 m/s²

Case 3: Object pushed up a rough slope

If a force pushes the object up the slope, friction acts down the slope (opposing the motion):

ma = P - mgsinθ - μmgcosθ (along the slope, upward positive)

Worked Example: A 5 kg block is pushed up a slope at 30° by a force of 60 N parallel to the slope. μ = 0.2. Find the acceleration.

R = mgcos30° = 5(9.8)(0.866) = 42.4 N

Friction = μR = 0.2 × 42.4 = 8.49 N (down the slope)

Weight component down slope = mgsin30° = 5(9.8)(0.5) = 24.5 N

F = ma (up slope positive): 60 - 24.5 - 8.49 = 5a

27.01 = 5a, so a = 5.40 m/s²

Connected Particles on Rough Surfaces

Worked Example: A 3 kg block on a rough horizontal table (μ = 0.4) is connected by a string over a smooth pulley to a 5 kg block hanging vertically. Find the acceleration.

For the 5 kg block: 5g - T = 5a ... (i)

For the 3 kg block: T - F = 3a

where F = μR = 0.4 × 3g = 0.4(29.4) = 11.76 N

So: T - 11.76 = 3a ... (ii)

Adding (i) and (ii): 5g - 11.76 = 8a

49 - 11.76 = 8a

a = 37.24/8 = 4.66 m/s²

T = 3(4.66) + 11.76 = 13.98 + 11.76 = 25.7 N

When Friction Prevents Motion

If the applied force or slope component is less than the maximum friction, the object does not move and friction takes whatever value is needed for equilibrium (less than μR).

Always check: is the applied force enough to overcome limiting friction? If not, the object is in equilibrium and friction ≠ μR.

Exam Tips

• Always find R first — friction depends on the normal reaction, and R is rarely just mg on a slope (R = mgcosθ on an incline).
• Friction is not always μR. It equals μR only at limiting equilibrium or during sliding. In static situations, friction adjusts to match the applied force.
• On a slope, resolve parallel and perpendicular to the slope, not horizontally and vertically.
• If a force is applied at an angle to the slope, it affects both the parallel and perpendicular equations — and therefore changes R, which changes friction.
• State whether the object is in equilibrium, on the point of sliding, or accelerating — this determines which friction value to use.
• The formula tanθ = μ at limiting equilibrium is a useful shortcut — derive it by dividing the parallel equation by the perpendicular one.