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Friction Models
Friction opposes relative motion (or the tendency for relative motion) between surfaces.
Static friction (\(f_s\)) prevents sliding: \[f_s \leq \mu_s N\]
- Adjusts from zero up to a maximum value
- Maximum occurs just before slipping begins
- \(\mu_s\) is the coefficient of static friction
Kinetic friction (\(f_k\)) acts during sliding: \[f_k = \mu_k N\]
- Constant value during sliding
- \(\mu_k\) is the coefficient of kinetic friction
- Generally \(\mu_k < \mu_s\)
Static friction is not always equal to \(\mu_s N\). It matches whatever force is needed to prevent motion, up to the maximum.
Interactive: Static vs Kinetic Friction
As applied force increases, friction responds:
Inclined Planes
On an inclined plane, weight must be resolved into components:
Parallel to slope (causes sliding tendency): \[W_{\parallel} = mg\sin\theta\]
Perpendicular to slope (determines normal force): \[W_{\perp} = mg\cos\theta\]
When no other perpendicular forces act: \[N = W_{\perp} = mg\cos\theta\]
Interactive: Force Resolution on a Slope
A block on a 30° incline with weight components:
Key insight: On a steeper slope, \(W_{\parallel}\) increases and \(W_{\perp}\) decreases.
Net Force on a Slope
For an object sliding down with kinetic friction:
\[F_{net} = mg\sin\theta - f_k = mg\sin\theta - \mu_k mg\cos\theta\]
\[a = g(\sin\theta - \mu_k\cos\theta)\]
Interactive: Sliding Down with Friction
A block accelerating down a slope (friction opposes motion):
Net force down slope = \(20.7 - 8.9 = 11.8\) N
Acceleration = \(11.8 / 5 = 2.4\) m/s²
Angle of Repose
The angle of repose is the maximum angle before an object begins to slide:
\[\tan\theta_{max} = \mu_s\]
At this angle, \(mg\sin\theta = \mu_s mg\cos\theta\), so slipping is about to begin.
Worked Examples
Example 1: Components on a slope
A 5.0 kg block rests on a 25° incline.
Solution:
Weight: \(W = mg = 5.0 \times 9.8 = 49\) N
Parallel component: \(W_{\parallel} = 49 \times \sin25° = 49 \times 0.423 = 20.7\) N
Perpendicular component: \(W_{\perp} = 49 \times \cos25° = 49 \times 0.906 = 44.4\) N
Normal force: \(N = W_{\perp} = 44.4\) N
Example 2: Sliding with kinetic friction
A 3.0 kg block slides down a 20° incline with \(\mu_k = 0.25\).
Solution:
Normal force: \(N = mg\cos\theta = 3.0 \times 9.8 \times \cos20° = 27.6\) N
Kinetic friction: \(f_k = \mu_k N = 0.25 \times 27.6 = 6.9\) N
Weight component down slope: \(W_{\parallel} = mg\sin20° = 3.0 \times 9.8 \times \sin20° = 10.1\) N
Net force: \(F_{net} = 10.1 - 6.9 = 3.2\) N down slope
Acceleration: \(a = \frac{F_{net}}{m} = \frac{3.2}{3.0} = 1.1\) m/s²
Example 3: Static friction threshold
A 4.0 kg block on a 15° incline has \(\mu_s = 0.40\). Will it slip?
Solution:
Force trying to cause slipping: \(W_{\parallel} = mg\sin15° = 4.0 \times 9.8 \times \sin15° = 10.2\) N
Maximum static friction:
- \(N = mg\cos15° = 4.0 \times 9.8 \times \cos15° = 37.9\) N
- \(f_{s,max} = \mu_s N = 0.40 \times 37.9 = 15.2\) N
Since \(W_{\parallel} (10.2\text{ N}) < f_{s,max} (15.2\text{ N})\), the block does not slip
Actual static friction: \(f_s = W_{\parallel} = 10.2\) N (just enough to prevent motion)
Example 4: Pulling up an incline
A 7.0 kg crate is pulled up a 25° incline at constant speed with \(\mu_k = 0.20\). Find the pulling force (parallel to slope).
Solution:
At constant speed, \(F_{net} = 0\)
Forces parallel to slope:
- Pull force F (up slope)
- Weight component (down slope): \(W_{\parallel} = 7.0 \times 9.8 \times \sin25° = 29.0\) N
- Kinetic friction (down slope, opposes motion): \(f_k = \mu_k N\)
Normal force: \(N = mg\cos25° = 7.0 \times 9.8 \times \cos25° = 62.2\) N
Kinetic friction: \(f_k = 0.20 \times 62.2 = 12.4\) N
Equilibrium: \(F = W_{\parallel} + f_k = 29.0 + 12.4 = 41.4\) N