Forces and Friction

The model. Friction is the contact force between two surfaces in the direction along the contact. It can take any value in the range

$0 \le F \le \mu R,$

where $\mu$ (mu) is the coefficient of friction (dimensionless) and $R$ is the normal reaction.

Three regimes:

1. Static, not at limit ($F < \mu R$). The particle is at rest; friction takes whatever value is needed to keep it in equilibrium.

2. Static, at limit / about to slip ($F = \mu R$). On the point of moving. Limiting friction.

3. Sliding ($F = \mu R$). Particle is moving; friction is at its maximum.

For exam purposes, regimes 2 and 3 are identical algebraically: set $F = \mu R$.

Direction: friction OPPOSES the direction of motion (or the tendency of motion).

Examples:

• Block sliding to the right $\to$ friction acts to the LEFT.
• Block on slope, about to slide DOWN $\to$ friction acts UP the slope.
• Block on slope, pushed UP and about to move $\to$ friction acts DOWN the slope.

Practical takeaway: in problems where a particle is moving (or about to), the formula $F = \mu R$ is used. In static-with-room-to-spare problems, $F$ is determined by the OTHER forces.

Setup. Particle of mass $m$ on a rough horizontal floor with coefficient of friction $\mu$.

Normal reaction: $R = mg$ (if no vertical applied force). With a vertical component $P_y$ in the applied force: $R = mg - P_y$ (if $P_y$ acts UP) or $R = mg + P_y$ (if down).

Maximum friction: $\mu R = \mu mg$.

Standard problem types:

1. Push to determine if it moves. Compare push $P$ with $\mu R$. If $P > \mu R$, it moves; if $P \le \mu R$, it stays.

2. Block sliding, find stopping distance. $a = -\mu g$ (deceleration). Use $v^{2} = u^{2} + 2as$ with $v = 0$:

$s = \dfrac{u^{2}}{2\mu g}.$

3. Push at an angle, find acceleration. Resolve push into components. Vertical: $R = mg - P\sin\theta$ (if $P$ pulls UP). Horizontal: $P\cos\theta - \mu R = m a$.

Worked example. Block of mass $4$ kg moving at $8$ m/s on rough floor, $\mu = 0.3$. Find stopping distance.

• $R = 4g = 39.2$ N. $F = 0.3 \cdot 39.2 = 11.76$ N.
• $a = -11.76/4 = -2.94\,\text{m s}^{-2}$ (deceleration).
• $0 = 64 - 2 \cdot 2.94 \cdot s \Rightarrow s = 64/5.88 \approx 10.88$ m.

Or directly: $s = u^{2}/(2\mu g) = 64/(2 \cdot 0.3 \cdot 9.8) \approx 10.88$ m.

Setup. Particle on a rough plane inclined at $\theta$ to the horizontal. Coefficient of friction $\mu$.

Choose axes along and perpendicular to the slope.

Weight components:

• Perpendicular (into slope): $mg\cos\theta$.
• Parallel (down slope): $mg\sin\theta$.

Perpendicular equilibrium:

$R = mg\cos\theta.$

Limiting friction:

$\mu R = \mu mg\cos\theta.$

Will it slide down? Compare:

• $mg\sin\theta$ — driving force (gravity along slope).
• $\mu mg\cos\theta$ — max friction (opposing motion).

Slides if $mg\sin\theta > \mu mg\cos\theta$, i.e. $\tan\theta > \mu$.

Stays if $\tan\theta \le \mu$.

The critical angle $\theta = \arctan\mu = \lambda$ is the angle of friction.

Acceleration when sliding down:

$a = g(\sin\theta - \mu\cos\theta).$

Acceleration when sliding up (driven by applied force, friction acts DOWN slope):

$a = \dfrac{P_{\parallel} - mg\sin\theta - \mu mg\cos\theta}{m},$

where $P_{\parallel}$ is the component of the applied force along the slope.

Worked example. Particle of mass $2$ kg on $30^{\circ}$ rough slope, $\mu = 0.2$.

• $R = 2 \cdot 9.8 \cdot \cos 30^{\circ} \approx 16.97$ N.
• $\mu R \approx 3.39$ N. $mg\sin\theta = 2 \cdot 9.8 \cdot 0.5 = 9.8$ N.
• $9.8 > 3.39$ — slides.
• $a = 9.8(\sin 30^{\circ} - 0.2 \cos 30^{\circ}) = 9.8(0.5 - 0.1732) \approx 3.20\,\text{m s}^{-2}$ down slope.

Definition. The angle of friction $\lambda$ is the maximum angle of incline at which a particle can rest in equilibrium on a rough surface without an applied force.

From the slide condition: at the critical angle, $\tan\theta = \mu$, i.e.

$\lambda = \arctan\mu, \qquad \tan\lambda = \mu.$

Geometric interpretation. At limiting friction, the total contact force $\sqrt{R^{2} + F^{2}}$ makes angle $\lambda$ with the normal to the surface. So the total contact force can be inclined up to $\lambda$ from the normal direction.

Worked example. $\mu = 0.3$: $\lambda = \arctan 0.3 \approx 16.7^{\circ}$. A particle on a $16.7^{\circ}$ rough slope is on the point of sliding.

Useful for range-of-$P$ problems on rough inclines: the applied force at the limit must point within $\lambda$ either side of the slope-normal, otherwise sliding occurs.

Three-force equilibrium with friction. When a particle is in equilibrium on a rough incline under an applied force, three forces act: weight, total contact force (at angle $\le \lambda$ from normal), applied force. The triangle of forces / Lami's theorem applies — with the contact force replacing the separate $R$ and $F$.

Setup. $A$ on rough horizontal table (coefficient $\mu$); $B$ hanging vertically; connected by light inextensible string over smooth pulley.

Force diagrams:

• $A$: weight $m_A g$ down, normal reaction $R$ up, tension $T$ horizontal toward pulley, friction $F$ along table OPPOSING motion (so directed AWAY from pulley).
• $B$: tension $T$ up, weight $m_B g$ down.

Equations:

• Vertical on $A$: $R = m_A g$.
• Friction: $F = \mu R = \mu m_A g$ (assuming motion).
• Horizontal on $A$ (toward pulley positive): $T - \mu m_A g = m_A a$ (1).
• Vertical on $B$ (down positive): $m_B g - T = m_B a$ (2).

Add:

$m_B g - \mu m_A g = (m_A + m_B) a \Rightarrow a = \dfrac{(m_B - \mu m_A) g}{m_A + m_B}.$

Pre-check 'does it move?': needs $m_B g > \mu m_A g$, i.e. $m_B > \mu m_A$.

Worked example. $m_A = 4$, $m_B = 3$, $\mu = 0.2$:

• $m_B g = 29.4$, $\mu m_A g = 7.84$. Moves ($29.4 > 7.84$).
• $a = (29.4 - 7.84)/7 \approx 3.08\,\text{m s}^{-2}$.
• $T = m_A a + \mu m_A g = 4 \cdot 3.08 + 7.84 \approx 20.16$ N.