Inclined planes

A block on a ramp can only move along the surface. Put one axis along the slope and the other perpendicular to it and that constraint becomes a number: a⊥ = 0, whatever else happens. Every unknown acceleration now sits in one equation, and the other equation is a pure force balance. Horizontal and vertical axes are not wrong, only expensive. On a frictionless 35.0° ramp the block accelerates at g sin 35.0° = 5.63 m s⁻² down the slope. Level axes report the same motion as 4.61 m s⁻² horizontally and 3.23 m s⁻² vertically, the normal force appears in both equations, and you must add by hand the condition that ties the two components together, vertical = horizontal × tan θ, before anything can be solved. Tilted axes cost one thing: weight becomes the only force off-axis, so it has to be resolved. The normal force, friction and a rope laid along the slope already lie on an axis.

Weight points straight down, and the perpendicular axis is tilted away from the vertical by exactly the ramp angle θ, because rotating a surface by θ rotates its normal by θ too. So the component into the surface is mg cos θ and the component along it, pointing downhill, is mg sin θ. If you cannot remember which one takes the sine, do not memorise: test. At θ = 0 the floor is level, nothing pulls along it and the whole weight presses in, so the along-slope part must be the one that vanishes. At θ = 90° the surface is a wall that supports nothing, and sin 90° = 1 hands all of mg to the along-slope direction: free fall. For a 15.0 kg crate on a 28.0° ramp, mg = 147 N splits into 147 × sin 28.0° = 69.1 N along the slope and 147 × cos 28.0° = 130 N into it, and √(69.1² + 130²) = 147 N rebuilds the weight. A steeper ramp is hard twice over: more of the weight pulls downhill, and less of it presses the surfaces together for friction to use.

With only weight and the surface acting, ΣF⊥ = 0 gives N = mg cos θ. That is the answer to one equation, not a law, and it changes the moment any other force has a perpendicular part. Take a 3.50 kg block on a 25.0° ramp, where mg cos θ = 31.1 N. Push it with a horizontal 30.0 N force directed into the slope: the push has a component 30.0 × sin 25.0° = 12.7 N pressing the block onto the surface, so N = 31.1 + 12.7 = 43.8 N, 41% more. Pull it instead with a 12.0 N rope held 35.0° above the surface: the rope lifts with 12.0 × sin 35.0° = 6.9 N and N drops to 24.2 N. At a tension of 54.3 N the normal force reaches zero, the block leaves the ramp and the model no longer applies. Because kinetic friction is μₖN, none of this stays inside the perpendicular equation: with μₖ = 0.30 the friction force is 9.3 N for the bare block, 13.1 N with the push and 7.3 N with the rope. A wrong N corrupts the along-slope equation as well.

A block at rest on a slope needs an up-slope force of exactly mg sin θ, and static friction supplies exactly that. The coefficient enters only as a limit, fₛ ≤ μₛN, so the method is a comparison, never a substitution. Demand: mg sin θ. Ceiling: μₛ mg cos θ. The block holds if the demand does not exceed the ceiling. For a 5.20 kg crate at 25.0° with μₛ = 0.60 the demand is 21.6 N and the ceiling 27.7 N, so it holds with 22% in hand, and the friction acting is 21.6 N. Divide the condition by mg cos θ and it becomes tan θ ≤ μₛ. Mass has cancelled and so has g: the tilt at which a resting object lets go, θₛ = arctan μₛ, belongs to the pair of surfaces alone. It is 16.7° for μₛ = 0.30, 26.6° for 0.50, 35.0° for 0.70 and 45° for 1.0. A tilting board with a protractor is therefore a friction meter, and a coin and a crate with the same finish slide off it at the same angle, on Earth or on the Moon.

Sliding turns the inequality into an equality, fₖ = μₖN, and replaces the question of size with a question of direction. Kinetic friction opposes the velocity, not the downhill pull. For a block sliding down a 32.0° ramp with μₖ = 0.30, friction points up the slope and a = g(sin θ − μₖ cos θ) = 9.81 × (0.5299 − 0.2544) = 2.70 m s⁻², about half the frictionless 5.20 m s⁻². Shove the same block up the ramp and friction swings round to point downhill, alongside the weight component: the acceleration is g(sin θ + μₖ cos θ) = 9.81 × (0.5299 + 0.2544) = 7.69 m s⁻² down the slope, against the motion. That is 2.8 times the downhill value, so the block stops quickly, and if tan θ exceeds μₛ it then returns slowly, arriving back with less speed than it left with. The sign in front of μₖ is therefore not part of a formula to remember. It is decided afresh each time, by asking which way the block is moving.

Tilt a board slowly with a block on it. Up to θₛ the acceleration is zero; at θₛ friction sits at its ceiling μₛN, and the instant the block breaks away it falls to μₖN. The net force is suddenly finite, so the block starts with a = (μₛ − μₖ) g cos θₛ. With μₛ = 0.50 and μₖ = 0.30, θₛ = 26.6° and the block leaves at 0.20 × 9.81 × 0.894 = 1.75 m s⁻², not from zero. Plotted against θ, the acceleration of a released block is flat, steps up, then climbs along g(sin θ − μₖ cos θ) towards g at 90°, where N and friction vanish together. Set μₖ equal to μₛ and the step closes. The same picture marks where the particle model stops. A uniform crate 0.40 times as wide as it is tall tips over its lower edge at arctan 0.40 = 21.8°, before a μₛ = 0.50 surface would let it slide. A ball or a cylinder rolls, and part of the downhill pull goes into spinning it up. And on a ramp that is itself accelerating, a⊥ is not zero, so N must be solved for again.