Lesson 3.4: Friction on Rough Surfaces

Introduction

In this lesson, we will explore the concept of friction on rough surfaces, a critical element in understanding how forces affect the motion of objects. We will learn about the friction model, including the coefficient of friction and the normal reaction force. By the end of the lesson, you should be able to analyze situations involving friction, identify whether a body is in equilibrium or about to move, and apply the necessary equations to solve related problems.

Learning Objectives

Understand the friction model: friction is at most $\mu R$, where $\mu$ is the coefficient of friction and $R$ is the normal reaction.
Recognize limiting equilibrium and the point at which a body is on the verge of moving.
Analyze the motion of a body on a rough horizontal surface and a rough inclined plane.
Apply the inequality $F \leq \mu R$ and use $F = \mu R$ in limiting equilibrium or when sliding.
Determine if a body remains in equilibrium or begins to move on a rough surface.

Section 1: The Friction Model

What is Friction?

Friction is a force that opposes the relative motion of two surfaces in contact. It plays a vital role in everyday activities such as walking or driving.

The Friction Formula

The frictional force ($F_f$) can be modeled as:

$$\ F_f \leq \mu R$$

Where:

$F_f$ is the frictional force.
$\mu$ is the coefficient of friction, which depends on the materials in contact.
$R$ is the normal reaction force, which acts perpendicular to the contact surface.

The frictional force can take on two forms: static friction (when the body is at rest) and kinetic friction (when the body is sliding). The coefficient of static friction ($\mu_s$) is typically greater than the coefficient of kinetic friction ($\mu_k$).

Worked Example 1: Static Friction

Problem: A block of mass $m = 10 \, \text{kg}$ rests on a horizontal surface. If the coefficient of static friction $\mu_s = 0.5$, determine the maximum force that can be applied to the block before it starts moving.

Solution:

First, we calculate the normal reaction force, $R$. Since the block is on a horizontal surface, the normal force is equal to the weight of the block:

$$R = mg = 10 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 98.1 \, \text{N}$$

Next, we calculate the maximum static frictional force:

$$F_{f_{max}} = \mu_s R = 0.5 \times 98.1 \, \text{N} = 49.05 \, \text{N}$$

Thus, the maximum force that can be applied to the block before it starts moving is $49.05 \, \text{N}$.

Section 2: Limiting Equilibrium

What is Limiting Equilibrium?

Limiting equilibrium occurs when an object is on the verge of moving. At this point, the applied force is equal to the maximum static frictional force, and any additional force will result in motion.

The Condition of Limiting Equilibrium

In limiting equilibrium, we have:

$$F = \mu_s R$$

Where $F$ is the applied force. This equation shows that at the threshold of motion, the force applied is equal to the maximum frictional force.

Worked Example 2: Limiting Equilibrium

Problem: A block of mass $m = 15 \, \text{kg}$ is on a rough horizontal surface with a coefficient of static friction $\mu_s = 0.4$. Determine the force required to start moving the block.

Solution:

Calculate the normal reaction force:

$$R = mg = 15 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 147.15 \, \text{N}$$

Now calculate the maximum static friction:

$$F_{f_{max}} = \mu_s R = 0.4 \times 147.15 \, \text{N} = 58.86 \, \text{N}$$

Therefore, the force required to start moving the block is $58.86 \, \text{N}$.

Section 3: Motion on a Rough Surface

Forces Acting on a Body

When a body is on a rough surface, several forces act upon it:

The weight ($W = mg$) acting downwards.
The normal reaction force ($R$) acting upwards.
The frictional force ($F_f$) acting in the opposite direction of the applied force.

Analyzing Motion

To analyze a body on a rough horizontal surface experiencing an applied force, set up the following equations based on Newton's second law, $F = ma$:

In the horizontal direction:

$$F - F_f = ma$$

In the vertical direction:

$$R - mg = 0$$

Motion on a Rough Horizontal Surface

When analyzing motion on a rough horizontal surface, it is essential to determine whether the body will slide or remain stationary based on the balance of forces and the coefficient of friction.

Worked Example 3: Motion Analysis

Problem: A box of mass $m = 20 \, \text{kg}$ is on a rough horizontal surface with a coefficient of kinetic friction $\mu_k = 0.3$. A horizontal force of $F = 60 \, \text{N}$ is applied. Will the box move?

Solution:

First, calculate the normal reaction force:

$$R = mg = 20 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 196.2 \, \text{N}$$

Now calculate the kinetic friction:

$$F_f = \mu_k R = 0.3 \times 196.2 \, \text{N} = 58.86 \, \text{N}$$

Now analyze the horizontal forces:

$$F - F_f = ma \Rightarrow 60 \, \text{N} - 58.86 \, \text{N} = 20 \, \text{kg} \cdot a$$

$$1.14 \, \text{N} = 20 \, \text{kg} \cdot a$$

Thus,

$$a = \frac{1.14}{20} \approx 0.057 \, \text{m/s}^2$$

Since the applied force exceeds the kinetic friction, the box will indeed move with an acceleration of approximately $0.057 \, \text{m/s}^2$.

Section 4: Motion on a Rough Inclined Plane

Analyzing Forces on an Incline

When dealing with an inclined plane, the forces acting on the body must be resolved into components. These can be separated into parallel and perpendicular components related to the surface of the incline.

Resolving Forces

The weight of the body can be resolved into:

Perpendicular to the incline: $W_{\perpendicular} = mg \cos(\theta)$
Parallel to the incline: $W_{\parallel} = mg \sin(\theta)$

The normal reaction force ($R$) acts perpendicular to the slope:

$$R = mg \cos(\theta)$$

The frictional force ($F_f$) is again $\mu R$ where $\mu$ is the coefficient of friction.

Worked Example 4: Motion on an Incline

Problem: A block of mass $m = 5 \, \text{kg}$ is on a frictional inclined plane ($\mu = 0.2$) inclined at $30^\circ$. Determine if the block will slide down the incline.

Solution:

Calculate the weight components:

Perpendicular to incline:

$$R = mg \cos(30^\circ) = 5 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times \frac{\sqrt{3}}{2} \approx 42.43 \, \text{N}$$

Parallel to incline:

$$W_{\parallel} = mg \sin(30^\circ) = 5 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times \frac{1}{2} = 24.52 \, \text{N}$$

Now, calculate the frictional force:

$$F_f = \mu R = 0.2 \times 42.43 \, \text{N} \approx 8.49 \, \text{N}$$

Compare the parallel component and the friction:

Gravitational force down the incline: $24.52 \, \text{N}$
Frictional force up the incline: $8.49 \, \text{N}$

The net force down the incline will be:

$$F_{\text{net}} = W_{\parallel} - F_f = 24.52 \, \text{N} - 8.49 \, \text{N} = 16.03 \, \text{N}$$

Since $F_{\text{net}} > 0$, the block will slide down the incline with a net force of approximately $16.03 \, \text{N}$.

Conclusion

In this lesson, we have covered the essential concepts of friction on rough surfaces. We discussed the friction model, limiting equilibrium, motion on rough horizontal surfaces, and motion on inclined planes. Understanding these principles allows us to analyze real-world situations where friction plays a crucial role. Through worked examples, we learned how to perform calculations that determine the state of motion for various objects under the influence of friction.

Study Notes

Friction opposes motion between surfaces in contact.
The maximum frictional force is given by $F_f \leq \mu R$.
Limiting equilibrium occurs when $F = \mu_s R$.
Analyze forces both vertically and horizontally for bodies in motion.
On inclined planes, resolve weight into components to apply friction accordingly.