00:02
All right, looks like you have an inclined plane.
00:05
We've got two boxes on the ramp.
00:07
They both have a different coefficient of friction and a different mass.
00:12
Unfortunately, the picture wasn't included, so i'm going to have to make a guess as to which one's in front.
00:18
I'm assuming that they're not stacked on top of each other.
00:22
If that's the case, i'll briefly explain that one, but i'm going to assume that box a is in front of box b.
00:30
I'm going to solve it in that fashion, since i didn't have the picture involved.
00:38
So first step to any type of second law of motion problems, you should start with the force diagram.
00:44
So i have a's force diagram right here, and then i have b's force diagram on the right.
00:54
So a's here, b is here.
00:59
Now notice that there's this double force going against box b.
01:08
It's the friction force.
01:11
Let me label them here because this is going to be a determining factor of what the net force is.
01:18
So we've got a friction force from b.
01:21
We also have the friction force from a that's affecting whether b gets to go down the ramp at the speed it would like to.
01:34
And then we also have box b's weight.
01:38
And then we have the normal force.
01:41
So i'm going to use the subscript b to keep these separate sense.
01:45
They both have a normal force and a weight.
01:48
So normal force of a and then the weight of a.
01:53
And then there's a friction force of b.
01:59
And this one, i'm going to just call it push force from b, f sub p, because i believe once we do the calculation that b's net force from its scenario is going to be greater than the friction force that box a can apply.
02:25
So we're going to end up with box b pushing on box a.
02:30
The reason that i'm thinking that is if we look, box b has less coefficient of friction.
02:36
And so that's going to make it, that's going to have in effect.
02:41
That box b could have slid down the ramp faster than it's going to if box a wasn't in play.
02:49
So that's going to have this little push.
02:51
So i added that.
02:54
So the second, the next step after we've decided what forces are involved is to find the net force.
03:02
And i'm going to solve it for box a, the acceleration on a.
03:06
Now, if this is the setup, box a and the year, going to accelerate together because box b has an acceleration of its own that would have been greater than box a's, but since it's behind, it's going to have the same acceleration.
03:21
So anyway, we can say that acceleration may be are going to be equal to each other.
03:27
So we could say that.
03:28
I'll just put that here.
03:31
So the net force on box a is going divided by the total mass of the system.
03:44
Let's see what we can do.
03:47
We've got to look at that fg parallel from a is going to try to bring it down the ramp, as well as the pushing force from box b, which i'm just going to call fp for right now, plus, or i should say minus, minus, minus, minus the friction force from box a.
04:27
Why i put box b there.
04:29
That should have been on my label as box a.
04:33
Sorry about that.
04:36
A's and b's here.
04:37
This is the friction force on a divided by massive a plus massive b.
04:53
Actually, i don't need mass of a.
04:55
It's just stuck behind.
04:58
Actually, i'll leave it when i'll leave it.
05:05
Because it is a system that's accelerating together.
05:10
It's causing a push though.
05:11
I'm thinking i'm going to have to get rid of it.
05:14
It keep going back and forth on that.
05:16
I want to make sure that we, and i'll keep thinking about that here while we work this problem.
05:24
So f2 parallel of a is going to be sine 20 times mass a, which we're going to let's just put some numbers in here since we're, so we've got massive a is 4 times 9 .8 plus the pushing force, which i'm going to have to do another calculation using the other force diagram.
05:53
Minus, what do we got? co -efficient of kinetic energy or kinetic friction on a is 0 .180 times fg perpendicular.
06:08
That's what the normal force is equivalent to in this scenario.
06:12
So we're going to say cosine 20 times 4 .9 .8.
06:21
Whereas this kind of friend, this is the perpendicular of fg, which is what's counteracting fn.
06:30
Imagine it looking like this.
06:32
I'll put this in red.
06:33
If you haven't done inclined planes before, this can be a little confusing.
06:38
If i break it into components, i have my fg, i have my fg perpendicular right here and that's canceling out the normal force from a.
06:55
And then i have my fg parallel, which is actually causing box a to slide down the ramp.
07:02
That's where these are coming from.
07:03
And the angle of the triangle is right here.
07:07
That's why it's sign is fg parallel and cosine is f2 perpendicular.
07:14
We can say the same thing about b as well, because it's a sign.
07:19
It looks like we're going to have to look at b's, b's components as well for this.
07:26
Because i'm still missing fp, which is going to be the pushing force on box a when we find the net force from just this force there.
07:38
So i'm going to do that next.
07:44
I'm going to find the net force, net force on b.
07:53
So i'm going to put this in red.
07:54
So i need that little piece right there.
08:01
That force of b is going to be fg parallel for b minus ffa minus ffb...