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A university laboratory places an aluminium block on a Teflon-lined wooden board. For its inclined plane stage, the pulley and block string are removed; a rope and hook connect the board to a crank. The block is placed gently near the sheet's middle without prior sliding. Smooth cranking raises the board; a $0^{\circ }$–$45^{\circ }$ protractor reads the smallest angle at which sliding begins. Four readings are taken without added load, then four each with $200$, $400$, and $800\mathrm{g}$ added symmetrically at two screws. Removing the string isolates gravity-driven onset; gentle placement and smooth raising avoid premature disturbance.^[ref]

Infer ${\mu }_s=\tan {\theta }_c$ under the assumptions of static friction on an inclined plane. Repetition tests onset scatter; added loads test coefficient constancy. No measured onset values are supplied here.

For a replication, additionally check the protractor zero with a level board, retain the same contact face and surface region, and observe whether tipping or rolling precedes sliding. These are proposed additional controls. A finite angle resolution and the disturbance needed to detect first motion limit the estimate; a slow ramp does not by itself establish perfectly rigid, dry, or direction-independent contact. At the model's ${\mu }_s=0.40$ threshold, differentiating the tangent function gives $d{\mu }_s/d\theta =1+{\mu }_s^2=1.16$, with $\theta$ in radians. A hypothetical $0.5^{\circ }$ measurement error would therefore correspond to approximately $0.0101$ in the inferred coefficient at first order. This sensitivity calculation is not an uncertainty reported by the laboratory.

First test rest. With no other tangential forces, the required uphill static friction is $f_s=mg\sin \theta$. Its maximum is ${\mu }_{s}N$, where ${\mu }_s$ is the coefficient of static friction. ThusThe tangent function is increasing on this angle range, and its inverse tangent determines the last permitted resting angle ${\theta }_c$. At equality, the ideal model still permits limiting rest; a larger angle excludes it. Below the threshold, setting $f_s={\mu }_{s}N$ would incorrectly impose the maximum static friction even when less is required. Static friction on an inclined plane and inclined-plane measurement of static friction develop this onset criterion further.

For a block already sliding downhill, uphill kinetic friction has magnitude $f_k={\mu }_{k}N$, where ${\mu }_k$ is the coefficient of kinetic friction. Taking downhill as positive, Newton's second law becomesEvery term in the force balance scales with mass; division by $m>0$ removes it. Doubling mass doubles both gravitational weight and the contact forces, while leaving this acceleration unchanged if the coefficients remain unchanged. A mass-independent result is a prediction of these assumptions, rather than a universal property of every ramp.

For an original numerical model, stipulate $m=3\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$, ${\mu }_s=0.45$ and ${\mu }_k=0.30$. The critical angle is ${\theta }_c\simeq 24.23^{\circ }$. At $20^{\circ }$, the normal force is $27.627\mathrm{N}$; rest requires $10.055\mathrm{N}$ of uphill static friction, below its $12.432\mathrm{N}$ bound. A resting block can stay there. If it is already sliding downhill, however, the kinetic friction is only $8.288\mathrm{N}$, giving downhill acceleration $a\simeq 0.5891m/{\mathrm{s}}^2$. Specifying the initial state matters.

At $30^{\circ }$, the downhill component of gravitational weight is $14.700\mathrm{N}$, while $N\simeq 25.461\mathrm{N}$ and the static friction bound is only $11.458\mathrm{N}$. Rest is impossible. During downhill sliding,Released from rest on a sufficiently long ramp, this ideal block covers its first $1.00\mathrm{m}$ in $t=\sqrt{2L/a}\simeq 0.9218\mathrm{s}$, reaching speed $v=\sqrt{2aL}\simeq 2.1697m/s$. As an independent energy calculation, the lost gravitational potential energy is $mgL\sin \theta =14.700\mathrm{J}$ and kinetic friction removes $f_{k}L\simeq 7.638\mathrm{J}$ of mechanical energy. The remaining $7.062\mathrm{J}$ equals $m{v}^2/2$ by the work-energy theorem.

At $\theta =\arctan {\mu }_k$, downhill sliding has zero acceleration and therefore constant velocity. Below that angle, a downhill-moving block slows; when it stops, test the static friction bound again. For uphill sliding, kinetic friction points downhill, so the signed downhill acceleration instead is $g(\sin \theta +{\mu }_k\cos \theta )$. The sign follows contact slip, and the downhill formula cannot be used unchanged after a reversal.