00:04
All right.
00:05
So we have a sled going up a hill.
00:14
This is a 15 degree hill and there's a sled.
00:38
Constant speed up a 15 degree hill.
00:43
Okay.
00:44
Well, the weight of the sled we're given is 60 newtons.
00:54
There's a rope and it's 35 degrees to the horizontal.
01:00
So the rope here is 35 degrees.
01:15
I'm going to call this alpha.
01:19
Didn't do a good job of drawing alpha to the horizontal.
01:27
Okay.
01:34
What's the coefficient of friction between the sled and the snow? all right.
01:39
So, the weight of the sled is going to have two components.
01:55
And furthermore, i need to figure out what the angles are on this.
02:10
If this is theta down here, i mean, phi down here, then this is going to be 90 minus phi, and then this is going to be phi again.
02:21
And so, the normal force, which would be here, is going to be the weight times the cosine of phi.
02:44
So that would be, i'm going to call this f, f, cosine, phi.
02:53
The force here, which i'll call f sub s, is going to be f sine of phi.
03:10
Now, if this angle is 35 degrees, then an angle with the slope is going to be 35 minus 15.
03:28
So i'm going to call this beta is going to be 35 minus 15.
03:33
That's 20 degrees.
03:36
So it's going to have two components.
03:43
And the components would be in the direction and perpendicular to the direction.
03:52
So he's pulling with a 20 -newton force.
04:09
So i'm going to write p equals, no, 20.
04:16
So p in the direction of the slope is going to be p, and then that's going to be the cosine of beta, which is the 20 degrees.
04:37
Okay.
04:39
Now there's no acceleration because it's a constant speed.
04:44
And therefore, in the slope direction, we have...
04:55
F sub s but we also have the frictional force so uh the pull which would be p cosine beta has to equal f sub s which is f sign phi it's strange that one of them is a sign and the other is cosine but yeah that makes sense um and then minus the friction force, or no, plus the friction force, which is mu times the normal force, which is f cosine phi.
05:52
Okay.
05:54
So i'm trying to figure out mu.
05:58
So i can begin by dividing f on both sides.
06:02
P over f cosine beta is going to be sine of five.
06:12
Plus mu cosine phi.
06:19
Okay.
06:23
Okay, good.
06:24
So, p over f cosine beta minus sine of phi over cosine of phi would be mu.
06:41
All right, let's go ahead and put this in a calculator.
06:45
Beta is 20.
06:48
Phi is 15.
06:58
Phi is 15, beta, 20.
07:06
Okay? t is 25, f is 60.
07:17
Now i need to put this in here.
07:20
P over f cosine beta minus sine phi over cosine phi and that would be mu, which i'm just going to write as you here.
07:52
0 .137.
08:10
Okay.
08:12
Now, he jumps on the sled and slides down the hill.
08:22
What's his acceleration? okay, so in part b, he's on the sled, he's going down the hill.
08:32
Again, we got the same phi, 15 degrees.
08:39
But he's the on the sled.
08:42
He's on the sled.
08:46
We don't know his mass, so that must not matter.
08:54
So let's draw this again...