How this instrument works
Weight points straight down, but a ramp cares about two other directions: along and into. Splitting one vertical pull between them is all this sheet does, and sine claims whatever runs downhill. Galileo built a career on that factor. Unable to time falling bodies with 1604 equipment, he rolled bronze balls down a polished groove cut into a twelve-cubit beam, diluting gravity by a known sine until seconds stretched far enough for a water clock to catch them. Two New Sciences printed those results in 1638, and every ramp calculation since inherits his geometry.
A ramp is a machine, and its advantage is exactly 1 ⁄ sin θ. Pitch a surface at 30° and a load presses downhill with half its weight, so restraining 100 kg costs 490 N rather than 981 N of straight lift. Nothing comes free, though: that ramp measures twice its own height, and 490 N dragged across two metres spends precisely as much energy as 981 N raised through one. Trading distance for effort while leaving energy untouched is what wheelchair approaches, ships' gangways, screw threads and funicular railways all quietly rely on.
Two assumptions hide inside one sine. Gravity acts alone here — no friction, rope, winch or wind — so Force along the slope reports what gravity supplies downhill, never what your shoulder must deliver. Shoving a crate upward also costs μ times Force into the slope, while holding it still may cost nothing whatsoever if grip already exceeds gravity's pull. Second, θ must genuinely be an angle. Highway signs quote grades, which are tangents: a 6% grade is 3.43°, and typing 6 into a degree box inflates downhill pull by roughly three quarters.
- Mass on the slope accepts grams, kilograms, tonnes or pounds; conversion happens before any arithmetic runs.
- Slope angle expects degrees by default, radians from its unit menu. Convert grades first — angle = arctan(rise ⁄ run), not grade itself.
- Read Force along the slope for gravity's downhill component, in newtons, kilonewtons, kgf or pounds-force.
- Read Force into the slope for what presses perpendicular; multiply by a friction coefficient to judge whether anything actually slides.
- Sweep Slope angle from 0° to 90° and watch both outputs trade places, meeting midway at 45°.
Worked example — 100 kg crate on a 30° loading ramp
A 100 kg crate waits on a loading ramp pitched at 30°. Type 100 into Mass on the slope and 30 into Slope angle. Sine of 30° is one half exactly, so Force along the slope returns 100 × 9.80665 × 0.5 = 490.3325 N — half that crate's 980.665 N weight, digit for digit. Force into the slope reports 100 × 9.80665 × cos 30° = 849.28 N, which is what your ramp decking carries.
Both readings repay a moment's thought. Restraining that crate takes 490 N, near enough 50 kgf or one determined adult, where lifting it vertically would demand 981 N. Energy, however, refuses to budge: raising it one metre costs 980.665 J whether you hoist it or push it two metres up a ramp. And because tan 30° is 0.577, any surface pairing whose static coefficient tops 0.58 parks that crate unaided — timber on timber often manages it, steel skids near 0.4 do not.
Questions
Is a 10% grade the same as a 10° slope?
No — grades are tangents, angles are angles. A 10% grade means one unit of rise per ten of run, which works out to arctan(0.10) = 5.71°, whose sine is 0.0995. Type 10 into a degree box instead and sine jumps to 0.1736, inflating downhill pull by about 74%. Convert first: θ = arctan(rise ⁄ run). Below roughly 10° both quantities agree closely enough that this habit survives unnoticed, which is exactly why it bites on steeper ground.
Does this tell me how hard I must push a load uphill?
Not by itself — gravity's share is all you get here. Pushing upward needs m·g·sin θ against gravity plus μ·m·g·cos θ against friction, so a 100 kg crate on a 30° ramp with μ = 0.3 demands 490 + 255 = 745 N, half again as much as this output alone. Rolling it back down, friction changes sides and subtracts instead. Take Force into the slope, multiply by your coefficient, then add or subtract according to direction of travel.
At what angle does a load begin to slide?
Wherever tan θ overtakes static friction — an angle of repose. Since Force along the slope divided by Force into the slope equals tan θ exactly, sliding starts where that ratio crosses μs, and mass vanishes from the comparison entirely: a paperclip and a piano let go at identical tilts on identical surfaces. Rubber on dry asphalt near 0.8 clings to about 39°, loose dry sand heaps itself at 32 to 35°, and PTFE on steel at 0.04 surrenders near 2°. Tilting a board until something slips measures μ with no instruments at all.
Why does a ramp save force but not work?
Because it stretches your path by precisely as much as it lightens your load. Mechanical advantage is 1 ⁄ sin θ, and a ramp reaching height h must be h ⁄ sin θ long — same factor, opposite sense. Drag 490.3325 N along two metres or hoist 980.665 N through one, and both come to 980.665 joules. Simple machines redistribute effort across distance; only friction alters energy totals, and friction only ever adds to them.
Which tilt produces the largest force along the slope?
Ninety degrees, where sine reaches one and Force along the slope equals full weight — at which point your ramp has become a cliff and holds nothing up. Force into the slope reads zero there, leaving friction nothing to grip with, so your load is simply falling. Both components trade continuously: equal at 45°, each showing 0.7071 of weight, and their squares always sum to weight squared, since sin²θ + cos²θ = 1.
Why does mass cancel out of so many ramp problems?
Because gravity supplies both downhill push and, through contact load, whatever grip resists it — and both scale with mass identically. Sliding condition tan θ > μs carries no m whatsoever; acceleration down a frictionless ramp is g·sin θ, mass-free again. Force is where mass reappears, which is why this sheet answers in newtons: decking, ropes and winches must hold 490 N for 100 kg and 4903 N for a tonne, even though both let go at exactly one tilt.