Forces on an inclined plane: splitting gravity along and across the slope

Forces on a slope

How steep can a slope get before a block starts to slide, and how fast does it go then?

Tilt the slope and change the mass and friction. Gravity splits into mg sin θ along the slope and mg cos θ into it; friction holds the block until tan θ passes μ, then a = g(sin θ − μ cos θ).

High school
Resolving forces
Normal force
Static vs kinetic friction
Newton's second law

\theta\,(^\circ)

m

2.0

\mu

0.50

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

$mg\sin\theta$: 6.70
$N = mg\cos\theta$: 18.42
$f_{\max} = \mu N$: 9.21
$f$: 6.70
$\tan\theta$: 0.364
$a$: 0.00

Blockstays put

Guided tour

1A gentle slope holds
A $2\,\text{kg}$ block on a $20^\circ$ slope with $\mu=0.5$ . Gravity, $19.6\,\text{N}$ , splits into $mg\sin20^\circ\approx6.70\,\text{N}$ down the slope and $mg\cos20^\circ\approx18.42\,\text{N}$ into it. Friction could hold up to $\mu N\approx9.21\,\text{N}$ , so it only has to supply $6.70\,\text{N}$ and the block stays put.
2The tipping angle
Steepen the slope. The pull $mg\sin\theta$ grows and the friction limit $\mu mg\cos\theta$ shrinks; they cross where $\tan\theta=\mu$ , at about $26.6^\circ$ . At $27^\circ$ the block has just started to slide, with $a\approx0.08\,\text{m/s}^2$ .
3Sliding at 40°
At $40^\circ$ friction is kinetic, $\mu mg\cos\theta$ , and the net force down the slope gives $a=g(\sin\theta-\mu\cos\theta)\approx2.55\,\text{m/s}^2$ .
4Mass cancels
Make the block four times heavier, $8\,\text{kg}$ . Every force grows fourfold (the pull is now $50.39\,\text{N}$ ), but so does the mass being pushed, and the acceleration stays $2.55\,\text{m/s}^2$ .

Explanation

1
Split gravity along the slope
On a slope the useful directions are along it and across it, not horizontal and vertical. Gravity $mg$ makes an angle $\theta$ with the normal to the slope, so it splits into
$G_\parallel=mg\sin\theta,\qquad G_\perp=mg\cos\theta$
The slope pushes back on $G_\perp$ with the normal force $N=mg\cos\theta$ ; only $G_\parallel$ is left to move the block.
2
Static friction only takes what it needs
Static friction is not $\mu N$ . It matches the pull, up to a limit of about $\mu N$ . The block stays put while $mg\sin\theta\le\mu mg\cos\theta$ , that is while
$\tan\theta\le\mu$
This tipping angle does not depend on the mass, which is how $\mu$ is measured in the lab: tilt the board until the block just slides.
3
Once it slides
Sliding friction is $\mu N=\mu mg\cos\theta$ , pointing up the slope. Newton's second law along the slope gives
$a=\frac{mg\sin\theta-\mu mg\cos\theta}{m}=g(\sin\theta-\mu\cos\theta)$
The mass cancels. On a smooth slope ( $\mu=0$ ) this is just $a=g\sin\theta$ .

Check yourself

Q01

A block slides down a smooth $30^\circ$ slope. Its acceleration is ( $g=9.8\,\text{m/s}^2$ )

Q02

A $2\,\text{kg}$ block rests on a $30^\circ$ slope with $\mu=0.8$ . The friction force on it is

Q03

A block slides down a rough slope. If its mass is doubled, its acceleration

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