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Force of friction keeping the block stationary

Block of wood kept stationary by the force of friction (Correction made in next video). Created by Sal Khan.
Video transcript
In the last video, we had a ten kilogram mass sitting on top of an inclined plane at 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 a perpendicular component, we got 49 times the square root of 3 N downwards That's 98 times this quantity over here, downwards But we said look! We don't see this block of ice 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 is the normal force of the wedge And these completely balance each other out in that normal, perpendicular direction And that is why this block is not accelerating either in that direction or in this direction over here But 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/component that is parallel to the surface of the plane And we figured that out to be 49N, it was essentially the weight of the block times the sin of this angle And we said look, if there is no other forces then it would be accelerated in this direction And to figure out the rate of acceleration you take the force in that direction divided by the mass of the block and you would get 4.9 m/s^2 Now let's say that wasn't happening. Let's say that you were to look at this system right over here and the block was just stationary And now, for the sake of argument, let's assume it is not ice on ice Let's assume that they are both made out of wood And now all of a sudden we have a situation where the block is stationary If it is stationary, what is necessarily the case? Well we already determined that if it is not accelerating in this normal direction there must be zero net forces on it But if it is stationary as a whole then there must be zero net forces in the parallel component too So there must be some force counteracting this 49 N that wants to take it down the slope So there must be some force counteracting the component of gravity that wants to accelerate it down the slope And the question is what is this force? We're dealing with a situation now where we're dealing with a stationary block, a block that is not accelerating So what is that force? I think you know from experience maybe what is the difference between a block of wood on top of wood and a block of ice on top of ice A block of ice on top of ice is much more slippery; there is no friction between ice and ice, but there is friction between wood and wood To make it a little bit more tangible, maybe we'd put some sandpaper on the 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 and the force of friction will always act in a direction opposite to the motion if there was not any friction or the potential acceleration if there was not any action So what is the force of friction in this case? Well, this block is completely stationary. It's not accelerating down the ramp The force of friction over here is going to be 49 N, upwards, up the ramp Now I want think about, this is something that can be determined experimentally as long as you have some way of measuring force, you can do this experimentally But the interesting question here is 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? Let's say you are able to experimentally determine that if you can apply another 1 N on this then all of a sudden, I can at least get the box to start accelerating down Not the rate which it would do naturally, but I can just start to nudge it down if I give it another push of one N in the parallel direction So what is the total force--so exactly one N So the total force at this point that's acting on it in order to just start to budge it I'll call this the budging force Remember here isn't a traditional class F sub B for the budging force The budging force is in the parallel direction If I'm applying 1 N in this direction and it already has 49 N due to the component of gravity in this direction, then my budging force is 50 N 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 to just start to overcome friction In this case, it's the budging force. That's a term that I made up And the 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 this, by the wedge on the block is the normal force, is 49 sqrt of 3 Maybe I should say 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 root of 3 N Over the magnitude of the normal force, and that is 49 sqrt of 3 We call this the coefficient of static friction We're gonna use this a little bit more deeply in other problems but it tends to hold true for different materials so that in the future if you have a different mass, you have a different incline but you have the same materials given the normal force, you can figure out the budging force You can 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 N over 49 square root of 3 N Let's get the calculator out So I have 50 divided by 40 times the square root of 3 Gives me .72--I'll just round to two significant digits--0.72 This is 0.72 And then you can use this information. This is the coefficient of 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 I guess the force just to overcome the force of friction, just kind of get right over that the most friction that can be applied by kind of the abrasiveness of the 2 things when the object is stationary I'll do a whole video on how this is different when an object is stationary to when it's moving A lot of times, they're very very very close, but for certain materials you have a very--at least, a noticeably different coefficient of friction when the object is stationery as opposed to when it is moving So I'll leave you there. In the next few videos, we'll use coefficient of friction or calculate coefficient of friction to do some more problems