Scene #

Description

Narration

1

The definition of apparent weight is on the screen. The narrator reads the definition out loud. The narrator draws a bathroom scale on a floor and adds a stick figure on top of the scale. Below the person a red arrow is pointing down with F-sub-g equals mass times gravity. The words actual weight are written next to the formula.

The narrator adds a red arrow pointing up extending from the stick figures head. At the top of this arrow he labels it F-sub-n. Next to this arrow he writes apparent weight.

At this point in the course we've examined the concept of weight. And apparent weight is a special case of that. Apparent weight is the magnitude of the normal force that would be exerted by a horizontal surface at a given moment. Now, that's a lot to unpack. So let's look at a really simple explanation of it. You can think of apparent weight being just what a bathroom scale would read for an object's weight if that bathroom scale were on a flat surface.

So let's say that we have a person standing on this bathroom scale. The person is exerting a downward force, F sub g, equal to their mass times gravity. Now, mass doesn't change. And on the surface of the Earth, gravity doesn't change. We call this the actual weight.

And as long as an object's mass and the acceleration of gravity aren't changing, actual weight is constant. Now, there's another force of play here in this situation. And that is, of course, the upward force-- the normal force that the scale is exerting on the person. That's F sub n. And usually, that's equal in magnitude to the force of gravity, but not always. And the normal force in this situation represents the apparent weight.

Now, in the vast majority of situations, apparent weight and actual weight will be the same number. But there are some situations where that's not the case. And we're going to look at a couple of those right now.

2

Example 1 is on screen. The narrator reads the problem out loud. He next describes how to draw a free body diagram and writes on screen what he is describing.

An 83 kilogram person is standing in an elevator, which is moving at a constant velocity. What is the person's apparent weight during this period?

Let's begin by drawing a free body diagram. On this person, we have the downward force of gravity, F sub g. And that's equal to mass times gravity-- the actual weight. And then, we have an unknown magnitude force upward. That's the normal force. And we don't know if that's greater or less than the force of gravity. So we're going to just leave that as a dotted line for right now.

3

The narrator writes down the givens in the problem. As he says them out loud he writes them on the screen in red. He then records the formula and writes out the steps to solve the problem in blue as he says the steps out loud.

Writing down what we know. We know acceleration is 0 meters per second squared. Because it says they're moving at a constant velocity. So acceleration is 0. And that means the net force is also 0. The net force is the sum of the upward normal force and force of gravity. Which is mass, 83 kilograms, times gravity, which is negative 9.81 meters per second squared.

So we get 0 newtons is equal to the normal force minus 814 newtons. Add 814 newtons to both sides, and you get the normal force is equal to 814 newtons. So in this situation, apparent weight is 814 newtons.

4

Example 2 is showing on screen and the narrator reads the entire problem out loud. The narrator draws a free-body diagram he describes what he is drawing as he draws it.

Now, that was an example where apparent weight and normal force were equal. Now, let's look at an example where that's not the case.

An 83 kilogram person is standing in an elevator, and the elevator is in freefall. What is the person's apparent weight during freefall?

Well, hopefully, this is an amusement park ride. Because if the elevator is in freefall, that is not a good sign. But let's begin by drawing our free body diagram. Again, this is the same situation where we have the downward force of gravity equal to mass times gravity. And we have an unknown normal force pointing upward, which is what we want to solve for.

5

In the middle of the screen using red the narrator writes the steps to solving the problem saying each step out loud as he writes it.

The acceleration in this case is equal to negative 9.81 meters per second squared, because the person on the elevator are in freefall. That means the sum of all forces here is equal to the object's mass, 83 kilograms, times their acceleration, negative 9.81 meters per second squared. And that's going to be the sum of forces, the upward normal force, and-- again-- 83 kilograms times negative 9.81 meters per second squared.

All right-- multiplying this all out, we get negative 814 newtons equals the normal force plus negative 814 newtons. If we add 814 newtons to both sides of this equation, we get that the normal force is equal to 0 newtons. And thus, apparent weight is 0 newtons. And this is why we say when an object is in freefall it is experiencing weightlessness. It's not that it doesn't have an actual weight. It's that it's apparent weight is 0 newtons, so they feel weightless.

6

The narrator reads example 2 out loud. In black he draws a free-body diagram as he describes it out loud. In red and blue he writes out the steps to solving the problem as he says the steps out loud.

In the next example, an 83 kilogram person is standing in an elevator. If the elevator is accelerating downward at a rate of 0.65 meters per second squared, what is the person's apparent weight during this acceleration?

Our free body diagram is going to be the same as the others. A downward force of gravity equal to ng and an unknown upward normal force, F sub n. Our acceleration in this case is, it's downward, going to be negative 0.65 meters per second squared.

That tells us that the sum of all forces on this is going to be equal to the mass, 83 kilograms, times the acceleration, negative 0.65 meters per second squared. And that is equal to the sum of forces, the upward force, the upward normal force, plus-- again-- 83 kilograms times negative 9.81 meters per second squared.

Multiply this out, and we get negative 54 newtons equals normal force minus 814 newtons. If we add 114 newtons to both sides of the equation, then we find the normal force is equal to 760 newtons. And thus, our apparent weight is 760 newtons.

Now, notice that this is less than our actual weight. And that's because of the downward acceleration of the elevator. You can experience this anytime you ride an elevator and it starts moving downward. You feel a little bit lighter. You can feel it even in your feet, just that your feet are touching the ground a little bit less.

7

The narrator reads the next problem out loud. In black he draws a free-body diagram as he describes it out loud. In red and blue he writes out the steps to solving the problem as he says the steps out loud.

One last example. An 83 kilogram person is standing in an elevator. If the elevator is accelerating upward at a rate of 0.45 meters per second squared, what is the person's apparent weight during this acceleration?

All right-- our free body diagram is again going to be the same. A downward force of gravity equal to ng and an unknown upward normal force. Acceleration in this case is equal to a positive 0.45 meters per second squared.

That means the sum of forces is going to be equal to mass times acceleration. Mass is 83 kilograms. Acceleration is 0.45 meters per second squared. So that's equal to the sum of our forces. We have the upward normal force plus mass times gravity, which is 83 kilograms, times negative 9.81 meters per second squared.

Multiplying this out, we get 37 newtons-- approximately-- is equal to the normal force minus 814 newtons. If we add 814 units to both sides of the equation, we get that the normal force is equal to 851 newtons. And this is greater than our actual weight.

And you'll notice that again when you ride an elevator or an airplane or an amusement park ride where you're accelerating upward. You feel heavier in your seat or on your feet in those situations. So in this situation, the apparent weight was 851 newtons.