Inclined planes and friction
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Inclined Plane Force Components
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Ice Accelerating Down an Incline
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Force of Friction Keeping the Block Stationary
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Correction to Force of Friction Keeping the Block Stationary
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Force of Friction Keeping Velocity Constant
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Intuition on Static and Kinetic Friction Comparisons
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Static and Kinetic Friction Example
Inclined Plane Force Components Figuring out the components of the force due to gravity that are parallel and perpendicular to the surface of an inclined plane
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- Let's say I have some type of a block here
- This block has a mass of 'm'
- So the mass of this block is equal to 'm'
- And it's sitting on this (you can view this as) an incline plane
- or a ramp or some type of wedge
- And we want to think about what might happen to this block
- We'll start thinking about the different forces that might keep it in place
- or not keep it in place
- and all of the rest
- So the one thing we do know is
- if this whole setup is near the surface of the Earth
- We'll assume that it is for the sake of this video
- There will be the force of gravity
- trying to bring or attract this mass
- towards the center of the Earth
- and vice versa the center of the Earth
- towards this mass
- So we're going to have some force gravity
- Let me start at the center of this mass right over here
- You're going to have the force of gravity
- The force due to gravity is going to be equal to
- the gravitational field near the surface of the Earth
- we'll call that 'g'
- 'g' times the mass
- Let me just write it as the mass times the gravitational field
- near the surface of the Earth
- and it's going to be downwards
- we know that, or at least
- towards the center of the Earth
- Now what else is going to be happening here?
- Well, it's a little bit confusing because
- you can't say the normal force is acting directly against
- this force over here
- Because remember the normal force acts
- perpendicular to a surface
- So over here the surface is not perpendicular
- to the force of gravity
- So we have to think about it a little bit differently
- than we would if this were sitting on level ground
- The one thing that we can do
- and frankly that we should do
- is maybe we can break up this force
- the force due to gravity
- We can break it up into components
- that are either perpendicular to the surface
- or that are parallel to the surface
- and then, we can use those to figure out
- what's likely to happen
- what are the potentially netting forces
- or balancing forces over here
- Let's see if we can do that
- Let's see if we can break this force vector
- the force due to gravity
- into a component that is perpendicular to the surface
- of this ramp
- And also, another component that is parallel
- to the surface of this ramp
- I'll do that in a different color
- That is parallel to the surface of this ramp
- This is a little bit of unconventional notation
- but we'll call this one over here
- the force due to gravity that is
- perpendicular to the ramp
- with that little upside T
- I'm saying it's perpendicular
- Because it shows a line that is perpendicular
- to this bottom line (this horizontal line)
- And this blue thing over here
- I'm going to call this the force
- or the part of force
- due to gravity
- that is parallel
- I'm just doing those two upward vertical bars
- to show something that is parallel
- to the surface
- So this is the component of force that is
- perpendicular
- component of force that is parallel
- So let's see if we can use a little bit of
- geometry and trigonometry
- given that this wedge is at a
- theta degree incline relative to the horizontal
- If you were to measure this angle right over here
- you would get theta
- So, in future videos we'll make it more concrete
- like 30 degrees or 45 degrees or whatever
- But let's just keep it general
- if this is theta
- let's figure out what these components of the
- gravitational force are going to be
- Well we can break out geometry over here
- This, I'm assuming is a right angle
- and so, if this is a right angle
- we know that the sums of the angles in a triangle
- will add up to 180
- so, if this angle, this angle, and this 90 degrees
- (Right angles are 90 degrees)
- add up to 180
- that means that this one and this one
- need to add up to 90 degrees
- or this is theta
- this angle right over here is going to be
- 90 minus theta
- Now, the other thing that you may or may not remember
- from geometry class is
- if I have two parallel lines here
- I'm going to assume this line is parallel to this line
- and then I have a transversal
- So a let's say I have a line
- that goes like this
- We know from basic geometry that
- this angle is going to be equal to this angle
- It's comes from alternate interior angles
- And, we prove it in the geometry module
- or in the geometry videos
- Hopefully, this makes a little bit of
- intuitive sense
- and we could even think about
- how these angles would change as
- the transversal changes and all of the rest
- But, the parallel lines makes
- this angle similar to that angle
- or actually, makes it identical
- it's congruent
- this angle is going to be the same measure
- as that angle
- So can we apply that anywhere over here?
- Well we have (and it's not obvious)
- This line is perpendicular to the surface
- of the Earth
- right over here that I'm knid of shading in blue
- and so is this force vector
- it is also perpendicular to the surface of the Earth
- So this line over here
- and this line over here in magenta
- are going to be parallel
- (I can even draw that)
- That line and that line are both parallel
- When you look at it that way
- you'll see that this big line over here
- can be viewed as a transversal
- Or you can have this angle and this angle
- are going to be congruent
- They're going to be alternate interior angles
- So this angle and this angle
- by the exact same idea here
- it just looks a little more confusing here
- because I have all sorts of things
- But this line and this line are parallel
- You can view this right over here as a
- transversal
- So this and this are congruent angles
- So this is 90 minus theta degress
- This too will be 90 minus theta degress
- 90 minus theta degrees
- Now, given that can we figure out this angle
- Well one thing, we're assuming that this
- yellow force vector right here
- We're assuming that it is perpendicular to the surface
- of this plane
- or perpendicular to the surface of this ramp
- So that's perpendicular
- This right here is 90 minus theta
- So what is this angle right here
- going to be equal to?
- This angle (let me do it in green)
- What is this angle up here going to be equal to?
- So this angle plus 90 minus theta plus 90
- must be equal to 180
- or this angle plus 90 minus theta
- (Let me write this down I don't
- want to do too much by head)
- So let's just call it x
- So x plus 90 minus theta
- plus this 90 degrees right over here
- needs to be equal to 180 degrees
- We can, let's see
- We can subtract 180 degrees
- from both sides
- So if you subtract 90 twice
- you're subtracting 180 degrees
- you get x minus theta is equal to 0
- or x is equal to theta
- so whatever the inclination of the plane is
- or of this ramp
- that is also going to be this angle right over here
- and the value to that is that
- now we can use our basic trigonometry to
- figure out this component and this component
- of the force of gravity
- and to see that a little bit clearer
- let me shift this force vector down over here
- the parallel component
- let me shift it over here
- and you can see the perpendicular component
- plus the parallel component
- is equal to the total force due to gravity
- and you should also see that this is a right triangle
- that I have set up over here
- this is parallel to the plane
- this is perpendicular to the plane
- and so we can use basic trigonometry
- to figure out the magnitudes
- of the perpendicular force due to gravity and
- the parallel force due to gravity
- So the magnitude
- (let's think about it a little bit)
- the magnitude of the perpendicular force
- due to gravity
- I should say
- the component of gravity that is
- perpendicular to the ramp
- the magnitude of that vector
- A lot of fancy notation
- but it's really just the length of
- this vector right over here
- so the magnitude of this
- over the hypotenuse
- of this right triangle
- Well what is the hypotenuse of this right triangle
- Well it's going to be the magnitude of the total
- gravitational force
- (I guess you can that)
- So you can say that is 'mg'
- We could write it like this
- but that is really the
- (Well I can write it like that)
- And so this is going to be equal to what?
- We have the
- (if we're looking at this angle right here)
- we have the adjacent over the hypotenuse
- Remember
- (Let's do this in a new color)
- Soh Cah Toa
- Cosine is adjacent over hypotenuse
- So this is equal to cosine of the angle
- so cosine of theta is equal to
- the adjacent over the hypotenuse
- so if you multiply both side by the magnitude
- of the hypotenuse
- you get the component of our vector
- that is perpendicular to the surface
- of the plane
- is equal to the magnitude of the force due to gravity
- the magnitude of the force due to gravity
- times the cosine of theta
- We'll apply this in the next video
- just so you can make the numbers a little more concrete
- sometimes just the notation makes it confusing
- you'll see it's actually pretty straight forward
- and then this second thing
- we could use the same logic
- we have, if we think about the parallel vector
- right over here
- the magniude of the force of the component
- of the force due to gravity that is parallel to the plane
- over the magnitude of the force due to gravity
- which is the magnitude of 'mg'
- that is going to be equal to what?
- Well we have, this is the opposite side of the angle
- and then, so then the blue is the opposite side
- or at least its length
- is the opposite side of the angle
- and then right over here
- this magnitude of 'mg'
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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