Main content

## MCAT

### Course: MCAT > Unit 8

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 the block stationary

Investigate the forces at play when a block of wood on a wedge is stationary. Understand the role of friction and the normal force in balancing the force of gravity. Learn how to calculate the coefficient of static friction and its implications in real-world scenarios.

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?

- Where is the first video that this video refers to?(20 votes)
- It's actually the next video in this section!

I think they are just out of order on accident :)(10 votes)

- Hello,

At around7:30, the video indicates that static friction is 49/(50 * 3^.5) = 0.72. However, this value calculates to ~ 0.59. Regards.(5 votes) - This video refers to a previous video, where is it..:)(2 votes)
- Curious as to why 1 N was arbitrarily put down to set the block into motion. How do we know that 1 N will actually move the block?? Could we have just guessed 2.. ? The coefficient of static friction is a constant so I'm just slightly confused.(1 vote)
- I believe he was just using that number randomly for this example. It doesn't have any real significance. He mentioned that in this scenario we found that value experimentally. It allowed us to figure out the "budging" force, as he calls it. It could have been any value really. He just needed a number for the example. This example just shows how the coefficient of static friction relates to the "budging" force and the normal force.

The coefficient of static friction is a constant, but remember this constant is different for different materials.

Hope this helps.(3 votes)

- Wouldn't the coefficient he calculated be the kinetic coefficient of friction since the block is moving 1N down the ramp?(1 vote)
- Do we have to leave nonperfect square roots as is or can we convert it to a decimal?(1 vote)
- Do as you see fit. I tend to maintain them as square roots just because it will often look cleaner and more manageable when plugged into equations.(1 vote)

- thank you Khan academy for the free lessons you take(1 vote)

## Video transcript

- [Voiceover] In the last video we had a ten kilogram mass sitting on top of an inclined plane and the plane had a 30 degree angle. And in order to figure out what would happen to this block, we broke down the force of gravity on this block into the components that are parallel to the surface of the plane, and perpendicular to the surface of the plane. And, for our perpendicular component, we got 49 times the square root of three Newtons downwards. It's 98 times this quantity over here, downwards. But we said, look, we don't see this block of ice. We don't see it accelerating downwards into this wedge because the wedge is supporting it. So there must be a counteracting force that the wedge is exerting on the block. And that counteracting force is the normal force of the wedge on the block of ice. And that is exactly opposite to the force of gravity in this direction. The normal force of gravity I guess you could say is the normal force of the wedge. And then these completely balance each other out in that normal direction, in that perpendicular direction. And that's why this block is not accelerating either in that direction, or in this direction over here. For the one component of the force of gravity that did not seem to have any offset, at least the way we set up the problem in the last video, is the force that is, or the component that is parallel. That is parallel to the surface of the plane. And we figured that out to be 49 Newtons. It was essentially the weight of the block times the sine of this angle. And similar, if there's no other forces, then it would be accelerated in this direction. And to figure out how the rate of acceleration, you take the force in that direction divided by the mass of the block, and you get 4.9 meters per second squared. Now let's say that that wasn't happening. Let's say were to look at this system right over here and the block is just stationary. And now, for the sake of argument, let's assume it's not ice on ice. Let's assume that they're both made out of wood. And now all of a sudden we have a situation where the block is stationary. What would, if it's stationary, what is necessarily the case? Well, we already said, we already determined that if it's not accelerating in this normal direction and in this perpendicular direction, there must be zero net forces on it. But if it's stationary as a whole, then there must be zero forces in this parallel component as well. So there must be some force counteracting this 49 Newtons that wants to take it down the slope. So there must be some force. There must be some force that is counteracting, that is counteracting that force, the component of gravity that wants to take it down, accelerate it down the slope. And the question is, what is this force? We're dealing a situation now where we're dealing with a stationary block. Where we're dealing with a stationary block. A block that is not accelerating. So what is that force? And I think you know, from experience maybe, what is the difference between a block of wood on top of a block of wood and a block of ice on top of a block of ice? A block of ice on top of a block of ice is much more slippery. There is no friction between ice and ice, but there is friction between wood and wood. And to make it maybe a little more tangible, maybe we'd put some sandpaper. Maybe we'd put some sandpaper on this surface over here. And then it becomes a little bit clearer. The force that is keeping this block from sliding down in this situation is the force of friction, is the force of friction. And the force of friction will always act in a direction opposite to the motion if there was not any friction. So what is the... Or the potential, I should say. The potential acceleration. I shouldn't say the motion. The potential acceleration if there was not any friction. So what is the force of friction in this case? Well, if this block is completely stationary, it's not accelerating down the ramp, the force of friction over here is going to be 49 Newtons. It's going to be 49 Newtons, but it's going to be upwards. Kind of up the ramp. 49 Newtons up, up the ramp. Now what I want to think about, and this is something that can be determined experimentally if you have blocks and ramps. Even if you don't have blocks and ramps. As long as you have some way of measuring force, you can do this experimentally. But an interesting thing, a question, an interesting question here is, well how much do I have to push on this block until it starts to move down the ramp? How much do I have to push on it? And let's say you were able to experimentally determine that if you can apply, if you can apply another one Newton. So let's say that if you can apply another one Newton on this. So above and beyond the parallel force, if you can apply another... I should do it small. If I can apply another one Newton. If I can apply another one Newton, then all of a sudden I can at least get the box to start accelerating down. Not at the rate at which it would do naturally, but I could just start to nudge it down. If I give it another push of one Newton in the parallel direction. So what is the total force that it is has is exactly one Newton? So the total force at this point that's acting on it in order to just start to budge it. The total force, I'll call this the budging force. You'll never hear this in a traditional class. F sub B for the budging force. The budging force that's in the parallel direction. Let's say that is, if I'm applying one Newton in this direction, and it already has 49 Newtons due to the component of gravity in this direction, then my budging force is 50 Newtons. And so an interesting thing that you can determine based on the materials that are coming in contact with each other, is just how much force you need just to start to overcome friction. In this case it's the budging force, and that's a term that I made up. And an interesting ratio, which tends to hold for given materials pretty well, is the ratio between the amount of force just to budge it and the amount of force between the two objects, between how much force they are exerting on each other. And in this case, the amount of force that is being exerted by the wedge on the block is the normal force. It's 49 square roots of three. So maybe I should say the magnitude. The magnitude of the budging force over the magnitude of the force that is putting these two things in contact. In this case it is 49 square roots of three Newtons. So let me write this, I'll write this over the magnitude of the normal force, of the normal force. And that is 49 square roots of three. We call this the coefficient of static friction. The coefficient of static friction. And we're going to use this deeply, or a little bit more deeply in other problems. But it tends to hold true for different material, so that in the future if you have a different mass, or maybe a different incline, but you have the same materials, given the normal force you could figure out the budging force. You could figure out exactly how much force you need to put if you know this, which you usually figure out experimentally. So what would be the value in this case? You have 50 Newtons over 49 square roots of three Newtons, so this is zero point seven two. And then you can use this information... So let me write this down. This is the coefficient, coefficient. Coefficient's another word that I have trouble spelling. Coefficient of static friction. Static friction. We call it the coefficient of static friction because this deals with the ratio of the force of friction relative to the normal force, or I guess the force necessary to just overcome the force of friction and just kind of get right over the most friction that can be applied by kind of the abrasiveness of the two things when the object is stationary. And I'll do a whole video on the difference between what how this is different when an object is stationary to when it's moving. A lot of times they are very, very, very close, but for certain materials you have a very, at least a noticeably different coefficient of friction when the object is stationary as opposed to when it is moving. So I'll leave you there and in the next few videos we'll use coefficient of friction, or calculate coefficients of friction to do some more problems.