Derivation

Friction: the Coulomb Model and the Inclined Plane

D-422 Home PU-101 Threads force · energy Depends on Galilean Invariance of Newton's Laws

Statement

For a rigid block of mass \(m\) resting on a rigid plane inclined at angle \(\theta\) to the horizontal, the normal reaction is \(N = mg\cos\theta\). The block remains at rest if and only if \(\tan\theta \le \mu_s\), where \(\mu_s\) is the coefficient of static friction. If \(\tan\theta > \mu_s\), the block slides down with acceleration \(a = g(\sin\theta - \mu_k\cos\theta)\), where \(\mu_k\) is the coefficient of kinetic friction. If the block is pushed up the incline, its acceleration while moving upward is \(a = -g(\sin\theta + \mu_k\cos\theta)\). The work done by kinetic friction over a sliding distance \(d\) is \(W_f = -\mu_k mg\cos\theta\, d\), which is dissipated as thermal energy.

Why it matters

The Coulomb friction model is the workhorse of introductory mechanics. It converts the invisible, distributed microscopic interactions at a contact surface into two simple scalar rules: a maximum static threshold and a constant kinetic drag. Once these rules are internalised, problems involving blocks on inclines, wedges, and stacked objects become tractable algebra rather than unsolvable integrals.

It also provides the canonical example of a non-conservative force. Unlike gravity or spring forces, friction depends on the path taken and the direction of motion, and it irreversibly converts mechanical energy into heat. Understanding this distinction is essential before tackling energy conservation in systems with friction.

Assumptions

The block is a rigid body.If the block deforms significantly (e.g., a soft rubber tire on asphalt), the contact area changes with load and speed, and the simple linear relationship \(f = \mu N\) fails; one must use viscoelastic models.

The plane is rigid and stationary.If the plane itself moves or accelerates, the normal force \(N\) is no longer simply \(mg\cos\theta\); additional terms from the plane's acceleration enter the force balance, and the friction force must be evaluated relative to the plane's motion.

Friction is independent of apparent contact area.This is an empirical approximation (Amontons' first law). At the microscopic level, the real contact area is proportional to the normal load, so the total friction force scales with \(N\). If the surfaces were perfectly flat and the contact area were fixed independently of load, this scaling would not hold.

The coefficient of kinetic friction \(\mu_k\) is independent of sliding speed.In real materials, \(\mu_k\) often depends on velocity (Stribeck effect), temperature, and wear. The Coulomb model assumes \(\mu_k\) is constant, which is a good approximation for moderate speeds and short timescales but breaks down in lubricated or high-speed regimes.

\(\mu_k < \mu_s\).This is typical for dry friction (keeping a block sliding is easier than starting it) though not universal. The formulas above do not actually need it; but when \(\mu_s\) clearly exceeds \(\mu_k\), or \(\mu_k\) falls with speed, sliding becomes jerky (stick-slip: squealing brakes, a bowed violin string) instead of steady.

Derivation

\[ \sum F_{\perp} = 0 \quad\Longrightarrow\quad N - mg\cos\theta = 0 \]

Resolve forces perpendicular to the incline. The block does not accelerate into or out of the plane, so the normal force \(N\) balances the perpendicular component of gravity. A

\[ N = mg\cos\theta \]

Solve for the normal reaction. This is the load pressing the block against the plane. A

\[ F_{\parallel} = mg\sin\theta \]

Resolve gravity parallel to the incline. This is the driving force that tends to pull the block down the slope. A

\[ f_s \le \mu_s N \]

Coulomb's static friction law: the static friction force adjusts to match the applied tangential force, up to a maximum value \(\mu_s N\). B

\[ mg\sin\theta \le \mu_s mg\cos\theta \]

For the block to remain at rest, the driving force \(mg\sin\theta\) must not exceed the maximum static friction \(\mu_s N = \mu_s mg\cos\theta\). B

\[ \tan\theta \le \mu_s \]

Divide both sides by \(mg\cos\theta\) (positive for \(0 \le \theta < 90^\circ\)). This is the condition for static equilibrium. B

\[ \theta_r = \arctan(\mu_s) \]

The angle of repose \(\theta_r\) is the critical angle at which the block just begins to slide. If \(\theta > \theta_r\), static friction is insufficient to hold the block. C

\[ f_k = \mu_k N = \mu_k mg\cos\theta \]

Once sliding begins, kinetic friction takes over. It is constant in magnitude and opposite to the direction of relative motion. A

\[ \sum F_{\parallel} = ma \quad\Longrightarrow\quad mg\sin\theta - f_k = ma \]

Newton's second law along the incline, taking down-slope as positive. Gravity drives the block down; kinetic friction opposes the motion. B

\[ a = g(\sin\theta - \mu_k\cos\theta) \]

Solve for the acceleration. Mass cancels, so the acceleration is independent of the block's mass. B

\[ \sum F_{\parallel} = ma \quad\Longrightarrow\quad -mg\sin\theta - f_k = ma \]

If the block is pushed up the incline, its velocity is up-slope. Kinetic friction opposes the motion, so it acts down-slope, in the same direction as the component of gravity. B

\[ a = -g(\sin\theta + \mu_k\cos\theta) \]

Solve for the acceleration (negative, indicating deceleration). The block slows down more rapidly when moving up than it accelerates when moving down, because both gravity and friction oppose the upward motion. B

\[ W_f = \int_{0}^{d} \vec{f}_k \cdot d\vec{r} = \int_{0}^{d} (-\mu_k mg\cos\theta)\, dx = -\mu_k mg\cos\theta\, d \]

Work done by kinetic friction over a sliding distance \(d\). The force is constant and opposite to the displacement, so the work is negative. This energy is dissipated as heat. C

\[ \Delta K + \Delta U = W_{\text{non-conservative}} = W_f \]

Work-energy theorem with non-conservative forces. The change in mechanical energy (kinetic + potential) equals the work done by friction, which is negative. B

Result

\[ \begin{aligned} &\text{Static:} \quad \tan\theta \le \mu_s \\ &\text{Sliding down:} \quad a = g(\sin\theta - \mu_k\cos\theta) \\ &\text{Sliding up:} \quad a = -g(\sin\theta + \mu_k\cos\theta) \\ &\text{Work by friction:} \quad W_f = -\mu_k mg\cos\theta\, d \end{aligned} \]

Reading. The block remains at rest if the incline angle is below the angle of repose \(\theta_r = \arctan(\mu_s)\). Once sliding, the acceleration down the slope is \(g(\sin\theta - \mu_k\cos\theta)\), and the acceleration while moving up is \(-g(\sin\theta + \mu_k\cos\theta)\). Friction does negative work, dissipating mechanical energy as heat.

Units check. \([g] = \mathrm{m/s^2}\), \([\sin\theta] = [\cos\theta] = 1\), \([\mu_k] = 1\). So \([a] = \mathrm{m/s^2}\). For work: \([\mu_k mg\cos\theta\, d] = \mathrm{N\cdot m = J}\). Consistent.

Limiting cases

Horizontal plane (\(\theta = 0\)). \(N = mg\), \(F_{\parallel} = 0\). The block remains at rest for any \(\mu_s > 0\). If pushed, \(a = -\mu_k g\) (opposing motion).
Vertical plane (\(\theta = 90^\circ\)). \(N = 0\), so \(f_s = f_k = 0\). The block falls freely with \(a = g\). Friction vanishes because there is no normal load.
Frictionless plane (\(\mu_s = \mu_k = 0\)). The block slides down with \(a = g\sin\theta\). No static threshold; any non-zero angle causes sliding.
Angle of repose (\(\theta = \theta_r\)). \(\tan\theta_r = \mu_s\). The block is on the verge of sliding. Any slight perturbation initiates motion.
Strong kinetic friction (\(\mu_k > \tan\theta\)). The down-slope formula gives \(a \lt 0\): a block that is already sliding down slows and stops, after which static friction holds it if \(\tan\theta \le \mu_s\). It never means a block released from rest starts moving up the slope; a block at rest with \(\tan\theta \le \mu_s\) simply stays put.

Breaks when

Lubrication or viscous regime. If a fluid layer separates the surfaces, friction becomes velocity-dependent (Stokes or turbulent drag), and the Coulomb model \(f = \mu N\) fails. The force may scale as \(v\) or \(v^2\) instead of being constant.
Very high sliding speeds. At high speeds, friction generates heat, softening or melting the surfaces, and \(\mu_k\) decreases with temperature. The assumption of constant \(\mu_k\) breaks down.
Strong adhesion or surface chemistry. For clean, flat surfaces in vacuum, adhesion forces can dominate over the normal load: friction persists even at zero load, so \(f\) is no longer proportional to \(N\) and the apparent \(\mu\) can exceed 1. The Coulomb model assumes dry, rough surfaces.
Deformable bodies. For soft materials (rubber, foam), the contact area depends on load and speed, and energy is dissipated through hysteresis. The simple linear \(f \propto N\) relationship does not hold.

Failure modes

Confusing static and kinetic friction. Using \(\mu_k\) to determine whether the block slides. The static threshold \(\tan\theta \le \mu_s\) determines onset of motion; \(\mu_k\) only applies once sliding has begun.
Assuming friction always opposes gravity. Friction opposes relative motion (or impending motion). When the block moves up the incline, friction acts down the incline, in the same direction as the component of gravity. When the block moves down, friction acts up the incline, opposing gravity.
Forgetting that \(N\) changes with \(\theta\). Writing \(N = mg\) instead of \(N = mg\cos\theta\). The normal force is the component of gravity perpendicular to the plane, not the full weight.
Treating friction as conservative. Assuming that the work done by friction over a closed path is zero. It is not: friction always does negative work, dissipating energy. \(\oint \vec{f}_k \cdot d\vec{r} < 0\) for any non-trivial path.

Discussion

The Coulomb model is empirical, not derived from first principles. It captures the essential features of dry friction: a static threshold, a constant kinetic drag, and independence from apparent contact area. These features arise from the microscopic roughness of surfaces, where only a small fraction of the apparent area is in true contact. The real contact area is proportional to the normal load, so the total friction force scales with \(N\). This is Amontons' first law, and it is why the model works so well for engineering purposes.

The distinction between static and kinetic friction is crucial. Static friction is a reactive force: it adjusts to match the applied tangential force, up to a maximum. Kinetic friction is a dissipative force: it is constant in magnitude and always opposes the direction of motion. This is why a block can remain at rest on an incline for a range of angles, but once it starts sliding, it accelerates (or decelerates) with a fixed acceleration determined by \(\mu_k\).

Friction is non-conservative because its work depends on the path. If a block slides down an incline and then back up, the work done by friction is negative in both directions, so the total work over the round trip is negative. This means mechanical energy is not conserved; it is dissipated as heat. This is in contrast to gravity, where the work over a closed path is zero, and mechanical energy is conserved.

At a deeper level, the Coulomb model is a phenomenological description of a complex many-body problem. The microscopic interactions involve asperity deformation, adhesion, and plowing, all of which depend on material properties, surface roughness, and environmental conditions. The model's success is a testament to the universality of the scaling laws that emerge from these complex interactions. However, it is important to recognize the model's limitations: it is an approximation that works well for dry, rigid, non-lubricated surfaces at moderate speeds, but fails in regimes where viscosity, adhesion, or deformation dominate.

Common misconceptions. Friction is not always opposite to the direction of motion; it opposes relative motion (or impending motion) between the surfaces. When a block is pushed up an incline, friction acts down the incline, in the same direction as the component of gravity. Also, friction is not a constant force; static friction adjusts to match the applied force, up to a maximum. And friction is not conservative; it dissipates energy, so mechanical energy is not conserved in systems with friction.

Worked examples

Example 1. A 5.0 kg block rests on an incline of angle \(30^\circ\). The coefficients of friction are \(\mu_s = 0.60\) and \(\mu_k = 0.40\). (a) Does the block slide? (b) If it is given a push up the incline, what is its acceleration while moving up? Take \(g = 9.8\ \mathrm{m/s^2}\).

\[ \tan\theta = \tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577 \]

Compute the tangent of the incline angle. A

\[ \mu_s = 0.60 > 0.577 = \tan\theta \]

Compare \(\tan\theta\) with \(\mu_s\). Since \(\tan\theta < \mu_s\), the block remains at rest. B

\[ a = -g(\sin\theta + \mu_k\cos\theta) \]

Use the acceleration formula for motion up the incline. A

\[ \sin 30^\circ = 0.5, \quad \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \]

Compute the trigonometric values. A

\[ a = -9.8(0.5 + 0.40 \times 0.866) = -9.8(0.5 + 0.3464) = -9.8(0.8464) \]

Substitute the values. A

\[ a = -8.29\ \mathrm{m/s^2} \]

Compute the acceleration. The negative sign indicates deceleration (opposing the upward motion). A

\[ \text{(a) Block remains at rest.} \quad \text{(b) } a = -8.3\ \mathrm{m/s^2} \]

Reading. The block does not slide because the incline angle is below the angle of repose. When pushed up, it decelerates at \(8.3\ \mathrm{m/s^2}\) due to both gravity and friction.

Units check. \([g(\sin\theta + \mu_k\cos\theta)] = \mathrm{m/s^2}\). Correct.

Example 2. A 2.0 kg block slides down a \(45^\circ\) incline with \(\mu_k = 0.30\). It starts from rest and slides a distance of 3.0 m. Find its speed at the bottom and the work done by friction. Take \(g = 9.8\ \mathrm{m/s^2}\).

\[ a = g(\sin\theta - \mu_k\cos\theta) \]

Use the acceleration formula for motion down the incline. A

\[ \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707 \]

Compute the trigonometric values. A

\[ a = 9.8(0.707 - 0.30 \times 0.707) = 9.8(0.707)(1 - 0.30) = 9.8(0.707)(0.70) \]

Substitute the values. A

\[ a = 9.8 \times 0.4949 = 4.85\ \mathrm{m/s^2} \]

Compute the acceleration. A

\[ v^2 = v_0^2 + 2ad = 0 + 2(4.85)(3.0) = 29.1 \]

Use the kinematic equation \(v^2 = v_0^2 + 2ad\). A

\[ v = \sqrt{29.1} = 5.4\ \mathrm{m/s} \]

Compute the final speed. A

\[ W_f = -\mu_k mg\cos\theta\, d = -0.30(2.0)(9.8)(0.707)(3.0) \]

Compute the work done by friction. A

\[ W_f = -0.30 \times 2.0 \times 9.8 \times 0.707 \times 3.0 = -12.5\ \mathrm{J} \]

Compute the numerical value. A

\[ v = 5.4\ \mathrm{m/s}, \quad W_f = -12.5\ \mathrm{J} \]

Reading. The block reaches a speed of \(5.4\ \mathrm{m/s}\) at the bottom. Friction does \(-12.5\ \mathrm{J}\) of work, dissipating energy as heat.

Units check. \([v] = \mathrm{m/s}\), \([W_f] = \mathrm{J}\). Correct.

Problems

A 3.0 kg block rests on a \(20^\circ\) incline. The coefficient of static friction is \(\mu_s = 0.40\). Does the block slide? If not, what is the magnitude of the static friction force? Take \(g = 9.8\ \mathrm{m/s^2}\).
Solution
\(\tan 20^\circ \approx 0.364\). Since \(\tan 20^\circ < \mu_s = 0.40\), the block remains at rest. The static friction force balances the component of gravity down the incline: \(f_s = mg\sin 20^\circ = 3.0 \times 9.8 \times 0.342 = 10.0\ \mathrm{N}\).
A 4.0 kg block slides down a \(30^\circ\) incline with \(\mu_k = 0.25\). Find its acceleration. Take \(g = 9.8\ \mathrm{m/s^2}\).
Solution
\(a = g(\sin\theta - \mu_k\cos\theta) = 9.8(\sin 30^\circ - 0.25\cos 30^\circ) = 9.8(0.5 - 0.25 \times 0.866) = 9.8(0.5 - 0.2165) = 9.8(0.2835) = 2.78\ \mathrm{m/s^2}\).
A 2.5 kg block is pushed up a \(25^\circ\) incline with \(\mu_k = 0.35\). Find its acceleration while moving up. Take \(g = 9.8\ \mathrm{m/s^2}\).
Solution
\(a = -g(\sin\theta + \mu_k\cos\theta) = -9.8(\sin 25^\circ + 0.35\cos 25^\circ) = -9.8(0.4226 + 0.35 \times 0.9063) = -9.8(0.4226 + 0.3172) = -9.8(0.7398) = -7.25\ \mathrm{m/s^2}\).
A 5.0 kg block slides down a \(40^\circ\) incline with \(\mu_k = 0.20\). It starts from rest and slides a distance of 4.0 m. Find its speed at the bottom and the work done by friction. Take \(g = 9.8\ \mathrm{m/s^2}\).
Solution
\(a = g(\sin\theta - \mu_k\cos\theta) = 9.8(\sin 40^\circ - 0.20\cos 40^\circ) = 9.8(0.6428 - 0.20 \times 0.7660) = 9.8(0.6428 - 0.1532) = 9.8(0.4896) = 4.80\ \mathrm{m/s^2}\). \(v^2 = 2ad = 2(4.80)(4.0) = 38.4\), so \(v = \sqrt{38.4} = 6.2\ \mathrm{m/s}\). \(W_f = -\mu_k mg\cos\theta\, d = -0.20(5.0)(9.8)(0.7660)(4.0) = -30.0\ \mathrm{J}\).
A block rests on an incline of angle \(\theta\). The coefficients of friction are \(\mu_s = 0.80\) and \(\mu_k = 0.60\). (a) Find the angle of repose. (b) If the block is released from rest at \(\theta = 50^\circ\), find its acceleration down the incline. (c) If the block is pushed up the incline with an initial speed of 5.0 m/s, find the distance it travels before stopping. Take \(g = 9.8\ \mathrm{m/s^2}\).
Solution
(a) \(\theta_r = \arctan(0.80) = 38.7^\circ\). (b) Since \(50^\circ > 38.7^\circ\), the block slides. \(a = g(\sin 50^\circ - 0.60\cos 50^\circ) = 9.8(0.7660 - 0.60 \times 0.6428) = 9.8(0.7660 - 0.3857) = 9.8(0.3803) = 3.73\ \mathrm{m/s^2}\). (c) When moving up, \(a = -g(\sin 50^\circ + 0.60\cos 50^\circ) = -9.8(0.7660 + 0.3857) = -9.8(1.1517) = -11.29\ \mathrm{m/s^2}\). Using \(v^2 = v_0^2 + 2ad\), with \(v = 0\), \(v_0 = 5.0\ \mathrm{m/s}\), \(a = -11.29\ \mathrm{m/s^2}\): \(0 = 25.0 + 2(-11.29)d\), so \(d = 25.0 / 22.58 = 1.11\ \mathrm{m}\).