# The Normal Force: Definition and Examples

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: Friction: Definition and Types

### You're on a roll. Keep up the good work!

Replay
Your next lesson will play in 10 seconds
• 0:00 What Is Normal Force?
• 0:40 Normal Forces on a…
• 4:46 Normal Forces on an…
• 5:44 Lesson Summary

Want to watch this again later?

Timeline
Autoplay
Autoplay
Create an account to start this course today
Try it free for 5 days!

#### Recommended Lessons and Courses for You

Lesson Transcript
Instructor: Matthew Bergstresser
The normal force is also called the contact force because it only exists when objects are touching. In this lesson, we will investigate what the normal force is and how to calculate it on flat and inclined surfaces.

## The Normal Force

Have you ever played a prank on someone, where they stand on a scale and you are behind them, and stealthily putting your foot on the scale to make them seem heavier? Your adding an extra force, in addition to the person's weight to the scale, and the scale responds by pushing back with the same force. This pushing back force is called the normal force, or contact force. Normal forces don't exist unless objects are touching and are always perpendicular to the surface(s). There is no general equation to calculate the normal force but there is a method to calculate the normal force. Let's get busy figuring this out.

## Normal Forces on a Flat Surface

Whenever dealing with forces, two major tools are required to analyze the situation: a free body diagram and Newton's second law, which is:

Î£F = ma (read as the net force equals mass times acceleration)

Let's look at a few situations where a mass is on a flat surface.

### Situation 1

In the first situation, a construction crane has lifted a 500kg crate full of construction equipment, which is on a platform, connected to the crane's cable. The mass is at rest (500kg); what is the normal force exerted by the platform on the mass? Let's pull the mass out of the scenario and draw a free body diagram including the forces acting on it.

This is the free body diagram for when the mass is at rest:

mg is the weight of the crate; mg stands for mass x the acceleration due to gravity, which is weight, and N is the normal force provided by the platform on the crate. Now, we use Newton's second law (Î£F = ma) or the net force equals mass times acceleration.

For the vertical forces on the mass, giving us:

The normal force minus the weight of the crate equals zero.

Since there is no acceleration, we can solve for N in one step, resulting in:

N = mg = (500kg) (9.8 m/S^2) = 4,900 newtons

### Situation 2

In the second situation, the construction worker calls for a crane to lift the 500kg crate to the top of the building. The mass accelerates at 1m/s^2. What is the normal force exerted by the platform on the mass while it's accelerating?

The free body diagram is the same as for situation 1:

Î£F = ma

Filling out the left side of Newton's second law, we get:

N - mg = ma

Solving for N results in:

N = (500kg)(1 m/s^2) + (500kg)(9.8 m/s^2)

N = 5,400 newtons

When the crane gets to the top of the building, the construction workers realize the wrong crate was delivered. They tell the crane operator to take that crate back to the ground. The crane lowers the mass towards the ground at 1 m/s^2. What is the normal force exerted by the platform on the mass? The free body diagram is still the same for this scenario and so is the expression we used in situation 2. We have a different acceleration:

(-2 m/s^2)

Entering the values, we get:

N = ma + mg

N = (500kg) (-1 m/s^2) + (500kg)(9.8 m/s^2)

N = 4,400 newtons

When the crate is on the ground, a fork lift pushes on it at a 45 degree angle with a force of 1,000 newtons to move it off of the platform. What is the normal force exerted by the platform on the mass? The free body diagram is different in this scenario:

f represents the applied force exerted by the crane.

We have to break down the applied force into its vertical component. Here is the new free body diagram:

Filling out the left side of Newton's second law, for this scenario, results in:

The normal force minus the weight of the crate minus the vertical downward force provided by the forklift

Plugging in the values, we get:

N = (500kg)(9.8 m/s^2)+(1000)(0.707)

N = 5,607 newtons

To unlock this lesson you must be a Study.com Member.

### Register for a free trial

Are you a student or a teacher?
What is your educational goal?
Back

Back

### Earning College Credit

Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.