00:01
Part a of this problem is asking to draw free body diagrams for both block a and block b.
00:06
For block a, we have a normal force upward.
00:12
We have a tension force to the right.
00:15
I'm going to denote by t1.
00:17
We have the weight of a downward.
00:20
And we have the friction force on a to the left.
00:25
For block b, which is sitting on the ramp, we have a tension force upward.
00:31
T2, it's different than t1.
00:33
We also have t1 downward.
00:36
Now this is the same as this t1 because they're connected to each other.
00:40
We have the normal force of b perpendicular to the ramp.
00:47
We have the friction force of b opposing the motion down here.
00:52
And we have the weight down here like this.
00:57
Now, of course, we can split this up into its components like this.
01:02
This becomes wb times cosine of 36 .9 degrees.
01:09
Because that's what this angle here is and this becomes wb times sign of 36 .9 and i believe that's the answer to part a there i think i've got all the forces for part b we want to figure out what t1 is the tension between blocks a and blocks b so the way we do that is we're going to first do the sum of the forces in the y direction which is equal to zero since there's no acceleration in the wide direction.
01:45
We're going to do it on a.
01:47
And just to keep it straight, that's a and that's b.
01:52
This implies that the weight of a equals the normal force because those are the only two forces in the wide direction and they're in the opposite directions.
02:01
Now we're going to do the sum of the forces in the x direction on a.
02:11
And again, it's equal to zero because although they're moving, they're not accelerating.
02:17
And so newton's second law there gives us that t1 is equal to the friction force on a.
02:27
But the friction force on a can be expressed as the coefficient of kinetic friction times the normal force.
02:34
We found the normal force here is equal to the weight of a.
02:39
Now these are both given in the problem, and so we can easily calculate this to be 8 .75 newtons, and that's the tension in the row between a and b.
02:54
Now in part c, we want to figure out what the weight of block c is.
03:01
And so the force diagram for block c looks like this.
03:05
We have tension t2, t2 because it's connected to block two, going up like that, and we have the weight of c pulling it down.
03:15
Newton's second law, the sum of the forces in the y, equals zero because there's no acceleration, implies that the tension in rope two is equal to the way of c.
03:25
And so if we can figure out what t2 is, we'll have the answer, because we're trying to solve for the weight of c.
03:31
So to figure out t2, we're going to look at block b, and we're going to take the sum of the forces along the ramp.
03:46
Equal zero, because there's no acceleration along the ramp.
03:50
This is, again, for block b gives negative friction force in b, if it's pointing in the negative direction, minus t1, minus the weight of block b times sine of 36 .9, plus t2.
04:10
Is equal to zero.
04:16
Solving for t2, this gives t2 is equal to t1 plus weight of b times sine 36 .9 plus friction force of b.
04:30
Now we want to figure out what this friction force of b is.
04:33
In order to do that, we're going to do the sum of the forces perpendicular to the ramp.
04:40
It's equal to zero because there's no acceleration in that direction.
04:43
And this equation implies that the normal force of block b is equal to the weight of b...