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!

Take Quiz Watch Next Lesson
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
Save Save Save

Want to watch this again later?

Log in or sign up to add this lesson to a Custom Course.

Log in or Sign up

Speed Speed

Recommended Lessons and Courses for You

Lesson Transcript
Instructor: Matthew Bergstresser

Matthew has a Master of Arts degree in Physics Education. He has taught high school chemistry and physics for 14 years.

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? You're 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:

ΣF = 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:

Free Body Diagram of an Object 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 (ΣFv = mav), or the net force equals mass times acceleration, for the vertical forces on the mass, giving us:


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, a 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:

ΣFv = mav

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

N - mg = mav

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:

(-1 m/s^2)

Entering the values, we get:

N = mav + 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 forklift 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:

Crane & Forklift Free Body Diagram

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:

Vertical Component of the Applied Force

To unlock this lesson you must be a Member.
Create your account

Register to view this lesson

Are you a student or a teacher?

Unlock Your Education

See for yourself why 30 million people use

Become a member and start learning now.
Become a Member  Back
What teachers are saying about
Try it risk-free for 30 days

Earning College Credit

Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 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.

To learn more, visit our Earning Credit Page

Transferring credit to the school of your choice

Not sure what college you want to attend yet? has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.

Create an account to start this course today
Try it risk-free for 30 days!
Create an account