00:01
So in this problem, we have these two blocks attached by a rope through a pulley, and we have some applied force here, and they're all on the ground.
00:14
So the first thing we do here is we need to see how the motion of these blocks relate, so we can draw through by diagrams that are linked together.
00:21
So if we have a force to the right here, we can see that the motion of the block one is going to be to the right.
00:27
Or i guess this is a two -kilogram block, so we can call it block two.
00:32
So what that means is with the rope carrying along, it's going to drag the rope on the bottom of this pulley here to the right.
00:41
The pulley is going to flip the motion around and it's going to continue like this.
00:46
So block one is going to move the opposite direction to the left based on how the rope works.
00:51
Because, of course, the rope translates the motion, velocity, and acceleration and everything, or the displacement of the velocity acceleration along its path.
00:58
And we use rope so we can say change the direction.
01:02
So we need to free by diagrams for each of these blocks.
01:05
So we can do physics on it.
01:06
Let's start with block one and see where we get.
01:09
And the important thing we're going to do here, though, is pick our motion in the positive x -axis.
01:13
So we're going to have plus x here to the right on block one, but plus x here to the left on block, or on block one is left, block two is right.
01:23
Pictures are all going to be right, though, so that we have equations that link together.
01:29
So for block one, we can just represent it as a point and put all the applied forces on it.
01:34
Now remember we have our, x or that's y upwards by convention but x is going to be to the left here or firstly we have the tension in the rope and then this block has a weight equal to its mass times gravity its mass is one and xirex2r.
01:54
And then we have a normal force which is from block one to the block two up to this block one here which is keeping this block from falling down and i'm going to call it n1 because we need to have other normal forces in this problem.
02:08
Now we saw the that the block 1 here is moving to the left, which means friction force is going to resist it to the right.
02:16
So we have a friction force here, which is equal to muke.
02:20
Now all the muks here are also the same, so we just have one muay, and it's also equal to the coefficient times normal force.
02:28
Here it's going to be n1.
02:31
So the first thing we should do is newton's second law sum of forces is equal to mass times acceleration, so that's how we can relate our forces to our acceleration here.
02:41
And we can break this into each axis.
02:43
So we can start with the y direction.
02:45
Some forces y is equal to m a y.
02:47
And what this is going to tell us is that there's no acceleration in the y because blocks just moving horizontally.
02:53
And it's going directly tell us that n1 minus one times g is zero.
02:58
So n1 is just g.
03:01
Now this is just in magnitude.
03:03
Of course, g is meters per second squared.
03:06
N1 should be a force.
03:07
So it should be newtons.
03:07
But we know this is one kilogram hidden here.
03:12
So we're not writing any units here, we just recognize that this is magnitude.
03:16
You need to correct the units.
03:20
But the more important part we want to look at is sum of forces in the x equal to m -a -x.
03:25
And this is where we're going to start solving the problem.
03:28
So we have tension in the positive direction.
03:30
Remember positive is still left here, minus friction.
03:33
We know that's mu -k -n.
03:34
And we know what n is.
03:36
It's g.
03:36
So it just equal to mu -k -g.
03:39
And this is equal to m times a because we have one acceleration here like we said the acceleration of each block is going to be the same so we have that equation we have two unknowns though tension and acceleration so we can kind of stop here because we can't go any farther we've run out equations on this block so let's look at the second block so there is the tension force which is also pulling to the right like this based on where the string is remember the strings can only work in tension also have our applied force the 20 newton's, and we said here that our axis is x and y.
04:22
So this block has a weight equal to two times gravity, but we also have newton's load law pairs we need to consider.
04:30
So this normal force on block one is coming from somewhere, it's coming from block two, and so the equal and opposite force is going to happen on block two.
04:37
So we also have a downwards force equal to n1, which we notice is equal to g.
04:42
The same way we have our friction, and this should be called friction 1 because it's on block 1...