00:01
In this problem, you have a block that's being pushed up a constant velocity, up the 24 -degree incline.
00:10
And the goal is to find what applied force it'll take to actually accomplish that.
00:17
On these problems, the key thing, the next thing to do is a free by diagram.
00:24
So here's our block.
00:26
We know the force f is up the incline, so that one's taken care of.
00:29
Normal force is perpendicular to the surface the block will be trying to go through.
00:35
Remember, it stops the block from going through that surface.
00:39
So it's perpendicular to that.
00:41
So here's the normal force, perpendicular to the incline surface.
00:46
Kinetic friction arises opposite to the motion.
00:49
Remember, it's going up the incline.
00:51
So fk will be down the incline.
00:54
And we always have weight, which is straight down.
00:58
Those are our forces.
01:01
What we need next is axes.
01:06
Simplest to do x along the incline, y perpendicular.
01:11
You don't have to.
01:12
And my x could be down the incline, my positive x.
01:15
It's a matter of choice.
01:16
But that's always best.
01:18
Try to do anything horizontal and vertical.
01:20
Just makes more work, much more work.
01:23
Now, we want to now look at newton's second law.
01:28
Now remember, in the second law, law, f -net is equal to m a, it's a vector equation, but we implement it as a component equation, f -net -x and f -net -y.
01:40
So let's look at f -net -x.
01:42
What does this mean? we're going to add up all the x components of the forces.
01:47
So let's look at our diagram.
01:49
The applied force f is only in positive x.
01:52
That means its x component is plus f.
01:56
So we'll just write f.
01:57
Remember, components tell you how to construct the vector.
02:01
I only got to give an x component to construct the applied force, and i got to draw it in the positive x direction.
02:07
That's it.
02:09
The connect friction is only the negative x direction.
02:12
So i basically say, draw, go fk units in the negative x direction.
02:17
So the x component is minus fk.
02:19
That's what it tells me.
02:21
That's what it tells me to do.
02:27
Now, normal force is only in y, so it has no x component, so it doesn't it right here.
02:32
Now, the weight now, we've got to break this up.
02:35
So we've got to draw a triangle.
02:37
Here is going to be the y component, and here is the x component.
02:43
Notice parallel to the axis.
02:45
No different than if you were vertical and horizontal.
02:49
So this is the y, this is w is mg, wy, notice is in the negative.
02:56
And this angle here is stated, and i'll show you that later, if you're not clear on that.
03:00
Wy, it's in the negative y direction.
03:02
I have to go in the negative to construct.
03:04
Negative y negative x so i'm going to write minus m g cosine cosine theta for w y w x is going to be minus m g sine theta remember sign is opposite over hypotenuse cosine is adjacent over ipotinous mg is the hypotenuse so there now we have all our components so let's put in w x here minus mg sine theta is equal to m a x but we are that is moving up the incline at constant speed.
03:38
That means vx, the x component of the velocity, is constant.
03:42
That means ax is zero.
03:46
So this is equal to zero.
03:50
So that's what we get from the x equation...