00:01
All right, so we're continuing with the same figure.
00:05
I am going to draw it again.
00:13
So we've got like a wedge here.
00:16
We've got phi in the corners.
00:19
Got m1 here.
00:22
We've got m2 here.
00:26
We've got a pulley up here.
00:30
Phi was 35.
00:35
M1 and m2, i forget exactly what they were.
00:42
So, oh, they were 3 .5 and 8, but i was correct about five.
00:56
Now, we're saying that now there is friction.
01:03
So, drawing a free body diagram of m1, we've got gravitational force, which is m1g, and it's going to have two components to it, this direction and this direction.
01:22
Then we've got the normal force, f1 normal, and we've got tension, but now we've got the frictional force, which is going to be mu times f1n.
01:49
Working, well, i think i'm going to let it go at that.
01:55
Now, m2 is going to have a gravitational force, which is going to be m2g.
02:15
Again, the gravitational force is going to have two components to it, parallel and perpendicular to the surface.
02:29
F2n, and then we've got tension.
02:35
The frictional force is going to be.
02:38
In the same direction as the tension for this one.
02:44
F2.
02:45
F, which is going to be mu times f2n.
02:53
Okay.
02:55
So, what are we trying to answer? what's the coefficient of kinetic friction and the tension if the acceleration is 1 .5 meters per second squared? so a equals 1 .5 meters per second squared.
03:24
We need to find mu and tension.
03:29
Okay.
03:31
So f1n is going to equal, i'm just summing the forces in the perpendicular direction, f1n is going to equal, and this is phi here, this is phi here, it's going to equal f1g, which is m1g times the sign of phi.
04:06
Now, summing the forces in the x direction, t, or in the direction of the slope, t minus mu f1n minus f1n minus ff1n minus fff.
04:33
F1g cosine phi is going to equal m1 times the acceleration.
04:52
Now let's look at the second diagram for m2.
04:59
F2n is going to be m2g sine phi.
05:07
T, wait, no, i don't want t first...