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### Course: MCAT > Unit 9

Lesson 4: Vector analysis and forces acting on an object- Normal forces questions
- Force of tension questions
- Forces on inclined planes questions
- Balanced and unbalanced forces
- Normal force and contact force
- Normal force in an elevator
- Slow sock on Lubricon VI
- Normal forces on Lubricon VI
- Inclined plane force components
- Ice accelerating down an incline
- Force of friction keeping the block stationary
- Correction to force of friction keeping the block stationary
- Force of friction keeping velocity constant
- Introduction to tension
- Introduction to tension (part 2)
- Tension in an accelerating system and pie in the face
- Intuition on static and kinetic friction comparisons
- Static and kinetic friction example

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# Force of friction keeping velocity constant

We delve into the world of friction, focusing on static and kinetic types, and their impact on forces on inclined planes. The discussion highlights how friction adjusts with applied force and its role in motion. We also learn how to calculate the coefficient of kinetic friction.

Visit us (http://www.khanacademy.org/science/healthcare-and-medicine) for health and medicine content or (http://www.khanacademy.org/test-prep/mcat) for MCAT related content. These videos do not provide medical advice and are for informational purposes only. The videos are not intended to be a substitute for professional medical advice, diagnosis or treatment. Always seek the advice of a qualified health provider with any questions you may have regarding a medical condition. Never disregard professional medical advice or delay in seeking it because of something you have read or seen in any Khan Academy video. Created by Sal Khan.

Visit us (http://www.khanacademy.org/science/healthcare-and-medicine) for health and medicine content or (http://www.khanacademy.org/test-prep/mcat) for MCAT related content. These videos do not provide medical advice and are for informational purposes only. The videos are not intended to be a substitute for professional medical advice, diagnosis or treatment. Always seek the advice of a qualified health provider with any questions you may have regarding a medical condition. Never disregard professional medical advice or delay in seeking it because of something you have read or seen in any Khan Academy video. Created by Sal Khan.

## Want to join the conversation?

- Is there a particular law of physics that states the reason behind kinetic friction being more "potent" than static friction?(2 votes)
- on a molecular level: during static friction the molecules sort of merge between each other since the surfaces are not exactly smooth (there are some impurities so the object's molecules wedge in). during dynamic friction the molecules do not have time to get stuck, they just slide by each other. therefore, there is less to resist when the molecules are already moving so sliding past each other than when they are stationary and sort of sunk into each other.(4 votes)

- How is the object moving down the ramp (with constant velocity) is all the forces are balanced?(2 votes)
- Does Sal mean to say newtons 2nd law at4:21?(1 vote)
- Why is it important to pull an object at a constant velocity to determine the force of friction?(1 vote)
- hi

can u help me with this problem, please becuse i dont understand the friction here.

what happend when we have friction in this problem.

here is the problem:

a toy car is pulled with a force of 10N for 5m.If the friction force between the car and the surface is 5N,what is the net work done on the car?

i am not understanding what effect friction does?(1 vote)- The net force is 5 N because you have to subtract 10N from 5 N it(1 vote)

## Video transcript

- [Voiceover] I wanna make a quick clarification to the last video and then think about what's friction up to when the block is actually moving. In the last video, we started off with the block being stationary. We knew that the parallel component of the force of gravity on that block was 49 newtons downward, down the slope. So, when the block was stationary we said there must be an offsetting force, and we said that's the force of friction and it must be 49 newtons upwards. So they completely net out in that direction. Now what we said is, we're going to keep applying a little more force until we can budge this block to start accelerating downwards. I said I applied, I kept applying a little bit more force, a little bit more force until I get to one newton and then the block started to budge. At that point, when it's started to budge, I'm applying this one newton over here, right over here, there's already 49 newtons of force or the component of gravity in this direction, so, combined, we're providing 50 newtons to just start budging it, to just overcome the force of friction. The one thing I want to clarify here is that this whole time, the force of friction was not constant at 49 newtons. When I wasn't messing with this block and the parallel component of the force was 49 newtons, then the force of friction was 49 newtons. When I started to press on it a little bit, apply a little bit of force, maybe I applied a tenth of a newton on top of that, then the force of friction was 49 and one tenth newton, because it was still providing enough force that this block was not moving. Then maybe I applied half a newton, and so the total force in the downward direction would have been 49 and a half newtons. But if it still was not moving, then the force of friction was still completely overcoming it. So the force of friction at that point must have been 49 and a half newtons. All the way up to the combined force in the downward direction being 49.999999 newtons, and then the force of friction was still 49.999999 newtons. All the way until I hit 50 newtons and then the block started to budge, which tells us that the force of friction, now, or at least the force of static friction, all of a sudden, now couldn't keep up and it started to accelerate downward. In that static scenario, the force of friction changed as I applied more or less force in this downward direction. Now with that out of the way, let's take a different scenario. Let me just redraw that same block, since all of this is getting messy. So we have the same block, as we said in the last video, we're assuming this is wood on wood. This is the wedge, this is the block right over here. We know that the component of gravity that is parallel to the plane right there is 49 newtons. We know that this is 49 newtons. We know the component of gravity that is perpendicular to the plane. We figured it out this two videos ago, is 49 square roots of three newtons, 49 square roots of three newtons. We know that this block is not accelerating in this normal direction, so there must be some force counteracting gravity in that direction, and that's the normal force of the wedge on the block. So that is going in that direction at 49 square roots of three newtons. Now, instead of assuming that this block is stationary, let's assume that it's moving with a constant velocity. So, now we're dealing with, we do that in a different color. So now we're dealing with a scenario where the block has a constant velocity. Constant velocity. For the sake of this video, we'll assume that that constant velocity is downward. So the constant velocity, V, is equal to, I don't know, let's say it is 5 meters per second down the wedge or down the ramp or I guess we could say in the direction that is parallel to the surface of the ramp. So it's in this direction, right over there. So that's the constant velocity. So what are all the forces at play? And be very careful here, there might be a temptation that says, "OK, you know there's a net force here, "we're moving so maybe that's the net force "that's causing the move." But remember, this is super important. This is Newton's first law. If you have a net force, if you have an unbalanced force, it will cause it to accelerate and we are not accelerating here. We have a constant velocity. We are not accelerating here. Not accelerating here. So, if you're not accelerating in that direction that means that the force in that direction must be balanced, so there must be some force acting in the exactly opposite direction that keeps this thing from accelerating downwards. It must be exactly 49 newtons in the opposite direction. As you can imagine, this is the force of friction. This right over here is the force of friction. The difference between this video and the last video, is last time friction was static, even at 49 newtons the box was stationary. You had to keep nudging it 'til you got to 50 newtons, then it started moving. Here we're just jumping into this picture where we see a box that's moving down the slope at 5 meters per second. We don't know how much force it took to overcome static friction, but we do know that there is some force of friction that is keeping this box from accelerating, that's keeping it at a constant velocity. That is completely negating the parallel component of the force of gravity, parallel to the surface of this plane. Given this, let's calculate another coefficient of friction, this is going to be the coefficient of kinetic friction, because now we are moving down the block. I'll do a video on why, sometimes, a coefficient of static friction can be different than the coefficient of kinetic friction. The coefficient of kinetic friction, ^we'll write it, so this is the Greek letter µ, ^and we put this little lower case k ^here for kinetic, moving friction. ^It's going to be equal to the force of friction, ^or I should say, the magnitude of the force of friction, ^over the normal force, or I should say, ^the magnitude of the normal force. You can derive this experimentally. One, if you just observed this whole thing going on and you knew the mass of the block so you knew the component of gravity that's going in this direction, if you knew this angle was 30° from the last situation, ^you could figure out this coefficient of kinetic friction. ^What's cool about this is this is, in general, going to be true for any two materials that are like this. Maybe this is a certain type of wood on a certain type of wood, or a certain type of sandpaper on a certain type of sandpaper, whatever you're talking about and then you can use that to make predictions if the incline was different or the mass was different or even if you were on a different planet or if someone was pressing down on this block that would change the normal force. ^Given this, right here, let's figure out, ^for the sake of doing it, ^the coefficient of kinetic friction here. ^The coefficient of kinetic friction, the force of friction ^here completely offsetting ^the parallel force of gravity, parallel to the surface, ^is 49 newtons. ^The normal force here, the force of contact ^between these two things, this block and this wedge ^is 49 square roots of three newtons. ^We get 1 over the square root of 3 is equal to 0.58 ^and there's no units here, because the units cancel out. It's a unit-less measurement. Now, the interesting thing here, is that the coefficient, the way I've set up this problem, the coefficient of kinetic friction is lower, if we assume the same materials, than the coefficient of static friction was. For some materials, they might not be that different, but for other materials, the kinetic friction can be lower than static friction. You never see a situation where static friction is lower, the coefficient of static friction, at least that I know of, is lower than kinetic friction, but you do see situations where the coefficient of kinetic friction is lower than the coefficient of static friction. Once something is moving, for some reason, and we'll theorize why that might be, for some reason, friction is a little less potent than when something is stationary. We can say this generally, that the coefficient of kinetic friction is less than or equal to the coefficient of static friction. It's a little bit easier, or friction provides a little less, or less than, or equal to the force when something is moving than when something is stationary. I'll think about that a little bit deeper in the next video.