98.066
N
49.033
N
84.928
N
84.928
N
25.478
N
23.555
N
2.3555
m/s²
0.5774
98.066
N
49.033
N
84.928
N
84.928
N
25.478
N
23.555
N
2.3555
m/s²
0.5774
The Inclined Plane Calculator analyzes forces acting on an object resting on or sliding down a tilted surface. Inclined planes are one of the six classical simple machines and appear everywhere in physics and engineering — from ramps and slides to roads winding up hillsides. Understanding the force decomposition on an incline is fundamental to Newtonian mechanics.
When an object of mass $$m$$ sits on a surface tilted at angle $$\theta$$ above the horizontal, gravity pulls it straight down with force $$W = mg$$. That weight vector resolves into two perpendicular components: one parallel to the surface ($$mg\sin\theta$$) driving the object downhill, and one perpendicular ($$mg\cos\theta$$) pressing it into the surface. The surface responds with a normal force $$N = mg\cos\theta$$, and kinetic friction opposes motion with $$f_k = \mu N$$.
This calculator computes every force in the system plus the resulting net force and acceleration. Engineers use these results when designing conveyor belts, loading ramps, vehicle braking on grades, and wheelchair-accessible slopes. Physics students encounter inclined-plane problems in virtually every introductory mechanics course, making this tool an invaluable study companion.
The calculator decomposes the weight force along and perpendicular to the incline, then applies Newton's second law:
Weight: $$W = mg$$ where $$g = 9.80665\;\text{m/s}^2$$.
Parallel component (drives object downhill): $$F_{\parallel} = mg\sin\theta$$
Perpendicular component: $$F_{\perp} = mg\cos\theta$$
Normal force: $$N = mg\cos\theta$$ (surface pushes back against the perpendicular component).
Friction force: $$f = \mu N = \mu\, mg\cos\theta$$
Net force along the incline: $$F_{\text{net}} = mg\sin\theta - \mu\, mg\cos\theta$$
Acceleration: $$a = g(\sin\theta - \mu\cos\theta)$$
If $$F_{\text{net}} > 0$$ the object accelerates downhill; if $$F_{\text{net}} < 0$$ friction is strong enough to keep it stationary (or the applied formula shows the net restraining force). The critical angle where the object just begins to slide satisfies $$\tan\theta_c = \mu$$.
A positive net force means the object accelerates down the incline. A negative value indicates friction exceeds the gravitational pull component, so the object remains stationary or requires an external push to slide. Compare $$\tan\theta$$ with $$\mu$$: when $$\tan\theta > \mu$$, sliding occurs. The magnitude of acceleration tells you how quickly the object picks up speed. For safety-critical designs (ramps, roads), keep the angle well below the critical angle to prevent uncontrolled sliding.
Inputs
Results
A 10 kg box on a 30° ramp with μ = 0.3 experiences a net 23.6 N downhill force, accelerating at 2.36 m/s².
Inputs
Results
Friction exceeds the gravitational pull — the 50 kg crate stays put on the 15° slope.
An inclined plane is a flat surface tilted at an angle to the horizontal. It is one of the six classical simple machines. By spreading the vertical lift over a longer distance along the slope, it reduces the force required to raise an object, trading force magnitude for distance traveled.
Apply Newton's second law along the incline: $$a = g(\sin\theta - \mu\cos\theta)$$. If there is no friction, the acceleration simplifies to $$a = g\sin\theta$$. A positive result means the object accelerates downhill; a zero or negative result means friction prevents motion.
The critical angle $$\theta_c$$ is the steepest angle at which the object remains stationary. It satisfies $$\tan\theta_c = \mu$$, so $$\theta_c = \arctan(\mu)$$. For example, with $$\mu = 0.5$$, the critical angle is about 26.6°.
No. On a frictionless incline the acceleration is $$a = g\sin\theta$$, which is independent of mass. All objects slide at the same rate regardless of weight, just as in free fall. Mass only matters when friction or air resistance are present.
The normal force equals the perpendicular component of the object's weight: $$N = mg\cos\theta$$. It is always less than the full weight $$mg$$ because part of the weight acts along the slope. As the angle increases, the normal force decreases.
Friction acts opposite to the direction of motion (or potential motion) along the surface. Its magnitude is $$f = \mu N = \mu mg\cos\theta$$. It reduces the net downhill force and therefore the acceleration. If friction is large enough, the object will not slide at all.
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