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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.

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 givesThe 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.

Before downhill contact slip, zero acceleration and Newton's second law requireThe required static friction points uphill. Substituting into the bound and dividing by the positive $mg\cos \theta$ givesThe tangent function increases on this range of angles, so the last admissible resting angle isThe 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.^[ref]

For the stipulated $m=2\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$ and ${\mu }_s=0.40$, the critical angle is approximately $21.8014^{\circ }$. At $\theta =15^{\circ }$, $N\simeq 18.932\mathrm{N}$ and $f_s\simeq 5.073\mathrm{N}$, comfortably below the bound ${\mu }_{s}N\simeq 7.573\mathrm{N}$. At $\theta =25^{\circ }$, maintaining rest would require $f_s\simeq 8.283\mathrm{N}$, but the bound is only $7.105\mathrm{N}$: rest is impossible. At ${\theta }_c$, $N\simeq 18.198\mathrm{N}$ and $f_s\simeq 7.279\mathrm{N}$. 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.