# Video: Normal Reaction Force

In this video we learn about the normal reaction force or normal force and how to calculate this force in examples.

14:25

### Video Transcript

In this video, we’re going to learn about the normal reaction force, what it is, what it means, and how to calculate it. As we get started, imagine that you, as a designer of unique aquarium experiences, have come up with a new invention. You’ve developed a clear thin material strong enough for people to walk and stand on that you can put over the tank of an aquarium. The question is, how many people can be standing on this surface at one time?

We can better understand this question through learning about the normal reaction force. The first thing to understand when it comes to this force, also called the normal force, has to do with its name. “Normal” doesn’t refer to words like “ordinary” or “standard” or “typical.” But actually it means perpendicular. This becomes more clear as we hear and learn phrases like normal to the surface.

Speaking of surfaces, the normal force always involves one. The normal force is a contact force, meaning it requires two objects to be touching. And it’s also a response force. It’s the response of a surface to the force an object in contact with the surface exerts on it. Perhaps the simplest example of the normal force applies to an object at rest on a flat surface. The object has a weight force acting on it, tending to pull it down towards the center of the Earth. The surface that the object rests on resists or responds to this weight force with a force of its own, called the normal force. We sometimes abbreviate it with a capital 𝑁.

In this scenario, the normal force is balancing out the weight force, so that vertically there is no acceleration of our object. But in general, the normal force is a response force to any kind of force, weight or not. Say, for example, that you took a block of wood and pressed it against a vertical wall with your hand. We could call the force that your hand exerts on the block 𝐹 sub 𝑃. And the normal force would be the response of the wall to that push.

The three conditions for the normal force are that it is always perpendicular to a surface. It’s a contact force, meaning that it requires the interaction of an object with a surface. And it’s a response force. It’s the response of a surface to the forces acting on an object against that surface. The normal force can apply for any type of shape or curve. Say we have a rollercoaster route where the coaster goes through a loop. At any point along that loop, we can draw in a normal force of the track acting against the coaster cars. The conditions are again that it’s perpendicular to the track surface at that point, that there’s contact between the track and the car. And it’s in response to the forces of the car on the track.

When we go to calculate the normal force in a given scenario, often we do that by inference. We infer the magnitude and direction of the normal force by the scenario we’re seeing. For example, say we have a block sliding down an inclined plane. We know there’s a normal force on the block from the plane acting perpendicular to the plane’s surface.

To actually calculate the magnitude of the force though, we can’t start with the normal force itself. We start with the force that it opposes, which is the component of the weight force on the block perpendicular to the plane. It’s by solving for that component of the weight force and knowing that the block is not accelerating perpendicular to the plane that we’re able to find out the normal force magnitude.

Often, as is the case with this inclined plane, we solve for the normal force magnitude based on the fact that our object is not accelerating perpendicular to the surface we’re considering. In other words, the forces balance out. They’re in equilibrium. But this isn’t always the case with the normal force.

Say that we had a perfectly circular track with a car that moves around it at a speed 𝑉. In this case, there is still a normal force pushing the car towards the center of the circle as it goes around. But there is no balancing force pushing it out from the center of the circle. This doesn’t mean that there being a normal force is not valid in this case. It just means that the object, the car, is accelerating.

We can consider that there are two tools we could call them for solving for normal force in different exercises. The first tool is a free body diagram. This is a picture of all the forces that are acting on an object. The second tool we often use is Newton’s second law of motion, that the net force acting on an object is equal to its mass times its acceleration. With these two tools in mind, let’s try a couple of normal force examples.

A horizontal force is applied to a block that has a weight of 1.50 kilonewtons. The force holds the block at rest on a plane inclined at 30.0 degrees above the horizontal. Assuming no friction, calculate the normal force on the block.

We can call this normal force capital 𝑁 and start out by drawing a diagram of the scenario. In this situation, our block sits on a plane inclined at 30.0 degrees above the horizontal. In total, there are three forces that act on the block. First, there’s its weight force pulling it straight down. We’re told we apply a horizontal force to the block we’ve called a capital 𝐹. And there’s also a normal force acting on the block from the inclined plane.

We’re told the magnitude of the weight force. But we don’t know 𝐹. And we want to solve for 𝑁. We are told that, under the influence of these three forces, the block isn’t moving. It’s staying still in one place on the plane. In our diagram, we can insert coordinate axes to define positive and negative direction. We’ll say that positive motion in the 𝑦-direction is perpendicular to the plane. And positive motion in the 𝑥-direction is up the plane parallel to it.

Our next task is to break up the forces that are not already aligned completely with one of these axes directions into components that are. First, we do that with our weight force, breaking it up into components that are in the 𝑥-direction and in the 𝑦-direction. We can know that the 30.0-degree angle we’ve been given, which we can call 𝜃, is equal to the angle in the upper-left corner of this triangle. If we do the same breakdown of our applied force 𝐹, we see it also can be divided into 𝑥- and 𝑦-components.

Our next step will involve writing out force balance equations in the 𝑥- and in the 𝑦-directions. In the 𝑥-direction, the two forces we have are 𝐹 times the cos of 𝜃 and negative 𝑊 times the sin of 𝜃. If we recall Newton’s second law of motion, that the net force on an object is equal to its mass times its acceleration, we can apply this to the forces in the 𝑥-direction of our scenario. We can write that 𝐹 cos 𝜃 minus 𝑊 sin 𝜃 equals the mass of our block multiplied by its acceleration in the 𝑥-direction.

We know though that because the block is motionless, this acceleration is zero. This means we can rewrite our equation and know that 𝐹 times the cos of 𝜃 is equal to 𝑊 times the sin of 𝜃. Or dividing both sides by the cos of 𝜃, 𝐹 is equal to 𝑊 times the tan of 𝜃. Both 𝑊, the weight force of the block, and 𝜃, the angle of the inclined plane, are given to us. So we’ve effectively solved for 𝐹, the applied force.

Without calculating it numerically, we can move on to looking at forces in the 𝑦-direction. Applying Newton’s second law to the forces in the 𝑦-direction, we can write that the normal force 𝑁 minus 𝑊 times the cos of 𝜃 minus 𝐹 times the sin of 𝜃 is equal to the block’s mass multiplied by its acceleration in the 𝑦-direction. Again, we know that that acceleration is zero, which means that 𝑁, the normal force magnitude, is equal to 𝑊 cos 𝜃 plus 𝐹 sin 𝜃.

If we substitute in for 𝐹, 𝑊 times the tangent of 𝜃, we now have an expression for 𝑁, what we want to solve for, in terms of known values 𝑊 and 𝜃. When we plug in 1.50 kilonewtons for 𝑊 and 30.0 degrees for 𝜃 and calculate this expression, we find that 𝑁 is 1.73 times 10 to the third newtons. That’s the magnitude of the normal force acting on the block.

Now let’s look at an example where the object experiencing a normal force is in motion.

A man with a mass of 75.0 kilograms measures his weight by standing on a bathroom scale. The man and the scale are inside an elevator that can move at different rates. What is the measurement of the man’s weight according to the scale if the elevator is accelerating vertically upward at a rate of 1.20 meters per second squared? What is the measurement of the man’s weight according to the scale if the elevator is moving vertically upward at a speed of 1.20 meters per second? What is the measurement of the man’s weight according to the scale if the elevator is accelerating vertically downward at a rate of 1.20 meters per second squared?

We’re told the man’s mass in this example is 75.0 kilograms. We’ll call that 𝑚. In each of these three examples, we have a man standing on a scale inside an elevator that’s either moving up or downward. If we define motion up as motion in the positive direction. And if we call 𝑊 sub 𝑠 the weight of the man as recorded by the scale in the accelerating or constantly moving elevator. Then we want to solve for that weight reading of the scale under three conditions. First, when the elevator is accelerating at positive 1.20 meters per second squared. Second, when it’s moving at a constant velocity of positive 1.20 meters per second. And lastly, we want to solve for the reading on the scale when the elevator’s acceleration is negative 1.20 meters per second squared.

What’s interesting about this scenario, looking back over to our diagram, is that if we were to draw the forces acting on the man, in all three cases, there are only two. There’s the gravitational force acting on the man, creating a weight force down on the scale. And then there’s the normal force of the scale, pushing up on the man. This normal force is equal in magnitude to the reading of the scale, 𝑊 sub 𝑠.

To get started on our first solution, we can recall Newton’s second law of motion, which tells us that the net force on an object is equal to its mass times its acceleration. In our case, that object is the man. And when the elevator is accelerating vertically upward at 1.20 meters per second squared, we can write that the normal force acting on the man minus his weight force is equal to his mass multiplied by his acceleration. Since the man is at rest with respect to the elevator, we know that his acceleration matches the elevator’s, 1.20 meters per second squared, in the positive direction.

Because we’re given the man’s mass, we also know the weight force he applies. And we know that the normal force exerted on the man is equal to the reading of the scale in this case, which is what we want to solve for. Recalling that the weight force of an object is equal to its mass times the acceleration due to gravity, we can rewrite our equation and then rearrange to solve for 𝑊 sub 𝑠.

If we treat 𝑔 as exactly 9.8 meters per second squared, then we can plug in for 𝑚, 𝑔, and 𝑎 and solve for 𝑊 sub 𝑠. When we do, we find a value of 825 newtons. That’s the weight of the man registered by the scale when the elevator accelerates upward 1.20 meters per second squared.

Next, we want to solve for the scale’s reading when the elevator is moving at a constant speed upward at 1.20 meters per second. In this case, looking back at our free body diagram of the man, we can write that the normal force acting on the man minus his weight force is equal to his mass times his acceleration, which is zero since the elevator is not accelerating. This means that the normal force is equal to the weight force. And we’ve said that the normal force is equal to the scale reading, 𝑊 sub 𝑠. So 𝑊 sub 𝑠 in this condition is equal to 𝑚 times 𝑔, or 75.0 kilograms times 9.8 meters per second squared. This is simply the weight of the man, 735 newtons.

Finally, we want to solve for the scale’s reading when the elevator is accelerating downward at a rate of 1.20 meters per second squared. Starting with the equation 𝑁, the normal force, minus 𝑊, the weight force, is equal to 𝑚 times 𝑎, we can substitute in 𝑊𝑠 for 𝑁 and rearrange so that 𝑊𝑠 is equal to 𝑚 times the quantity 𝑔 plus 𝑎. When we plug in for these values, this time we have a value of negative 1.20 for 𝑎. And when we calculate 𝑊 sub 𝑠, we find it’s equal to 645 newtons. That’s the scale’s reading when the elevator accelerates downward. We see from these three results how the scale’s reading depends on the acceleration of the elevator.

Let’s summarize what we’ve learned about the normal reaction force. We’ve seen that the normal reaction force, or normal force for short, is a contact force that acts in response to other forces. The normal force direction is always perpendicular to the surface that supplies the force. And we go about practically solving for the normal force using two tools: a free body diagram and Newton’s second law of motion.