A2-5. Frictional Forces

Understanding Friction

Friction is a contact force that acts parallel to the surfaces in contact, opposing relative motion or the tendency of relative motion between them. The magnitude of friction depends heavily on the nature of the contact surfaces (roughness, material properties) and the normal force, but it does not depend on the surface area of contact.

1. Static Friction ($f_s$)

Definition: The force that opposes the tendency of relative motion when the object is at rest relative to the contact surface.

  • Direction: Opposite to the direction the object would move if there were no friction.
  • Magnitude: Static friction is a variable force. It perfectly matches the applied external force ($f_s = F_{applied}$) to keep the net force at zero, up to a certain limit.
  • Maximum Static Friction ($f_{s(max)}$): The exact moment before an object begins to slide, static friction reaches its maximum value. This value is directly proportional to the normal force ($N$).
    $$f_{s(max)} = \mu_s N$$
    Where $\mu_s$ is the coefficient of static friction.

2. Kinetic Friction ($f_k$)

Definition: The force that opposes the actual relative motion between surfaces once the object has started sliding.

  • Direction: Exactly opposite to the direction of the object's relative velocity.
  • Magnitude: Once sliding begins, kinetic friction is approximately constant and is generally slightly less than the maximum static friction.
    $$f_k = \mu_k N$$
    Where $\mu_k$ is the coefficient of kinetic friction. ($\mu_k < \mu_s$)

Friction vs. Applied Force Profile

Applied Force (F) Friction (f) 0 fs(max) = μsN fk = μkN Static Region (v = 0) Kinetic Region (v > 0) fs = F

When the applied force $F$ is less than $f_{s(max)}$, the object is at rest, and the friction perfectly matches the applied force. The moment the applied force exceeds $f_{s(max)}$, the object breaks free and begins to move, and the friction drops to a constant kinetic friction value $f_k$.

Example 1: Range of Force for Equilibrium on an Incline

Problem: An object of weight $W$ rests on a fixed incline angled at $\theta$ to the horizontal. The coefficient of static friction between the object and the incline is $\mu_s$. If an upward force $F$ is applied parallel to the incline to keep the object in equilibrium, what is the required range for $F$?

Solution:

  1. Resolve the weight: The component parallel to the slope is $W \sin\theta$, and the perpendicular component is $W \cos\theta$.
  2. Determine Normal Force and Max Friction: The normal force is $N = W \cos\theta$. Therefore, the maximum static friction is $f_{s(max)} = \mu_s N = \mu_s W \cos\theta$.
  3. Minimum limit of F: To prevent the object from sliding down, $F$ and upward friction must balance the downward weight component. At the limit: $$F + f_{s(max)} \ge W \sin\theta \implies F \ge W \sin\theta - \mu_s W \cos\theta$$
  4. Maximum limit of F: To prevent the object from being pushed up the incline, $F$ must be counteracted by both the downward weight component and downward friction: $$F \le W \sin\theta + f_{s(max)} \implies F \le W \sin\theta + \mu_s W \cos\theta$$

Answer: The force must be within the range: $$W \sin\theta - \mu_s W \cos\theta \le F \le W \sin\theta + \mu_s W \cos\theta$$

⚠ Friction Causing Acceleration (Stacked Blocks)

Friction does not always "slow things down." In cases of relative motion between stacked objects, friction can act as the propelling force.

Imagine Block A rests on top of Block B. If you pull Block B forward with a strong force $F$, and Block A moves forward with it (slipping slightly backwards relative to B), the only horizontal force acting on Block A is the kinetic friction exerted by Block B. This friction acts in the forward direction on A!

$$f_{kA} = \mu_k m_A g \implies a_A = \dfrac{f_{kA}}{m_A} = \mu_k g$$

In this scenario, kinetic friction is the very force causing Block A to accelerate forward.

Example 2: Stacked Cartons and Relative Sliding

Problem: Two cartons A and B, each with a mass of 10.0 kg, are stacked on a smooth, frictionless floor (A is on top of B). The coefficient of static friction between A and B is $\mu_s = 0.40$, and the coefficient of kinetic friction is $\mu_k = 0.30$. A horizontal force $F$ is applied to pull the bottom carton B. ($g = 10\text{ m/s}^2$)

  1. What is the maximum force $F$ that can be applied to move them together quickly without causing A and B to slide relative to each other?
  2. If a force $F = 90\text{ N}$ is applied, what is the acceleration of carton A?

Solution:

  • (1) Maximum Force for No Relative Sliding:
    When A and B move together to the right, the force accelerating A is the static friction $f_s$ exerted by B on A. The maximum acceleration A can have without slipping is when static friction reaches its maximum:
    $$f_{s(max)} = \mu_s m_A g$$
    $$a_{A(max)} = \dfrac{\mu_s m_A g}{m_A} = \mu_s g = 0.40 \times 10 = \mathbf{4.0\text{ m/s}^2}$$
    To achieve this acceleration for the entire system (A + B moving as one unit), the external force $F_{max}$ must be:
    $$F_{max} = (m_A + m_B) \times a_{A(max)} = (10 + 10) \times 4.0 = \mathbf{80\text{ N}}$$
  • (2) Acceleration of A when F = 90 N:
    Since $90\text{ N} > 80\text{ N}$, the force exceeds the maximum limit, meaning the cartons will slide relative to each other. The friction between them becomes kinetic friction.
    $$f_{kA} = \mu_k m_A g$$
    The acceleration of carton A is now purely dictated by this kinetic friction:
    $$a_A = \dfrac{f_{kA}}{m_A} = \mu_k g = 0.30 \times 10 = \mathbf{3.0\text{ m/s}^2}$$

Special Application: Measuring $\mu_s$ with an Inclined Plane

A classic experimental method to find the coefficient of static friction ($\mu_s$) between two materials is to place a block on an adjustable incline and slowly raise the angle until the block just barely begins to slip.

θ N = W cosθ fs W sinθ W cosθ W = mg

Derivation:

  1. While the object is completely at rest:
    The forces parallel to the incline are balanced: $f_s = W \sin\theta$
    The forces perpendicular to the incline are balanced: $N = W \cos\theta$
  2. Inequality of Static Friction:
    We know that static friction cannot exceed its maximum value: $f_s \le f_{s(max)} = \mu_s N$
    Substituting the force equations: $W \sin\theta \le \mu_s (W \cos\theta)$
    Dividing by $W \cos\theta$ yields: $\tan\theta \le \mu_s$
  3. At the critical angle of slipping:
    If we slowly increase the incline angle $\theta$ until the exact moment the block is just about to slide, the static friction has reached its absolute maximum limit. At this critical angle:
    $$\tan\theta = \mu_s$$

By simply taking the tangent of the critical angle where sliding begins, we can experimentally determine the coefficient of static friction without needing to measure mass or exact forces!