Lesson 2.5: Friction, Drag and Terminal Velocity

Introduction

Welcome, students! In this lesson, we're going to explore the fascinating topics of friction, drag, and terminal velocity. 😃 Our objectives today are to understand the concepts of static and kinetic friction, motion on rough surfaces, drag forces, and how these relate to terminal velocity. This lesson will equip you with the knowledge to apply these principles practically!

Learning Objectives

By the end of this lesson, you should be able to:

• Understand static and kinetic friction and the coefficient of friction.
• Analyze motion on rough horizontal and inclined surfaces.
• Explain drag forces and their dependence on speed.
• Comprehend terminal velocity as the balance of weight and drag.
• Resolve forces on an inclined plane including friction.

Understanding Friction

Friction is the force that opposes the relative motion of two surfaces in contact. It plays a crucial role in our daily lives. Imagine sliding a book across a table. The force you exert to move the book is opposed by the friction between the book and the table. There are two main types of friction:

1. Static Friction ($f_s$): This is the frictional force that needs to be overcome to start moving an object.
2. Kinetic Friction ($f_k$): This occurs when an object is already in motion.

The formula for static friction is:

$$ f_s \leq \mu_s \cdot N $$

where:

• $f_s$ is the static friction force
• $\mu_s$ is the coefficient of static friction
• $N$ is the normal force (the perpendicular force exerted by a surface)

For kinetic friction, the equation is:

$$ f_k = \mu_k \cdot N $$

where:

• $f_k$ is the kinetic friction force
• $\mu_k$ is the coefficient of kinetic friction

Example of Friction

Suppose you want to push a heavy box across the floor. The box has a normal force of 100 N pressing down due to its weight, and the coefficient of kinetic friction between the box and the floor is 0.4.

• Calculate the kinetic friction:

$$ f_k = \mu_k \cdot N = 0.4 \cdot 100 \text{ N} = 40 \text{ N} $$

This means that you must apply a force greater than 40 N to move the box!

Motion on Rough Surfaces

When an object moves along a rough surface, friction is acting against its motion. The net force can be expressed as:

$$ F_{\text{net}} = F_{\text{applied}} - f_k $$

As per Newton’s second law ($F = ma$), we can relate this to the acceleration ($a$) of the object:

$$ ma = F_{\text{applied}} - \mu_k \cdot N $$

In this equation, any unbalanced forces affect the object's acceleration or deceleration. If the applied force exceeds the frictional force, the object will accelerate!

Example of Motion on Rough Surfaces

If you push a sled with a force of 60 N on a rough surface where the coefficient of kinetic friction is 0.3 and the sled has a normal force of 50 N, calculate its acceleration.

1. Calculate the friction:

$$ f_k = 0.3 \cdot 50 \text{ N} = 15 \text{ N} $$

1. Find the net force:

$$ F_{\text{net}} = 60 \text{ N} - 15 \text{ N} = 45 \text{ N} $$

1. Use Newton's second law to find acceleration ($a$):

$$ ma = F_{\text{net}} $$

Assuming the mass ($m$) of the sled is 10 kg, we calculate:

$$ 10a = 45 $$

$$ a = 4.5 \text{ m/s}^2 $$

So, the sled will accelerate at $4.5 \text{ m/s}^2$!

Understanding Drag Forces

Drag is a force that opposes the motion of an object through a fluid (like air or water). It's similar to friction but acts on objects moving through a medium. The drag force depends on several factors, including the object's speed, shape, and the density of the fluid. The formula for drag force ($f_d$) is often given as:

$$ f_d = \frac{1}{2} C_d \cdot

ho $\cdot$ A $\cdot$ v^2 $$

where:

• $C_d$ is the drag coefficient (depends on the shape of the object)

ho is the density of the fluid

• $A$ is the cross-sectional area
• $v$ is the velocity of the object

Example of Drag Force

Consider a car moving through air with a drag coefficient of 0.3, a cross-sectional area of 2.5 m², and the air density of 1.2 kg/m³. If the car travels at a speed of 25 m/s, the drag force would be calculated as:

$$ f_d = \frac{1}{2} \cdot 0.3 \cdot 1.2 \cdot 2.5 \cdot (25)^2 $$

This results in

$$ f_d = \frac{1}{2} \cdot 0.3 \cdot 1.2 \cdot 2.5 \cdot 625 = 28.125 \text{ N} $$

The drag force opposing the car's motion is 28.125 N.

Terminal Velocity

Terminal velocity occurs when the drag force equals the weight of the object, resulting in a net force of zero. At this point, the object falls at a constant speed. The conditions for reaching terminal velocity can be expressed as:

$$ f_d = mg $$

where:

• $f_d$ is the drag force
• $m$ is the mass of the object
• $g$ is the acceleration due to gravity (approximately $9.81 \text{ m/s}^2$)

When an object reaches terminal velocity, there will be no further acceleration.

Example of Terminal Velocity

Let’s analyze a skydiver with a mass of 70 kg. The weight ($W$) of the skydiver is:

$$ W = mg = 70 \cdot 9.81 \approx 686.7 \text{ N} $$

At terminal velocity, the drag force must equal the weight:

$$ f_d = 686.7 \text{ N} $$

This means the skydiver will fall with a constant speed when the drag force becomes equal to 686.7 N.

Conclusion

In this lesson, we explored the concepts of friction, motion on rough surfaces, drag forces, and terminal velocity. We learned that friction opposes motion, and drag acts within fluids, affecting object speed. Both forces play essential roles in how objects move. With this knowledge, you are better equipped to analyze real-world situations involving motion!

Study Notes

• Friction opposes the motion of surfaces in contact.
• Static friction must be overcome to start moving an object, while kinetic friction acts during motion.
• The drag force depends on speed, shape, and fluid density.
• Terminal velocity is achieved when drag equals weight, prompting constant speed.
• Forces can be resolved on inclined planes, affected by friction and the angle.