Inclined Planes

Place a marble on a flat table. It just sits there. Tilt the table even slightly and the marble starts rolling. Tilt it more — it rolls faster.

Gravity hasn’t changed. It still pulls straight down with the same force. What changed is how that downward pull gets split between two jobs: pulling the marble along the slope (making it slide) and pressing the marble into the slope (making it stay in contact).

Inclined-plane problems show up on nearly every MCAT, and they all reduce to one move: split the weight vector ($mg$) into a piece along the ramp and a piece perpendicular to the ramp. Master that one move and every incline problem becomes plug-and-chug.

Test your prediction below. Raise the angle until the block lets go, then drag the mass slider up and down: every force arrow grows and shrinks with the mass, but the slip point never moves.

Setting Up the Problem

The trick to inclines is to tilt your coordinate axes so one axis runs along the slope and the other runs perpendicular to it. This is unusual — every other problem uses x = horizontal, y = vertical — but here it pays off massively. Why? Because the object only moves along the slope. If your axes are aligned with the slope, “motion direction” and “x-axis” are the same thing, and the math collapses to one dimension.

Once you tilt the axes, gravity is the one force that no longer points along an axis. Gravity always pulls straight down, regardless of which way you’ve drawn your axes. So you have to split gravity into two components:

The Normal Force on an Incline

On a flat surface, the normal force equals the full weight: $N = mg$. On an incline, the surface only has to balance the perpendicular component:

$N = mg\cos\theta$

This means the normal force is always less than $mg$ on an incline. The steeper the slope (bigger $\theta$), the smaller $\cos\theta$, the smaller $N$. At $\theta = 90°$ (a vertical wall), $N = mg \cos 90° = 0$ — a vertical wall provides no support to an object that isn’t held against it.

Frictionless Incline

On a frictionless incline, the only force along the slope is $mg\sin\theta$. Apply Newton’s second law along that direction:

$ma = mg\sin\theta$

The mass cancels out: $a = g\sin\theta$.

Sanity checks (these are great for confirming you wrote the formula right):

• At $\theta = 0$ (flat surface): $a = g \sin 0° = 0$. No acceleration on a flat floor. ✓
• At $\theta = 90°$ (vertical drop): $a = g \sin 90° = g$. Free fall. ✓
• Anywhere in between: $a$ is between 0 and $g$. ✓

Incline With Friction

When friction is in play, it acts along the slope, opposite to the direction of motion (or attempted motion).

Object sliding down the ramp. Gravity pulls it down the slope; friction pushes it up the slope:

$\text{Net} = mg\sin\theta - f_k = mg\sin\theta - \mu_k(mg\cos\theta)$

$a = g(\sin\theta - \mu_k\cos\theta)$

Object pushed up the ramp. Both gravity and friction now point down the slope (friction always opposes motion):

$a = g(\sin\theta + \mu_k\cos\theta) \text{ (this is a deceleration as the object slows on the way up)}$

When Does an Object Stay Put?

An object on a ramp doesn’t slide as long as static friction can balance the down-slope pull of gravity:

$\mu_s N \geq mg\sin\theta$

Substitute $N = mg\cos\theta$:

$\mu_s mg\cos\theta \geq mg\sin\theta \quad\Rightarrow\quad \mu_s \geq \tan\theta$

Mass cancels. The object slips at the critical angle where $\tan\theta = \mu_s$. Below that angle, friction can hold it. Above it, friction can’t keep up and the object starts sliding.

Worked Example

A 10 kg box sits on a 30° ramp. The coefficient of kinetic friction is 0.2. What is the box’s acceleration down the ramp? (Use $g = 10$ m/s²; $\sin 30° = 0.5$; $\cos 30° = 0.87$.)

• $a = g(\sin\theta - \mu_k\cos\theta) = 10(0.5 - 0.2 \times 0.87) = 10(0.5 - 0.174) = 10(0.326) \approx 3.3$ m/s².

Compare to a frictionless 30° ramp: $a = g\sin 30° = 5$ m/s². Adding friction shaved off about 1.7 m/s² of acceleration. As a sanity check: the answer is positive, so the box does slide down (friction wasn’t strong enough to hold it). If you’d gotten a negative number, friction would be strong enough to keep the box stationary, and the actual acceleration would be zero.