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Ice accelerating down an incline

Explore the physics of an ice block sliding down an icy incline. Understand the forces at play, including gravity and the normal force, and how they contribute to the block's acceleration. Learn to calculate these forces using trigonometry and Newton's second law. Created by Sal Khan.

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  • male robot hal style avatar for user Ruben
    Maybe I missed it in the video, but why do the Fg(parallel) and the Fg (perpendicular) not ad up to 98 N? shouldn't they? Because all we did was split up the gravity vector into 2 vectors, so if we add them it should be the same, no?.. Thank you.
    (24 votes)
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    • blobby green style avatar for user Jeffrey Wiedl
      Actually they do add up to 98! :-) These are vectors, so adding them depends on the angle between them, not just their values. If they were pointed the same way, then they would add up to the original value, but they are at 90 degrees to each other.
      Explanation:
      Look at the screenshot of the video (you don't have to replay it to see). There is a right triangle with the hypotenuse as the force from gravity (98) and the two component values for the perpendicular force and the parallel force. If you use the Pythagorean theorem with the components, you get 98 for the magnitude of the force due to gravity vector.
      Details:
      Fgparallel = 98 sin (30)
      Fgperpendicular = 98 cos (30)
      These are the legs of a right triangle.
      The hypotenuse then is Ftotal so...
      c^2 = a^2 + b^2
      Ftotal^2 = Fgparallel ^2 + Fgperpendicular ^2
      Ftotal^2 = (98sin(30)) ^2 + (98cos(30)) ^2
      Ftotal^2 = (98 * 1/2) ^2 + 98 * (sqrt(3)/2) ^2
      Ftotal^2 = 2401 + 7203
      Ftotal^2 = 9604 (take the square root of both sides)
      Ftotal = 98!
      (70 votes)
  • old spice man green style avatar for user Tushar
    Hey there Sal, it was a great video and i looove your website but after watching this, I had a question in my mind.
    It the force acting on the block parallel to the slope is 49 N and the force due to gravity is 98 N, where does the remaining 49 N go?
    Is the remaining 49 N the force that is being counteracted by the normal force on the block?
    thx
    (7 votes)
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    • male robot hal style avatar for user Sheridan Teasel
      Hi Tushar, you are on the right lines but remember that forces are vectors so they have direction and cannot be simply added or subtracted like that unless they are acting in the same or opposite directions. What Sal did is to decompose the 98N into two components: the 49N acting parallel to the surface of the slope and the 49(3^0.5)N which is the force that the block exerts perpendicular to the slope. And as you say that component is exactly equal and opposite to the normal force which the slope exerts on the block. But you cannot do 98-49N because they are vectors in different directions. Please remember to be very wary of doing that with vectors.

      All the best,
      Sheridan
      (16 votes)
  • blobby green style avatar for user UdyAgg
    Hello!
    This video has helped me understand the basics of forces on the inclined plane (THANK YOU!) but I have a question:
    From to , why is the Cossin calculated on the y component and the sin calculated on the x component? Please explain.

    Udyant Aggarwal
    (5 votes)
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    • female robot ada style avatar for user Angelica Chen
      Understandably, you seem to have gotten turned around! The cosine is used in this case because what we have is an angle on a right triangle. Based on this angle, we want to know the the side beside it, which we will then compare to the hypotenuse.

      Basically, cos and sin should not be thought of as x and y (even though in the unit circle they can be used this way). To keep yourself less confused, the sine of an angle is the ratio of the leg directly opposite to it over the hypotenuse, while the cosine is the leg beside the angle over the hypotenuse. Trying to rotate and reflect everything into terms of x and y will just get you doing extra work.

      If you still have questions, I suggest you draw a bunch of triangles and turn them around to see what I mean on your own.
      (5 votes)
  • starky ultimate style avatar for user Ronit Danti
    Does ice itself actually have negligible friction? Isn't it because melting ice creates a surface layer of water that enables it to move smoothly?
    (Sorry, this isn't really related to the video in itself, but I've been having this doubt for a while)
    (6 votes)
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  • blobby green style avatar for user Maureen Burchell
    If a body is held stationary on an incline by more than just friction, would the frictional forcre be regarded as static or dynamic?
    (3 votes)
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  • aqualine ultimate style avatar for user MathBunnyRabbit
    How does the surface area of an object affect the sliding frictional force present while pulling it across a surface?
    (5 votes)
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    • duskpin ultimate style avatar for user Michael
      It also depends on what you mean by "surface area." An increase in surface area in the sense that the surface is bumpier or rougher would increase the frictional force (since it increases the coefficient of friction) but just increasing the surface area in the sense that the surface of contact is greater would generally have minimal impact on the frictional force.
      (1 vote)
  • blobby green style avatar for user Allison
    What would happen if kinetic friction was involved in the problem but it was still accelerating (rather than being stationary in the next video)?
    (4 votes)
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  • blobby green style avatar for user nickvanraden
    can you explain why dropping things of different mass reach the ground at the same time but sliding things of different mass down an incline do not?
    (4 votes)
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  • leafers ultimate style avatar for user Pranav Krishna
    Mass changes with the speed of the object ? [Remember m'=m/(1-(v^2/c^2)) ]
    Does this make mass a scalar quantity?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user steffen bakken
    How does normal force relate to liquids?
    (2 votes)
    Default Khan Academy avatar avatar for user

Video transcript

Let's say that I have a ramp made of ice. Looks like maybe a wedge or some type of an inclined plane made of ice. And we'll make everything of ice in this video so that we have negligible friction. So this right here is my ramp. It's made of ice. And this angle right over here, let's just go with 30 degrees. And let's say on this ramp made of ice, I have another block of ice. So this is a block of ice. It is a block of ice, it's shiny like ice is shiny. And it has a mass of 10 kilograms. And what I want to do is think about what's going to happen to this block of ice. So first of all, what are the forces that we know are acting on it? Well if we're assuming we're on Earth, and we will, and we're near the surface, then there is the force of gravity. There's the force of gravity acting on this block of ice. And the force of gravity is going to be equal to-- it's going to be in the downward direction, and its magnitude is going to be the mass of the block of ice times the gravitational field times 9.8 meters per second squared. So it's going to be 98 newtons downward. So this is 98 newtons downward. I just took 10 kilograms. Let me write it out. So the force due to gravity is going to be equal to 10 kilograms times 9.8 meters per second squared downward. This 9.8 meters per second squared downward, that is the field vector for the gravitational field of the surface of the earth, I guess is one way to think about it. Sometimes you'll see the negative 9.8 meters per second squared. And then that negative is giving you the direction implicitly because the convention is normally that positive is upward and negative is downward. We'll just go with this right over here. So the magnitude of this vector is 10 times 9.8, which is 98 kilogram meters per second squared, which is the same thing as newtons. So the magnitude here is 98 newtons and it is pointing downwards. Now what we want to do is break this vector up into the components that are perpendicular and parallel to the surface of this ramp. So let's do that. So first, let's think about perpendicular to the surface of the ramp. So perpendicular to the surface of the ramp. So this right over here is a right angle. And we saw in the last video, that whatever angle this over here is, that is also going to be this angle over here. So this angle over here is also going to be a 30-degree angle. And we can use that information to figure out the magnitude of this orange vector right over here. And remember, this orange vector is the component of the force of gravity that is perpendicular to the plane. And then there's going to be some component that is parallel to the plane. I'll draw that in yellow. Some component of the force of gravity that is parallel to the plane. And clearly this is a right angle, because this is perpendicular to the plane. And this is parallel to the plane. If it's perpendicular to the plane, it's also perpendicular to this vector right over here. So we can use some basic trigonometry, like we did in the last video, to figure out the magnitude of this orange and this yellow vector right over here. This orange vector's magnitude over the hypotenuse is going to be equal to the cosine of 30. Or you could say that the magnitude of this is 98 times the cosine of 30 degrees newtons. 98 times the cosine of 30 degrees newtons. And if you want the whole vector, it's in this direction. And the direction going into the surface of the plane. And, based on the simple trigonometry-- and we go into this in a little bit more detail in the last video-- we know that the component of this vector that is parallel to the surface of this plane is going to be 98 sine of 30 degrees. Sine of 30 degrees. Sine of 30 degrees. And this comes straight out of this magnitude, which is opposite to the angle over the hypotenuse. Opposite over hypotenuse is equal to sine of an angle. And we did all the work over here. I don't want to keep repeating it. But I always want to emphasize that this is coming straight out of basic trigonometry, straight out of basic trigonometry. So once you do that, we know the different components. We can calculate them. Cosine of 30 degrees is square root of 3 over 2. Sine of 30 degrees is 1/2. That's just one of those things that you learn and you can derive it yourself using 30-60-90 triangles, or actually even equilateral triangles. Or you could use a calculator. But it's also one of those things that you memorize when you take trigonometry. So no kind of magical trick I did here. And so if you evaluate this, 98 times the square root of 3 over 2 newtons, tells us that-- let me write it in that same orange color-- the force, the component of gravity that is perpendicular to the plane. And this kind of implicitly gives us this direction, it's perpendicular to the plane. But the force component of gravity that's perpendicular to the plane is equal to 98 times square root of 3 over 2. 98 divided by 2 is 49. So it's equal to 49 times the square root of 3 newtons. And its direction is into the surface of the plane, or downward or, let me just write, into surface of plane. Surface of the plane, or the surface of the ramp. And it's in this direction over here. And I have to do this because it's a vector. I have to tell you what direction it's going in. And the component of the force of gravity that is parallel. The component of the force of gravity that is parallel, I drew it down here, but I could shift it up over here. It's the same exact vector. The component of gravity that is parallel to the surface of the plane is 98 times sine of 30. That's 98 times 1/2, which is 49 newtons. And it's going in that direction, or parallel to the surface of the plane. Parallel, I always have trouble spelling parallel. Parallel to-- don't even know if I spelled it right-- surface of the plane. So what's going to happen here? Well, if these were the only forces acting on it. So if we had a net force going into the surface of the plane of 49 square roots of 3 newtons. If this was the only force acting in this dimension or in the dimension that is perpendicular to the surface of the plane, what would happen? Well, then the block would just accelerate. At least just due to this force it would accelerate downward. It would accelerate into the surface of the plane. But we know it's not going to accelerate. We know that there's this big wedge of ice here that is keeping it from accelerating in that direction. So at least in this dimension, there will be no acceleration. When I talk about this dimension, I'm talking about in the direction that is perpendicular to the surface of the plane. There will be no acceleration because this wedge is here. So the wedge is exerting a force that completely counteracts the force, the perpendicular component of gravity. And that force. You might guess what it's called. So the wedge is exerting a force, just like that, that's going to be 98 newtons upward. The wedge is going to be exerting a force that is 49 square roots of 3, because this right here is 49 square roots of 3 newtons into. And so this is 49 square roots of 3 newtons out of the surface, out of the surface. And this is the normal force. It is the force perpendicular to the surface that essentially, you could kind of view as the contact force that the, in this case, that the surface is exerting to keep this block of ice from accelerating in that direction. We're not talking about accelerating straight towards the center of the earth. We're talking about accelerating in that direction. We broke up the force into kind of the perpendicular direction and the parallel direction. So you have this counteracting normal force. And that's why you don't have the block plummeting or accelerating into the plane. Now what other forces do we have? Well, we have the force that's parallel to the surface. And if we assume that there's no friction-- and I can assume that there's no friction in this video because we are assuming that it is ice on ice-- what is going to happen? There's no counteracting force to this 49 newtons. 49 newtons parallel downwards, I should say parallel downwards, to the surface of the plane. So what's going to happen? Well, it's going to accelerate in that direction. You have force is equal to mass times acceleration. Force is equal to mass times acceleration. Or you divide both sides by mass, you get force over mass is equal to acceleration. Over here, our force is 49 newtons in that direction, parallel downwards to the surface of the plane. And so if you divide both by mass, if you divide both of these by mass. So that's the same thing as dividing it by 10 kilograms, dividing by 10 kilograms, that will give you acceleration. That will give you our acceleration. So acceleration is 49 newtons divided by 10 kilograms in that direction, in this direction right over there. And 49 divided by 10 is 4.9, and then newtons divided by kilograms is meters per second squared. So then you get your acceleration. Your acceleration is going to be 4.9 meters per second squared. And maybe I could say parallel. That's two bars. Or maybe I'll write parallel. Parallel downwards to the surface. Now I'm going to leave you there, and I'll let you think about another thing that I'll address in the next video is, what if you had this just standing still? If it wasn't accelerating downwards, if it wasn't accelerating and sliding down, what would be the force that's keeping it in a kind of a static state? We'll think about that in the next video.