Inclined Plane Calculator
Parallel (N)
49
Perpendicular (N)
84.9
How it works
An inclined plane reduces the force required to raise a load by distributing the work over a longer distance. For a load on a frictionless incline at angle θ: required force = W × sin(θ), where W is the weight. Normal force = W × cos(θ).
**Mechanical advantage** The ideal mechanical advantage (IMA) of an inclined plane = 1 / sin(θ) = hypotenuse / height. A ramp with a 5° slope has IMA = 11.5 — pushing a 1000 N load requires only 87 N of force (ignoring friction). The trade-off: you push over a longer distance. Work input = Work output for ideal machines.
**Effect of friction on inclined plane force** With friction (coefficient μ): force to push up = W × (sin θ + μ × cos θ). Force to push down = W × (sin θ - μ × cos θ). If μ × cos θ > sin θ (i.e., tan θ < μ), the load won't slide down under gravity — self-locking. This explains why steep ramps need brakes while gentle ones don't.
**Components of weight on incline** Weight has two components on an incline: parallel to surface (W × sin θ, causing motion) and perpendicular to surface (W × cos θ, causing normal force). This decomposition applies to any inclined surface problem: snow on a roof, fluids in tilted containers, vehicles on grades.
**Vehicle grade calculations** Road grades are expressed as rise/run (not angle). A 6% grade means 6 feet of rise per 100 feet of horizontal distance. Angle ≈ arctan(0.06) = 3.43°. Extra force for a 2000 kg vehicle on a 6% grade: 2000 × 9.81 × sin(3.43°) = 1178 N ≈ 265 lb_f. This explains why steep grades require lower gears.
Frequently Asked Questions
- Weight component along the slope: F_parallel = W × sin(θ). Normal force: F_normal = W × cos(θ). For braking requirement: brakes must provide F_parallel to prevent rolling. For a 2000 kg vehicle on a 15% grade (angle ≈ 8.5°): F_parallel = 2000 × 9.81 × sin(8.5°) = 2909 N ≈ 655 lbf. The parking brake must exceed this force. Also check tire-to-road friction: if μ < tan(θ), the vehicle slides even with locked wheels — rare on paved roads but relevant for ice or loose gravel.
- A screw thread is a helical inclined plane wrapped around a cylinder. The helix angle (lead angle) of the thread determines whether it self-locks or backdrives. Self-locking condition: lead angle < friction angle (arctan μ). Standard threads (60° V-thread, coarse pitch): lead angle ≈ 2–4°, friction angle with steel ≈ 8–12° → self-locking. This is why bolts stay tightened without locknuts on most applications. Acme threads and ball screws have higher lead angles for efficient power transmission but may not self-lock, requiring brakes or detents on vertical axes.
- Ideal mechanical advantage (IMA) = hypotenuse/height = 1/sin(θ). Actual mechanical advantage (AMA) = load/effort (measured). Efficiency η = AMA/IMA. With friction, effort to push up = W(sin θ + μcos θ), so AMA = W/[W(sin θ + μcos θ)] = 1/(sin θ + μcos θ). Efficiency = sin θ / (sin θ + μcos θ). For 10° slope, μ = 0.3: efficiency = sin10°/(sin10° + 0.3cos10°) = 0.174/(0.174 + 0.295) = 37%. Steeper angles and lower friction improve efficiency.
- A wedge is two inclined planes back-to-back. The mechanical advantage depends on the included angle — a thin wedge has high IMA and can generate enormous force from a small input. Splitting a log: the wedge converts downward hammer force to horizontal splitting force multiplied by the IMA. A door stop wedge stays in place because: the floor reaction force on the inclined surface has a horizontal component that pushes the wedge forward, compressing against the door, while friction prevents it from sliding out. The thinner the wedge, the harder it is to remove (higher IMA, more self-locking).