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All right, hello.
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In this question, we're going to discuss the block on a ramp.
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So i've drawn a ramp here.
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It's got some angle theta.
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And we're going to have three forces acting on this.
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We're going to have its weight.
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And its weight pulls straight down, regardless of how the incline is.
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The normal force acts normal to the plane.
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So what that means is that if we extend it down, that's at a right angle there.
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So it's pointing completely normal to the ramp.
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And that's going to be our normal force.
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And then we have a force of static friction.
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If there was no friction, this would obviously slide down the ramp.
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So we have a force of friction, in this case, static friction, holding that up.
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Now, we're told to define our axes like so, where x is going down the ramp and y is normal to the ramp.
00:43
The reason that's convenient is because these two forces are now just in the x or y directions.
00:48
However, our force of gravity is kind of kiddywampus pointing in not just the negative y direction.
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So the way we get around that is we can go ahead and we can break this vector down into its components.
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We have these two components.
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This is going to be our x component.
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And this is going to be our y component.
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And now, this angle here is the angle theta.
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And that's because of similar triangles.
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But a good way to remember it is let's consider what fgy and fgx are going to be.
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So if we have fgy, well, that's going to be our force of gravity, which is going to be mg times the cosine of theta.
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Now, let's think about what happens if theta is equal to 0.
01:37
That means that we have no ramp.
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It's just a flat surface.
01:41
And so we'll have this situation here.
01:43
We just have a block sitting on a table, basically.
01:46
And we'll have a natural weight going down and our overforce going up.
01:50
Well, if theta is equal to 0, this term, cosine of theta, is going to be cosine of 0.
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And cosine of 0 is 1.
01:59
So our force of gravity in the y direction will just be mg, which is just our force of gravity.
02:05
As a mental exercise, you can say, well, if this was not our angle theta, this would be the sine of theta here.
02:12
And if that was the sine of theta, then when theta equals 0, we would have no weight, which does not correspond to reality.
02:21
So because this angle here is theta, fgy is going to be mg cosine theta.
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And the other component is going to be the sine of theta component.
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And that's the x component, which is ultimately what is going to be, even though it's drawn down here, we can translate that up here.
02:35
So it's going to be pulling down the incline.
02:39
So with that in mind, we can go ahead and do a sum of the forces in the x and y directions.
02:44
Look in the x direction.
02:45
Like we said, we have a component of our weight pointing down and to the right.
02:49
And then that is being opposed by our force of friction, in this case, our static friction...