Lessons · dynamics

Down a ramp: Newton's second law with friction

Regime 1 — the algebra is the calculus, evaluated. Step the algebra, step the calculus, and watch the algebra formula fall out of the integral. SymPy proves the two registers agree.

Regime 1 · algebra is calculus, evaluated Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

A block is released from rest on a $\theta = 30°$ incline with kinetic friction $\mu = 0.2$ . Resolving forces along the slope gives a constant acceleration $a = g(\sin\theta - \mu\cos\theta)$ — and crucially, the mass cancels. From there the motion is constant-acceleration kinematics: $v = at$ and $x = \tfrac12 at^2$ , which are just the integrals of that constant $a$ . Drag the angle and the friction coefficient and watch all three panels steepen together.

Acceleration

a = g(\sin\theta - \mu\cos\theta)

3.268 m/s²

Frictionless acceleration

g\sin\theta

5 m/s²

Speed after 2 s

6.536 m/s

Distance in 2 s

6.536 m

The acceleration is constant — integrate it once for the velocity

v(t) = \int_0^t a\,dt' = at

Because $a = g(\sin\theta - \mu\cos\theta)$ does not change with time, its integral is just $at$ . The velocity rises in a straight line — its slope is $a$ .

check $v = \int a\,dt$ ; check $x = \int v\,dt$ ; show $v^2 = 2ax$ ; check the $\mu=0$ limit $g\sin\theta$

✓ The velocity is the integral of the (constant) acceleration: $v = \int_0^t a\,dt' = at$ . [structural]
✓ The position is the integral of the velocity: $x = \int_0^t v\,dt' = \tfrac12 at^2$ — the memorized formula is the quadrature. [structural]
✓ The velocity–position relation falls out: $v^2 = 2ax$ (no time needed). [structural]
✓ With no friction ( $\mu = 0$ ) the acceleration is the bare gravity component $g\sin\theta$ . [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “A heavier block slides down the ramp faster, because gravity pulls on it harder.”

It does not — two blocks of different mass slide down the same ramp with the same acceleration. Gravity does pull harder on the heavier block ( $mg\sin\theta$ is larger), but that same larger mass is harder to accelerate (the $m$ in $F = ma$ ), and friction also scales with the normal force $\mu mg\cos\theta$ . Every term carries a factor of $m$ , so it cancels: $a = g(\sin\theta - \mu\cos\theta)$ depends only on the angle and the friction coefficient. This is the same reason all objects fall at the same rate — Galileo's insight, on a ramp.

Modeling assumptions — author-asserted, disclosed not discharged

A rigid block on a planar incline; kinetic friction acts (the block is already moving), with a constant coefficient $\mu$ and the normal force $N = mg\cos\theta$ .
$g = 10$ m/s² as a magnitude; the block is released from rest and the incline is long enough that it keeps sliding (we take $\tan\theta > \mu$ , so $a > 0$ ).

The stacked graph, fully annotated

A static rendering (Matplotlib) at the default parameters — the interactive version is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):