00:02
So for this problem, it's asking us to find the acceleration of the center of this spool, as well as the magnitude and direction of the friction force.
00:11
So the first thing we want to do is set up our equilibrium equations, which i've written down here.
00:18
And so we first want to sum the forces in the x direction.
00:22
And that will give us t minus mg sine theta minus f equals m .a.
00:28
And so we can solve this equation for f by just adding f to the other side, subtracting m .a.
00:37
And that'll give us f equals 200 minus 3a.
00:47
And so that's after we plug in the given values, which are up here in green.
00:53
The mass is 30 kilograms.
00:56
We know g and we know theta is 20 degrees.
00:59
And so that's our force from the first summation equation.
01:03
So next we can sum forces in the y direction.
01:07
And that gives us this equation, n minus m g cosine theta equals 0.
01:11
And so again, we can solve for n.
01:15
We're going to get that n equals 276 .6 newtons.
01:26
And now we can sum the moments giving us this equation here.
01:31
Negative t times the inner radius plus f times the outer radius equals the inertia, which is mass times the radius, of gyrations squared times alpha, which is our angular acceleration.
01:45
And so we can again solve this.
01:46
We can get 60 plus 0 .45 times f.
02:00
This is going to be equal to 2 .27 alpha.
02:10
And so we also know that a equals r times alpha in this equation down here.
02:16
So we can substitute that in by dividing r.
02:19
We get alpha equals a over r.
02:21
So we can sub that in.
02:23
And we also have an expression for force up here, 200 minus 3a.
02:27
And so we can substitute both of those in...