= Static friction on an inclined plane
{title2=$\tan\theta_c=\mu_s$}
An <inclined plane> tilted slowly from horizontal turns a <coefficient of static friction> into a critical <angle>. Assume a nonrotating block on a stationary, rigid, dry surface, with <gravitational weight> and the contact <forces> as its only <forces>. Take $0\le\theta<\pi/2$ to be the <angle> between the uphill surface direction and horizontal. Let $\mathbf e_t$ be a unit <vector> pointing uphill and $\mathbf e_n$ a perpendicular unit <vector> pointing outward from the surface. Resolving the <gravitational weight> gives
$$
\mathbf W=-mg\sin\theta\,\mathbf e_t-mg\cos\theta\,\mathbf e_n.
$$
The <angle> between the downward vertical and the inward normal $-\mathbf e_n$ is also $\theta$. Thus the inward component uses <cosine> and the downhill component uses <sine>. These are components of one <force>, not two additional <forces>.
\Image[/static-friction-incline-onset-geometry.png]
{title=<Inclined plane> <force balance> and the geometry of impending downhill <contact slip>}
{description=Original diagram at the stipulated critical <angle> $\theta_c=\arctan(0.40)\simeq21.80^\circ$. Left: the three actual <forces> on a resting block. Right: a separate decomposition of <gravitational weight> into downhill and inward-normal components; the dashed diagonal represents their sum. The two panels must not be combined into a diagram with five applied <forces>.}
Before downhill <contact slip>, zero <acceleration> and <Newton's second law> require
$$
f_s=mg\sin\theta,\qquad N=mg\cos\theta.
$$
The required <static friction> points uphill. Substituting into the bound and dividing by the positive $mg\cos\theta$ gives
$$
mg\sin\theta\le\mu_smg\cos\theta
\quad\Longleftrightarrow\quad
\tan\theta\le\mu_s.
$$
The <tangent function> increases on this range of <angles>, so the last admissible resting <angle> is
$$
\theta_c=\arctan\mu_s,\qquad \tan\theta_c=\mu_s.
$$
The <inverse tangent> takes a <dimensionless quantity> to an <angle>. At equality the ideal model still permits a limiting resting state; increasing $\theta$ further excludes it. At a smaller <angle>, $f_s/N=\tan\theta$ is the demand ratio, not the <coefficient of static friction>.<Friction (Physics 183, Experiment 5, Fall 2004)>[Printed p. 1, theory and Figure 2; p. 5, Part III; worksheet p. 2, Table 3]{ref}
For the stipulated $m=2\,\mathrm{kg}$, $g=9.8\,\mathrm{m/s^2}$ and $\mu_s=0.40$, the critical <angle> is approximately $21.8014^\circ$. At $\theta=15^\circ$, $N\simeq18.932\,\mathrm N$ and $f_s\simeq5.073\,\mathrm N$, comfortably below the bound $\mu_sN\simeq7.573\,\mathrm N$. At $\theta=25^\circ$, maintaining rest would require $f_s\simeq8.283\,\mathrm N$, but the bound is only $7.105\,\mathrm N$: rest is impossible. At $\theta_c$, $N\simeq18.198\,\mathrm N$ and $f_s\simeq7.279\,\mathrm N$. Notice that the limiting tangential <force> here is smaller than the horizontal $7.84\,\mathrm N$ limit because tilting reduces the <normal force>.
Cancellation of <mass> makes this relation useful: the ideal critical <angle> tests a contact coefficient without needing a calibrated <force> gauge. Doubling the block's <mass> doubles both demanded <static friction> and its bound, leaving $\theta_c$ unchanged if $\mu_s$ is unchanged. This is a testable model prediction. It need not survive a change in surface condition, a direction-dependent coefficient, appreciable deformation, or vibration.
The derivation also assumes continued contact and that sliding occurs before tipping or <rolling without slipping>. The resultant contact <force> must act within the supporting footprint so that the block can satisfy <torque> balance; a tall block may tip first. No extra <tension>, imposed tangential pull, or lift is included. In a slowly raised apparatus, <quasistatic process> means that motion of the board contributes negligible inertial <forces> while locating the threshold. Once downhill <contact slip> begins, a constant-<velocity> sliding <angle> would concern <kinetic friction>, not the onset of <static friction>.
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