00:01
So in this problem, we have a train carrying two cars.
00:03
And so to see what this looks like, so we can then draw for a bi diagrams from it, we have one each after the other, so the locomotive, and then each rail car behind it.
00:19
And then there's going to be a coupler in between.
00:22
And when we look at part c, what is the force exoded by the locomotive on the first rail car? so we know this is going to be rail car 1, and then that would leave the other rail car to be rorockard 2.
00:31
They both have mass big m and the locomotive mass small m.
00:40
So what we want to do is find a couple things, the tension in each of these couplers.
00:45
So we're going to call them t and t1.
00:50
And then the locomotive to keep this acceleration is going to have to, well, so the force exited on the track by the locomotive.
01:00
So of course with the reaction reaction pair, there's going to be some force on the the locomotive by the track, so i guess we could draw it on the wheels.
01:11
Some force we can call force f, which is going to allow us to keep the train accelerating at one meter per second squared.
01:23
And so with this we can start drawing for body diagrams.
01:26
If we kind of imagine though, what's the forces on locomotive f? we have two unknowns, tension one and force f.
01:34
Same thing for our car one, we have the both tensions that we need to solve for.
01:38
If we look at car 2, though, we only have one unknown, which is tension.
01:43
So we're going to start there, which is kind of how the problem guides you, looking at part a.
01:51
So now we can draw a free by diagram for our car 2.
01:56
It has some weight, which is big m times g, and a resulting normal force to keep the car on the rail.
02:05
We see we don't really care about that, though.
02:07
And then there is some tension force in the coupler.
02:11
So if we couple it, if you imagine it's a string, for some kind of string, the tensions have to be towards the middle of it, and they're equal and opposite direction on each end.
02:23
And so that's how we're going to model these tensions and what model these couplers.
02:28
So we've resulted in some free by diagram here.
02:31
We'll put an axis on x and y.
02:35
So then we can look at the x direction in newton's second law, so some of forces in the x direction is equal to big ma because we have our, that's the mass of this car we're considering.
02:50
This is for car two.
02:53
And we know what acceleration is, so we want to see what some forces is.
02:55
We only have one force tension, which is in the positive direction by our axis.
03:00
And so we can simply plug in the mass of the car was 50 times 10 to 3 kilograms or 50 tons, 50 metric tons, 50 ,000 kilograms.
03:11
And the acceleration was is one meter per second squared.
03:18
I'm going to put the unit on the other one, we leave it off.
03:20
So it says the entire train is an acceleration of one meter per second, which is just saying specifically that this train is all coupled.
03:27
It's moving as a unit with a constant acceleration.
03:30
If it didn't say the entire train, you'd still have to assume that because otherwise you're missing information.
03:36
But we see we're just multiplying by one here.
03:37
So tension is equal to 50 times 10 to 3.
03:42
And we have kilograms times meters per second squared.
03:45
Squared.
03:47
Yep, so that is the unit for newton.
03:50
So tension in the first coupler is 50 times 10 to 3 newtons...