KAHIBARO
Table of Contents
Basic idea of motion on a slope
An inclined plane is a surface that makes an angle with the horizontal. Many mechanics problems become clearer when motion happens on a slope, because gravity acts vertically downward while the object is constrained to move along the surface.
The special feature of an inclined plane problem is that the weight of the object must be separated into two parts, one perpendicular to the plane and one parallel to the plane. Once this is done, Newton's second law can be applied along convenient directions.
Choosing axes along the plane
For inclined plane problems, it is usually best to choose one axis parallel to the surface and one axis perpendicular to it. This avoids unnecessary algebra, because the motion often occurs only along the plane.
If the plane makes an angle $\theta$ with the horizontal, then the weight $mg$ can be resolved into:
$$mg\sin\theta$$
parallel to the plane, directed down the slope, and
$$mg\cos\theta$$
perpendicular to the plane, directed into the surface.
For a block on an incline of angle $\theta$:
$$W = mg$$
$$W_{\parallel} = mg\sin\theta$$
$$W_{\perp} = mg\cos\theta$$
These components are the starting point for almost every inclined plane problem.
Normal force on an inclined plane
The normal force is perpendicular to the plane. If the block does not accelerate away from the plane or into it, then the net force in the perpendicular direction is zero.
So, for a simple incline with no other perpendicular forces,
$$N = mg\cos\theta$$
This result is very important because friction often depends on the normal force.
On a simple incline, if there is no acceleration perpendicular to the surface,
$$N = mg\cos\theta$$
Do not confuse this with $mg$. The normal force is smaller than the full weight unless the surface is horizontal.
Motion on a frictionless incline
If there is no friction, the only force along the plane is the component of gravity down the slope. Applying Newton's second law parallel to the plane gives
$$mg\sin\theta = ma$$
so the acceleration is
$$a = g\sin\theta$$
This means that a steeper incline gives a larger acceleration.
If $\theta = 0^\circ$, then $a = 0$, which matches a flat surface. If $\theta = 90^\circ$, then $a = g$, which matches free fall.
For motion on a frictionless incline,
$$a = g\sin\theta$$
This acceleration does not depend on the mass of the object.
Inclined planes with friction
If friction is present, it acts along the plane and opposes the direction of actual motion or attempted motion.
The friction forces are:
$$f_s \le \mu_s N$$
for static friction, and
$$f_k = \mu_k N$$
for kinetic friction.
Since on a simple incline $N = mg\cos\theta$, the kinetic friction becomes
$$f_k = \mu_k mg\cos\theta$$
If the object slides down the plane, friction acts up the plane. Then the net force parallel to the plane is
$$mg\sin\theta - f_k = ma$$
so
$$mg\sin\theta - \mu_k mg\cos\theta = ma$$
and therefore
$$a = g\left(\sin\theta - \mu_k \cos\theta\right)$$
If instead the object is being pulled up the plane and is moving upward, friction acts downward along the plane, adding to the downhill forces.
When an object remains at rest
An object can remain at rest on an incline if static friction is large enough to balance the downhill component of gravity.
For equilibrium along the plane,
$$f_s = mg\sin\theta$$
But static friction can only adjust up to its maximum value:
$$f_{s,\max} = \mu_s N = \mu_s mg\cos\theta$$
So the object can remain at rest only if
$$mg\sin\theta \le \mu_s mg\cos\theta$$
which simplifies to
$$\tan\theta \le \mu_s$$
This gives an important condition for slipping.
A block will not slip on an incline if
$$mg\sin\theta \le \mu_s mg\cos\theta$$
or equivalently,
$$\tan\theta \le \mu_s$$
If $\tan\theta > \mu_s$, static friction is not enough, and the block starts to slide.
Comparing common cases
The main incline situations can be summarized clearly.
| Situation | Force along plane | Acceleration |
|---|---|---|
| Frictionless, sliding down | $mg\sin\theta$ | $a = g\sin\theta$ |
| With kinetic friction, sliding down | $mg\sin\theta - \mu_k mg\cos\theta$ | $a = g(\sin\theta - \mu_k\cos\theta)$ |
| At rest | static friction balances $mg\sin\theta$ | $a = 0$ |
| Pulled up plane with friction | applied force must overcome gravity component and friction | depends on applied force |
Example of a simple incline calculation
Suppose a block of mass $2.0\ \text{kg}$ is on a frictionless incline at $\theta = 30^\circ$. Its acceleration is
$$a = g\sin 30^\circ$$
Using $g = 9.8\ \text{m/s}^2$ and $\sin 30^\circ = 0.5$,
$$a = 9.8 \times 0.5 = 4.9\ \text{m/s}^2$$
The normal force is
$$N = mg\cos 30^\circ$$
$$N = 2.0 \times 9.8 \times 0.866 \approx 17.0\ \text{N}$$
This example shows the two main ideas, gravity is split into components, and Newton's law is applied along chosen axes.
A useful force diagram
Common mistakes
A common mistake is to use $N = mg$ on a slope. This is only true on a horizontal surface when there are no other vertical forces. On an incline, the normal force is usually $mg\cos\theta$.
Another common mistake is to reverse the sine and cosine components. For an incline angle measured from the horizontal, the component parallel to the plane is $mg\sin\theta$, and the perpendicular component is $mg\cos\theta$.
Students also sometimes assign friction in the wrong direction. Friction always opposes the relative motion, or the tendency of motion, between surfaces.
Always check these three points in incline problems:
$$N \neq mg \text{ in general}$$
$$W_{\parallel} = mg\sin\theta,\quad W_{\perp} = mg\cos\theta$$
Friction opposes motion or attempted motion along the plane.
Final perspective
Inclined planes are important because they turn a vertical gravitational force into a force that partly causes motion and partly presses the object against a surface. The method is always the same, choose axes along and perpendicular to the plane, resolve the weight into components, include friction if present, and then apply Newton's second law along each axis. Once this structure is understood, many mechanics problems become much easier to solve.