Solving Multi-Force Problems on Inclined Planes

To solve a multi-force problem on an inclined plane, choose axes parallel and perpendicular to the slope, resolve the weight into components mg sin θ (down the slope) and mg cos θ (into the slope), then apply Newton’s second law along each axis: N = mg cos θ perpendicular to the slope, and ΣF = ma along it. In brief: the normal force is N = mg cos θ when no other perpendicular forces act, and friction is f = μN. Down the slope with kinetic friction, a = g(sin θ − μ_k cos θ). An object stays at rest if tan θ ≤ μ_s.

Ramps, hills, roofs and ski slopes are inclined planes, and the weight of an object acts partly along the surface and partly into it. This guide gives a step-by-step method, discusses cases with friction and an applied force, compares the cases in a table, works through three problems and corrects common mistakes.

What is the method?

Draw a sketch and a free-body diagram of the object, with weight (vertical), normal force (perpendicular to the surface), friction (along the surface) and any applied force.
Tilt the axes: x along the slope (positive down the slope, or in the direction of motion), y perpendicular to the slope.
Resolve the weight: the component along the slope is mg sin θ, and the one perpendicular to the slope is mg cos θ (θ is the angle of the slope with the horizontal).
Perpendicular direction: there is no acceleration, so N = mg cos θ (plus or minus any perpendicular component of an applied force).
Friction: f_k = μ_kN when sliding; f_s ≤ μ_sN at rest. It acts opposite to the (possible) sliding.
Along the slope: ΣF = ma. Solve for a, or for an unknown force.

What are the common cases?

Frictionless slope: a = g sin θ down the slope, independent of mass.
Sliding down with friction: a = g(sin θ − μ_k cos θ). If μ_k = tan θ, the acceleration is zero (constant velocity).
Sliding up with friction (after being pushed): friction and the weight component both act down the slope, so the deceleration is g(sin θ + μ_k cos θ).
At rest: the friction is static, f_s = mg sin θ, and the object does not slide if mg sin θ ≤ μ_smg cos θ, that is tan θ ≤ μ_s. The critical angle θ_c = tan⁻¹ μ_s.
Pushed or pulled up by a force parallel to the slope: ΣF = F − mg sin θ − f.

Cases compared

Acceleration of an object on a slope at angle θ (positive down the slope)
Case	Acceleration
Sliding down, no friction	g sin θ
Sliding down, kinetic friction	g(sin θ − μ_k cos θ)
Sliding up, kinetic friction (deceleration)	−g(sin θ + μ_k cos θ)
Pulled up by force F parallel to slope	(F − mg sin θ − μ_kmg cos θ) ÷ m, up the slope

Forces on the 10 kg box on the 30° slope, drawn to one scale. The normal force (84.9 N) balances the weight component into the slope, mg cos θ. The dashed component along the slope, mg sin θ = 49.0 N, pulls the box down the slope and is opposed by friction.

Worked example: a box on a 30° slope

A box of mass 10 kg is on a slope inclined at 30°, with μ_k = 0.20 and μ_s = 0.70. Use g = 9.8 m/s², sin 30° = 0.500 and cos 30° = 0.866.

Components of the weight: mg = 98 N. Along the slope: 98 × 0.500 = 49.0 N. Perpendicular: 98 × 0.866 = 84.9 N. So N = 84.9 N.
Does it slide? The maximum static friction is 0.70 × 84.9 = 59.4 N, which is more than 49.0 N. Since tan 30° = 0.577 < 0.70, the box stays at rest, held by a static friction force of 49.0 N.
Sliding down (given a push, so that it moves): kinetic friction = 0.20 × 84.9 = 17.0 N. Net force = 49.0 − 17.0 = 32.0 N. a = 32.0 ÷ 10 = 3.2 m/s² down the slope. Check with the formula: 9.8 × (0.500 − 0.20 × 0.866) = 9.8 × 0.327 = 3.2 m/s².
Pulled up the slope by 100 N parallel to the slope: the friction now acts down the slope. Net force up the slope = 100 − 49.0 − 17.0 = 34.0 N. a = 3.4 m/s² up the slope.
Speed after 4.0 m down: from rest, v² = 2as = 2 × 3.2 × 4.0 = 25.6, so v = 5.1 m/s.

How was motion on slopes studied?

Simon Stevin used the inclined plane in the 1580s to demonstrate the parallelogram of forces in the equilibrium of a chain on a wedge. Galileo used inclined planes in the early 1600s to dilute gravity, slowing down falling motion so that it could be measured with the instruments of the time, and found that the acceleration down a slope is a fraction g sin θ of free-fall acceleration.

Which mistakes are common?

Mistake 1: Swapping sine and cosine

With θ measured from the horizontal, the component along the slope is mg sin θ. Check the limit: at θ = 0° it must be zero.

Mistake 2: Using N = mg on a slope

N = mg cos θ, which is less than mg.

Mistake 3: Putting friction in the wrong direction

Friction opposes the sliding. When the object moves up the slope, friction is directed down the slope.

Mistake 4: Using the slant length as the height

The vertical drop is s sin θ for a path length s.