https://youtu.be/2PrWL6VHDmw
PHYS 1101: Lecture Thirteen, Part Five
So here’s our first example. At an airport, luggage is unloaded from the plane into the three cars of the luggage carrier. The acceleration of the carrier is .12 meters per second squared, and we can ignore friction. The coupling bars are very light, says here they have negligible mass. Each cart with luggage has a mass of 39 kilograms. What’s the tension in the coupling bar B?
Okay, do I have to consider this is multiple objects to solve the problem? Well, coupling bar B is the bar here that’s connecting Car 2 to Car 1. All of these objects are accelerating the same in the positive x direction; let’s call it to the right. So, the objects are all moving the same, they’re not, there isn’t any motion between them, but it satisfies that second bullet that I pointed out.
I’m being asked about a force between these objects, namely the coupling bar. Whatever pull the coupling bar is putting on Car 1, it’s going to have the same magnitude of pull on Car 2. I can think of this coupling bar as a rope, in the tensions in that bar and that tension is going to mean the same pull on Car 1 as on Car 2. So I do have to consider them as multiple objects.
I’m going to have to think about Newton’s Third Law as I work the problem. Sometimes it helps to decide what objects to focus on by very early on in the problem, deciding what it is you need a value for.
So, beside here I’m going to write a note right away that I’m after what I’m going to call Tension 2. I’m after the tension in coupling bar 2, that’s the pull to the right on Car 2, it’s another way, it’s the pull to the left on Car 1. This is the variable I’m going to use to represent the tension in that coupling bar. Okay, given that’s the focus, this is a force between these two objects.
I’m going to start working with Newton’s Second Law for one of these objects. Which one to pick? Well, perhaps your eyes are first drawn to Car 1. Let’s flesh it out, let’s walk through Newton’s Second Law applied to Car 1 and see if that gets us what we need, the tension in the coupling car. It’s very common in these Third Law problems, or these more involved problems, that your first approach might not be the most efficient, or the way to go. You’ll see what I mean by that in a minute.
We’re going to start out focusing on Car 1, but I’m going to have you soon realize and appreciate that, you know what, we need to backtrack. Focusing on Car 2, which also experiences this force, is going to be a better way to go for us.
But let’s start out assuming we, naively, our eye was drawn to Car 1. We knew we needed to start and focus on one of these two objects, because these are the, those that are experiencing the force that we’re after. So for Car 1, my acceleration is to the right, that’s going to be my positive x direction. And it has a value of, I’ll jot that down over here, 0.12 meters per second squared. So that’s the consideration for the left side of my equation.
Let’s draw the free body diagram from Car 1 to think about the f net, the net forces. So, here’s my car, drawn as this red object that I’m focused on. What are the forces? I’ve got the weight of the car, I have the normal force. And going around the object, I’ve already got the push of the ground up on the right side for Car 1. I have the coupling bar that’s attached to the jeep. I’m going to call that Tension 1; I’ll jot that in my picture above here. As I continue around, nothing touching the top, when I get to the back end of the car, I have Tension 2, due to this rope or the coupling bar at the end.
I intentionally drew this force to be a little bit larger than that, so I have an fnet consistent with my picture. So, as always, when I apply Newton’s Second Law, it’s really these equations that I mathematically think about. The vertical and the horizontal acceleration m forces and are they consistent with Newton’s Second Law?
These, this motion in these objects are not accelerating up down, only horizontally, so my y equation, which just tells me what I already know, namely that my forces up down have to balance. So that’s my visual check in equating the normal force and the gravity force.
All the action’s happening horizontally, so it’s the a(x) equation that I would look to. Well whatever is causing this .12 meters per second squared acceleration, that’s a result of the balance of the positive tension to the right, minus the tension force that’s to the left, divided by m. This is the value I’m after, and I know the number for the acceleration, I know the mass of each car, but I don’t know the tension in this coupling bar that’s due to the jeep pulling on it.
Now when you run into a situation like that, my instincts then tell me, well, because this is a force between this object and the jeep, let’s now do, apply Newton’s Second Law to the jeep. Well, the jeep is going to have to have some larger force to the right that overcomes the same T1 that’s pulling on the jeep to the left.
That force, technically comes from the interaction of the tires with the ground, and it’s the friction force of the ground that accounts for that. But I don’t know anything about those details to solve that, so I kind of run into a dead end here. I’m going to jot down here, “Not enough info about forces on the jeep to apply Newton’s Second Law to the jeep and therefore solve for T1.”
Okay, so my reaction, then, when I run into a dead end like that is to go back and reconsider my initial thought, which was to focus on Car 1. Let’s go over and look at Car 2. It also involves this tension force, T2, and therefore may be the route to our answer.
So, for Car 2, it has the same acceleration, let me draw the free body diagram for it. It also has a weight that’s balanced by the normal force. It has a pull to the right of T2. But then, as I go around the cart, I realize there’s a cable on the back end of Car 2, I’m going to call that T3.
Again, I drew it so the T3 was smaller than T2, so I end up with my net force consistent with the acceleration. Focusing again on the a(x) equation, I know my .12 meters per second squared has to be equal to plus T2 minus T3 divided by m. That’s the variable I need to solve for, I have my acceleration number, I have the mass.
Again, I’ve got another force here I don’t know. But now, looking at T3, in order to determine a value for that, I can look to Car 3 and see if that gives me a path to solving for the value of T3, and it does.
So let’s move over and work on Car 3 for a while, Car 3 is the object. It has to have, again, the .12 meters per second squared to the right acceleration. Its forces look very similar to what I’ve seen all along, the normal force balancing the force due to gravity.
As I go around this car, now the only tension forces I see that I have is T3 to the right. That gives me an equation that my .12 meters per second squared is equal to, plus T3 divided by the mass, that’s the only horizontal force that I have. And I can now solve for T3, doing a little algebra to get a number for it.
I have 39 kilograms times my .12 meters per second squared, to give me 4.68 newtons. Now, with that number I can go back. Let’s go back up here to this equation for Car 2 and plug in our, we now have a value for T3.
So let me rewrite that equation there for Car 2. It was that a(x) was equal to T2 minus T3 divided by m. Do a little algebra here to solve for the variable that we want, T2. Moving my m to the other side, I’m then going to add T3 to both sides so I have m times a(x) plus T3 now on my left. Everything is canceled on the right, so that equals T2, or T2 is equal to max plus T3 and I’m ready to solve the problem.
Let me point out that for T3, I certainly could plug in the number that I have, and that’s what most of you will do. Let me also suggest that instead of plugging in this number, I could have waited to calculate it and I, let me just plug in these variables m times a(x).
So the tension between the last car and Car 2 is only one of the car’s mass times the acceleration, but the tension between the second car and the first car, think of it as not only supporting the load of the last car, but that tension also has to provide the acceleration to the second car. They all, they both weight the same, so I end up with twice, there’s twice the tension between Car 1 and 2 compared to Car 2 and 3. That leaves me with roughly 9.36 newtons.