https://youtu.be/s7pneQSYMeM
PHYS 1101: Lecture Twelve, Part Five
Okay, our last example will help you out with one of your homework problems, very similar.
We have a scenario where a car is traveling at 20 meters per second and it stops in a distance of 50 meters. Assume that the deceleration is constant. The coefficient of friction between the passenger and the seat are static coefficient of 0.5, and a kinetic coefficient of friction of 0.3. Will a 70 kilogram passenger slide off the seat if not wearing a seat belt?
Let me sketch a couple of things first, just to solidify in my mind this scenario that’s playing out before me. A person’s in a car and the motion diagram shows deceleration, slowing down, coming to a stop.
Let’s make it headed to the right in a positive-X direction. My acceleration vector has to point opposite to get the slowing down, in order for the slowing down to occur. Let me jot down here, “Positive X to the right”.
Here’s my picture. Here’s my seat and here’s the person sitting in the seat. So there is the motion diagram for the car, and here’s just a sketch of the person sitting in the seat. Will the 70-kilogram passenger slide off the seat if not wearing a seat belt?
Okay, what’s the object? This is a good example where it’s less obvious what object you need to focus on. But you need to pick the right object and rigorously follow through the problem-solving steps with or for that object.
The strongest hint of all is often buried in the question you’re supposed to answer, where it directly asks you for the value of something. It’s, in this case, not a number, but will something happen, or not.
But what’s it asking about? It’s the 70-kilogram person. So the object, make it the person, make it the passenger. Now, everything we do, everything we think about is for the passenger.
So Question 18, what objects do you focus on? Some people think it’s the car. Is it the car? The person? The road? The wheels? What should you focus on?
The second problem-solving step, after you decide the object, is to think about the acceleration that that object’s undergoing. This is another tricky question to answer, for a problem like this. Because we really have two scenarios.
The acceleration will be two different things, depending on whether the person slides off the seat or not, and that’s what we’re trying to figure out. So how do we pick one? What’s the right “A” to pick?
The approach to a problem like this is to pick one of the scenarios, make that assumption, make that assumption that the acceleration will have the value consistent with that picture. Follow through logically with your mathematics, based on that assumption. And then, whatever the math predicts, does that match what would happen in real life? Or what does that tell you about real life?
So by that I mean, pick one of the scenarios. Either you picture that the person does slide off the seat, or that they don’t slide off the seat. My eye is drawn to picturing that the passenger doesn’t slide off the seat, and follow through and see where that leads us.
I’m drawn to that, because if I imagine that the person does stay stuck to the seat, then I know that the motion diagram that represents the car’s motion will also represent the person’s motion. And I’ve already got that drawn; I already have that visualized. So let’s go with that.
Let’s assume that this person also experiences this motion diagram. And so, must be decelerating. Therefore, this person must have a net force in the backward direction. I have to have Fnet, on this person, if this is the acceleration that results.
So here I’ve explicitly written it down. We’re going to assume that the person doesn’t slide off the seat, and see what it gets us. I know, then, that the motion diagram I drew above matches, or is also then what I have for the person. Let me copy that over here, again.
Step three, then, let’s think about all of the forces on the person, to be sure they match. Or what does it take for them to match this acceleration picture?
Well, for the person, I’m going to draw my dot, and I’m going to start drawing forces. I know that there’s the weight of the person. Now, as I go around, what else is touching the person? Well, I didn’t draw the floor. Their feet probably are sitting on the floor.
The biggest effect is that the seat is supporting most of their weight. So I’m going to just simplify that all of the effects of the floor are really here in the seat. This will have a small effect.
So I’m touching the seat. A surface can be responsible for two types of forces, a normal force and a frictional force. Let me sketch those in, normal force. Let me jot down that there’s the weight.
Now, looking at this person, I need to think about friction. In fact, if I look at my motion diagram, where I’ve got the surface only causing a normal force, I know there has to be another force, because I don’t yet have anything to explain an Fnet to the left.
I’m missing a force. What am I missing? A surface can provide a frictional force in addition to the normal force. What direction would it be? Frictional forces always oppose the motion. Or in the case of static friction which is what I’m considering, because the person is not sliding on the seat, it opposes the tendency to motion.
Well how do I check which direction that is? Imagine all of a sudden the seat were very slick. Would the person move to the right, or to the left?
Well, if you think about, from the perspective of this seat, if the seat were slick, the person would move to the right. Or, technically what would happen is, the seat would slide out from under the person. The seat would go to the left and the person would go to the right, therefore looking like the person’s going to the right.
So my friction has to oppose that. So instead of the person going to the right, they don’t, so my friction force has to be to the left.
Okay, Question 19. Now that I have a picture of all the forces, and something that’s consistent with my acceleration, and therefore the direction of Fnet, think about what is the direct cause of the person’s deceleration? What’s the direct cause of this Fnet?
Then, step four, we’re ready to think about any additional trigonometry we may have to do. Looking at our free body diagram, and our direction, our picture of A, all of these vectors already look purely vertical or horizontal. So there is no trig to do.
We’re ready for step five. We’re honing in now on the specific variables that we have, and what we need to solve for. So we know some kinematic information in this problem.
We know an initial velocity. We know the displacement, this 50 meters. That’s the distance over which this car and passenger come to a stop. We know the mass of our object. And I wrote down here the coefficient of static friction.
We were given both, but it’s the static coefficient I’m going to care about, because I have chosen the scenario that the passenger stays stuck in the seat. That this static friction between the person’s pants and the seat is what’s responsible for the Fnet, and therefore, the deceleration, the slowing down of the person and bringing them to a stop, as the car comes to a stop. So all I need to worry about is the static coefficient.
Well what do I want? That deserves some careful thought, here too. In the end I’m after answering this question of whether the person slides off the seat or not. Do they stay stuck to the seat?
The static friction is what’s causing this deceleration. So at the heart of it, I need to ask, is the static friction large enough, or can it be large enough, to provide this much deceleration?
So here’s how I suggest we go after answering that. Let’s first calculate what this value of A is. And then, based on this person’s mass, we can figure out how large Fnet has to be. That tells us how large the static friction has to be, because it is Fnet. It is the only force horizontally.
Once we have a value for the requirements on the size of the static friction force, what about if we then go compare it to the maximum possible value of the static friction force? If we compare those two numbers, then we should be able to say right away if this logical assumption, this assumption we started out with, is true.
That means, if we conclude that the static friction required to cause this deceleration is less than the maximum value, then the person will stay stuck in their seat.
If we learn that the static friction would have to be larger than is physically possible, larger than that possible maximum value, then the scenario we’ve picked of the person staying stuck won’t pan out. The person will slide off. So that’s what we’re going to do.
So let’s break it up into these two exercises. The first exercise is figuring out what magnitude for FS we need to cause this value of A. Then we’re going to go see how that number compares to the maximum value for the static friction.
So in order to answer number one first, we’re comparing and seeing the connection between the static friction and a. Which means, we’re going to work with a(x)is equal to the sum of the forces horizontally, divided by m.
What is that for our problem? I only have one horizontal force. It’s negative, in the negative direction. So whatever the value of a(x) is, it has to be minus f(s) over m. This is my focus now. I’m trying to solve for this. What does this value have to be to give us the acceleration that we needed? Quiz question 20 has you calculating what a(x) is, based on that kinematic information.
This is one-dimensional motion. This is a Chapter 2 exercise, so everyone should be able to do that, solve that for a(x). Report that to three digits.
Once you do that, you’ll then have a number for a(x). You know the mass of the passenger. You can do you algebra and you can solve for the value of a(x), the static friction.
So I show you, here, I still leave it up to you to get the number for a(x). But when I solved for it, plugged it in times the mass, I end up with a static friction that has a value of 280 Newtons.
Okay, that’s what it takes to slow the person down and bring him to a stop. That’s the force directly responsible for bringing them to a stop. Can the static friction be that big? That’s our second exercise, or step number two, that we need to pursue.
What is the maximum possible value? Well, from our list of forces, you go down and you see that the static friction force can, at most, be the coefficient of static friction times the normal force.
Okay, the coefficient was given to us; we have a number for that. We’re trying to figure out this maximum possible value. So we need to know the value of f(n) in order to finish that up.
Well what sets f(n)? F(n), again, is the normal force; it reflects the load that the surface is supporting, in this case the seat, the car seat. And the load is the person. F(n) appears in our free body diagram as this upward force, of course, that visually we see balances the weight, the force of gravity on the person.
The object, this person, if he stays stuck in the seat, has an acceleration to the left that’s purely horizontal, no acceleration in y. So the a(y) component equation, for Newton’s Second Law, becomes very simple. A(y) has to be 0; the mass drops out. So all of the vertical components have to add up to 0.
That tells us what we see visually in our free body diagram. The upward force, f, has to equal the weight, which is mg. Then to mathematically plug that into this equation, or express that, Question 29 asks you to determine, as you’re trying to determine f(n), what should you substitute in for the numerator? The sum of all Y forces.
So let me make that note, or that comment, to generalize. That whenever either a(x) is 0, or a(y), that right away tells you that our general Newton’s Law equations become 0 equals all of the horizontal components. Or, if a(y) was 0, we have that Newton’s Law becomes 0 equals the sum of all of the vertical force components.
This says they balance. Applying that to our problem then, we have in the process of trying to determine f(n), we’re using a(y) is equal to the sum of the y forces. There is no up-down acceleration. So I have this case that 0 is equal to the sum of all the y components.
That then gives me that 0 is equal to, I only have two Y components. It’s plus f(n), because the normal force points up. And it’s minus weight, which has a magnitude of m times g. And that’s minus, because the force due to gravity points down.
Adding mg to the other side, because I’m trying to solve for f(n), tells me that for this case, the normal force does just equal the force due to gravity.
Many people, say in a high school physics class, have only looked at very simple problems where the only up-down forces on the object are the force due to gravity, down, and then the surface force, pointing up, or the normal force, up.
So f(n) equals mg. And because of that, there’s a tendency to memorize, or just to assume, that f(n) is always equal to the force due to gravity. That is not the case.
If there are any other forces involved in the vertical direction, the normal force will have a different value. It won’t be just mg. It’s whatever it takes to balance out all of the forces, so we end up with 0 for our acceleration perpendicular to the surface.
So Question 23 is for you to go through and follow through on the math. What is the maximum value for the static friction? Go through now and plug numbers in.
And then the last question is, after you’ve done that, compare this maximum possible resistance force that the friction can provide, so to speak, to keep the person in the seat.
This maximum f(s), see how that compares to the value of f we concluded we had to have in order to slow the person down, bring him to a stop, as the car does, therefore keeping the person, or having the person stay in their seat.
See how they compare, and then answer for yourself. Is the person going to slide off the seat? Yes, or no.
That then brings us to the end of Lecture 12. We just went over the three examples and I gave you the thorough overview of what we’ve covered in Chapter 4, up to date, to try to solve f equals ma, for each of those problems.