00:02
All right, so the figure that your question is referring to was not included with your question.
00:10
So i cannot answer your question specifically.
00:14
However, i'm going to create a problem that is most likely very similar to the one that you're talking about.
00:22
And then hopefully my setup will make sense and you'll be able to solve your problem here.
00:28
So i'm assuming that there is a sled because it talks about it.
00:33
And that sled's going to have some mass.
00:38
I'm going to just call it m.
00:42
And then on that mat or on that sled are a set of dogs, each with their own rope.
00:53
So i'm going to say this here.
00:57
So there's a pull force each rope, which is being called tension, tension one, tension two.
01:12
And one of the tensions was given, it's 150, but the other one is unknown.
01:22
And it doesn't say anything about angles, but i assume there is.
01:26
So i'm going to make it simple, though.
01:28
I don't know how complex the angles are for you, if they're equal to each other, different angles.
01:36
So i'm going to make my angles different.
01:40
I'm going to make one 20 degrees and the other one 3rd.
01:43
30 degrees.
01:49
Now we still need to know more information to be able to find the tension are they traveling as the slide traveling at constant speed.
01:58
Is it moving at an accelerator? as does it have an acceleration in some sort? so we need more information than what we have here.
02:09
So i assume it's in the problem itself or on the picture.
02:14
So what i'm going to do is i'm going to say my acceleration on the system, which is going to be a vector that sends it this way.
02:23
And we'll just call it a diagram.
02:27
But we'll make up a number here.
02:33
Actually, i'll just leave it in a day.
02:35
We're going to solve this as if it's unknown variables.
02:40
And we'll just have it set up for you to be able to solve using the information that you have.
02:47
So the first step that we need to look at is how do we find the net force? the net force in this scenario is equal to the mass of the system times the a.
03:10
Now, because we have a two -dimensional force on both ropes, we're going to need to look at this in terms of what parts are acting on the sled, in which parts are just trying to counteract each other.
03:30
So we're going to assume that the net force is sending the slide straight forward.
03:36
And neither of these two dogs is out performing the other one in terms of sending the sled what would look like downwards on our diagram to the south or to the north.
03:51
It's just going straight to the east as we look down on this slide.
03:56
So knowing that, that means that the y component or the north -south component must have a net 4 sub -zero.
04:12
And so if we can say that, then we can look at our diagram here and say that the sign of the angle 2 times f -t -2, the tension in the second string, must equal the sign of the angle 3.
04:51
Three times the tension, not three, one, sorry, times the tension in the first rope.
05:08
Now that may be enough to solve this problem, but most likely it's not because, well, it may be, it may be enough because we would know if we know that there's no acceleration in the north -south, then we know this is true, and we do have enough information if we know the angles to salt for f ft2 right now.
05:37
So we could just rearrange this to say ft2, the tension in the second string, which is what's being asked for, must be equal to the angle of the rope on the first row times the tension in the first row, which was given at 150 divided by sine theta 2.
06:12
But i don't think it was that easy because they gave us this coefficient kinetic correction.
06:18
So i'm thinking that they want us to look at it in terms of the x component instead.
06:24
So i'm ready to make the assumption, well, wasn't enough information to be able to solve straight up in this fashion right here.
06:30
Maybe the angles are unknown, or one of the angles is unknown.
06:40
So if we look from the x component, what we do know is that all of the acceleration is in the x component.
06:48
So the acceleration of this map, is the full acceleration on our system, which we'll just call it a sub x.
07:05
And what does that have to be equal to? that has to be equal to this tug -of -war match.
07:10
Maybe we could do a little force diagram over off to the side of this mass here.
07:22
So i'm looking at the force diagram above the system.
07:29
Like i'm watching it from a drone here...