00:01
Okay, so here we've got this block sliding down this trough.
00:05
And we can kind of treat this as a normal sliding problem to start where we have the downward weight force that is acting straight down, and this is going to be split into parallel and perpendicular components.
00:19
We're going to have a friction force that is up the ramp, and then we're going to have a normal force that is perpendicular to the ramp.
00:30
So if we write the forces acting on this thing, we're going to have the mass of it times its acceleration equals the parallel component of the weight force, so mg sine a theta, minus that friction force.
00:48
And what is this friction force? well, the friction force is going to be equal to mu kinetic friction times the normal force.
00:59
However, this is complicated because we're actually going to have two different normal forces acting on this.
01:06
We're going to have a combined normal force, and we're going to have a friction force due to each side of the trough.
01:23
And this coefficient of kinetic friction is given to us in terms of the trough itself.
01:28
So we are going to have this equation here where we have twice the friction force.
01:34
We're going to have a little bit of friction over here.
01:36
We're going to have some friction over here.
01:38
So now we've got our friction forces, 2 mu kinetic friction times the normal force acting on each side.
01:48
Well, now we need to figure out what that is.
01:52
For that, we need to look at what is the combined normal force here? well, the combined normal force, and let's just call that fnc for combined, that's going to be equal to the vertical part of each of these because their horizontal parts are going to cancel each other out.
02:15
And so we're going to get two times their components cosine of theta, where this theta, that's not actually theta.
02:32
That's 45 degrees.
02:36
So cosine of 45 degrees.
02:38
Well, 2 times cosine of 45 degrees, that's going to give us square root of 2.
02:45
So we'll just call this square root of 2 times this normal force there.
02:50
Well, this is the same as this.
02:52
So i can then plug and solve that for that.
02:57
We're going to get this over square to 2, or square to 2, fn, c, over.
03:04
2.
03:05
Plug this into this.
03:06
The 2s are going to cancel...