# Physics Tutorial: Types of Forces II. Resistive Forces (Frictional Force. Drag). Terminal Velocity

In this Physics tutorial, you will learn:

• The meaning of resistive forces
• Which resistive forces are more relevant in daily life?
• What are the factors affecting resistive forces?
• How to calculate the frictional force between two surfaces in contact?
• Where does drag differs from the frictional force?
• What is terminal velocity and in which conditions it is observed?
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4.3Types of Forces II. Resistive Forces (Frictional Force. Drag). Terminal Velocity

## Introduction

What happens when you throw a ball along a horizontal surface? Will it continue move incessantly or it eventually will stop?

Is it easier to walk in water or on a dry road? Why?

These and other similar questions will be answered in this tutorial. Hence, it is important to please read it carefully.

## What are Resistive Forces? Types of Resistive Forces

It is clear that the answer of the first question in the Introduction paragraph is "the ball will eventually stop after a while." We know this from practice. But what causes the ball to stop? Is there any resistive force acting in the opposite direction of motion that slows down and eventually stops the moving object?

The answer is YES. The ball is under the effect of two resistive forces, which oppose its motion. They are:

1. Frictional force. This is a force produced at the contact area between the ball and the ground. Frictional force has an intermolecular nature. This means that molecules of the two objects in contact resist to the change of structure driven from the friction between them. As a result, the object loses energy during the motion and eventually it stops because all the initial energy of motion (Kinetic Energy) converts into other forms such as heat energy etc. (look at the Physics tutorial "Work and Energy. Types of Energy" and Physics tutorial "Kinetic Energy. Work - Kinetic Energy Theorem."
2. Drag. This is a resistive force produced when an object moves inside a fluid (a liquid or a gas). The object tries to displace the fluid's molecules during the motion to make place to itself. As a result, a resistive force is produced by the fluid, which tries to maintain its original position and structure. In the special case, when drag is produced by the air while resisting to a moving object moving through it, this resistive force is known as "air resistance." Therefore, we can say: "Air resistance is a kind of drag."

Look at the figure below.

In the figure, there are two resistive forces, air resistance and frictional force, which try to oppose the moving force. Therefore, the moving force must be at least equal to the frictional force (giving that air resistance does not exist when an object is at rest) to make it move.

## Factors Affecting Resistive Forces

Let's start with the frictional force. When the surfaces in contact are smooth, the object slides easier and therefore, it goes farther. On the other hand, it is very difficult to make an object slide on a rough surface. This mean the smoothness or roughness of surfaces in contact is an important factor that affects the motion.

Mathematically, this factor is represented through the friction coefficient, μ. This is a dimensionless constant that varies between 0 and 1. Thus, if μ = 0, this means there is no friction produced (the surfaces are ideally smooth). On the other hand, when μ = 1, the surfaces in contact are the roughest possible. If we enlarge the view of the abovementioned rough surfaces, they will look like in the figure.

It is obvious that it is very difficult to move horizontally the object in the figure as the gaps in the irregular surfaces in contact hamper the sliding process.

Machinery constructors use to lubricate with oil the parts that are more exposed to friction. Lubrication reduces the friction as the liquid fills the gaps on rough surfaces. Therefore, these gaps become less evident.

Another factor affecting the friction is the weight of the object. Thus, heavier the object, more it presses the ground and therefore it becomes more difficult to move it horizontally. Suppose there is a sheet of paper on a table. It is very easy to pull it horizontally along the table. But if you put your hand above the sheet, it becomes more difficult to pull the sheet horizontally as you are exerting an extra downward force on it.

Since when dealing with the frictional force we are interested in the friction caused by the environment to the object, we replace the weight with its counterpart, i.e. the normal force N exerted by the ground on the object. It has the same magnitude as the weight but opposite direction.

Summarizing the above factors and expressing the result mathematically, we obtain the equation below for the magnitude of frictional force f:

|f| = μ × |N|
= μ × |W|
= μ × |m × g × cos θ|

The angle θ is relevant only when the object is on an inclined plane (slope). When the object slides along a horizontal plane, the angle θ to the horizontal direction is equal to zero. Therefore, cos 00 = 1 and therefore this part can be removed from the formula.

On the other hand, when we press an object from above and then we try to slide it along the surface, we have an extra force to add in the calculation of Normal Force N. Let's explain all the abovementioned things through an example.

### Example 1

A 8 kg object rests on a horizontal surface as shown in the figure.

What is the friction coefficient between the object and the horizontal plane if the minimum horizontal force needed to make the object move is 25 N?

### Solution 1a

Since the minimum force to make the object move is 25N, this value also represents the frictional force to be surpassed. Therefore, we have: f = 25 N. Look at the figure.

On the other hand, since the plane is horizontal, we have θ = 0° so cos θ = cos 0° = 1.

Hence, we have

|f| = μ × |N| = μ × |m × g × cos θ|

Thus,

μ = |f|/|m × g × cos θ| = 25 N/8 kg × 9.81N/kg × 1 ≈ 0.32

An additional 20 N force is exerted vertically down on the object as shown in the figure.

What is the minimum horizontal force needed now to make the object move?

### Solution 1b

The additional force is added to the gravitational force as both act vertically down. The resultant gives the weight. We will use the magnitude of this new weight to find the frictional force because (as stated earlier) weight is numerically equal to the normal force N.

We have:

|f | = μ × |N|
= μ × (m × g + Fextra )
= 0.32 × (8 × 9.81 + 20)
= 31.5 N

The minimum force used to make the object move must be at least equal to the frictional force. Therefore, we have for the magnitude of the new minimum force:

|Fmin| = |f| = 31.5 N

If the same surface and object are placed on a 150 slope as shown in the figure, what is the minimum pulling force to make the object slide downhill? Take cos 150 = 0.966 and sin 150 = 0.259.

### Solution 1c

Now, we must consider the angle θ in the calculation of the frictional force. We have

Fmin = f = μ × m × g × cos 150
= 0.32 × 8 × 9.81 × 0.966
= 24.26 N

Remark! Frictional force does not depend on the size of the areas in contact. Indeed, no area exist in the formula of frictional force.

As for the drag, one of the factors affecting it, is the density of fluid in which the object is moving. Denser the fluid, harder the motion through it.

Another factor (unlike in the frictional force) affecting the drag's value is the surface of contact. This means sharper the object moving through a fluid, easier its motion. This is the reason why car designers pay particular attention to the aerodynamic form of the car. When the aerodynamic form is perfect, the car moves easier through the air and as a result, it saves fuel. Look at the figure below:

In the first figure, the air turns back due to the snaggy shape of the car. As a result, a region with denser air is created in front of the car, which makes its motion more difficult. On the other hand, in the second figure most of the air slides aside or above the car. As a result, the car moves easier and faster.

The formula of drag includes many other elements besides the ones mentioned above such as coefficient of viscosity, Stock's force etc. Therefore, we have not considered as appropriate to show it here, as this goes beyond the scope of this course.

## Frictional Force and Drag Acting Simultaneously at the Same Object. Terminal Velocity.

When an object is at rest, there is neither frictional force nor drag produced. When the object starts moving, the frictional force is the first resistive force that appears in the process. When objects move at low velocities, the air drag is negligible. It becomes considerable only when the velocity increases at certain values that create air currents in the opposite direction of motion. In this regard, we can say that frictional force has a fixed value that depends only on the surfaces roughness and the object's weight, while drag is variable, depending of the moving velocity.

When air drag (resistance) reaches such a value that when added to the frictional force they balance the moving force, the object cannot speed up anymore. It continues to move at constant velocity - the value of the last velocity before the equilibrium was reached. This constant velocity produced when moving force equals the sum of frictional force and drag, is known as "terminal velocity." It depends on the mass and the surface of object.

For example, a parachute falls down at low terminal velocity because it has a large surface. As a result, it blocks more air during its way and falls down slower than a free falling human. Therefore, parachute is considered as an important life-saver.

### Example 2

The engine of a 1200 kg car produces 8000N moving (driving) force. The friction coefficient between the car's tires and the asphalt is 0.6. What is the air resistance if the car is moving at terminal velocity?

### Solution 2

The situation is described in the figure below.

If we denote the air resistance (drag) by D and the moving force by F, then the equation to be used in this case is:

F = f + D

Rearranging, we obtain

D = F - f
= F - μ × m × g
= 8000 N - 0.6 × 1200 kg × 9.81 N/kg
= 8000 N - 7063.2 N
= 936.8 N

## Whats next?

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