00:01
Okay, so to solve this problem, we need to do a net force equation for each of the masses and a net torque equation for the pulley.
00:13
So i'm going to start with m1.
00:15
The net force acting on m1 will be m1a, and that will equal the tension force, which we'll call t1 in this string right there, minus the friction force.
00:30
So there will be friction force.
00:32
It's going to be mu.
00:34
It should be a mu.
00:36
M1g.
00:37
So mu times the normal force.
00:39
Now i'm going to solve this for the tension force because we're going to want that later when we get to the torque and i'm going to want to substitute that in.
00:47
So i'm going to do that right now.
00:49
We can add mu m1g to both sides.
00:52
And so we'll get m1a plus mu times m1g.
00:59
We can plug numbers in.
01:00
Later.
01:02
Next we will look at m2 and so i'm going to change to green.
01:08
M2a, the net force acting on m2, will be equal to the weight force down the ramp, the parallel part of the weight force, so m2 g times the sign of 30 degrees, and then we'll have the tension force, which will call t2, up the ramp, and then also there'll be a friction force up the ramp, which will be mu times m2g cosine of 30 degrees now sign of 30 degrees is one half so we'll plug that in in a second we are going to do the same thing where we're going to solve this for t2 and t2 will equal m2g sine 30 so that's one half so m2g over 2 minus m2 we'll just plug in mu and 2g cosine 30 degrees, which is square to 3 over 2.
02:11
So let's just plug that in, actually, since we're doing sign of 30.
02:16
So square to 3 over 2.
02:18
And then we are going to subtract m2a from both sides, so minus m2a.
02:25
And we could factor some stuff out there, but we will get to that later.
02:29
And now let's look at the net torque.
02:32
So on the pulley, the net torque will be i times alpha, and that will equal the torque due to tension 1.
02:41
Torc due to a tension 2 minus tension 1.
02:44
So i'm going to choose the clockwise direction to be positive since that's the direction of motion.
02:50
And so we're going to have r -t2 minus r -t1.
02:55
And then we can substitute for i, if this is a cylinder, it's 1 -5m, capital, r squared is the rotational inertia and alpha will be the linear acceleration over r and if we divide both sides by r we get r squared there and then we'll have tension two which we're going to substitute to this whole thing in for so we have m2g minus mu and 2g square root three all over two and then minus m2a and then minus m2a then we're going to subtract to t1, which is this right here...