1

concept

## Intro to Incline Plane with Friction

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Hey, guys. So in previous videos, we've seen how to solve inclined plane problems, and we've seen friction problems. So we're really just gonna combine those two concepts together in this video because sometimes you're gonna run to problems where you have objects on rough inclined planes meaning inclined planes with friction. So let's go ahead and just jump right into the problem here. There's really nothing new. So we've got this 10 kg block that's on a ramp at 37 degrees. We have our coefficient of static and kinetic friction, and in part of what we want to do is calculate the friction force that's acting on the block when it's released. So in part of what we want to do is figure out. What is this friction Force F. It could be static or kinetic, so to do that, to start things off, we're gonna need a free body diagrams. Let's get started here. We've got this block, there's going to be an MG that's downwards, right? And then we have no applying forces or tensions, but we are going to have a normal forest that that points perpendicular to the ramps surface, and then finally we have some friction now whether the block actually starts to move down the ramp or is trying to move down the ramp. That friction force wants to oppose that motion, and so it's going to act up the ramp regardless of whether it's static or kinetic. So that's what we're trying to find here. Now remember that when we draw a free body, diagrams for inclined planes were gonna tilt our coordinate system so that the incline so that the X axis points down the ramp like this. So basically, this is a plus X plus y. So when we do that, remember, RMG basically just get split up into its components. So we have M g y. And then we have MG X. All right, so those are MG components here. So now what we have to do is we have to determine what kind of friction we're using, or we do what we have in this problem, whether it's static or kinetic friction. So that brings us Step Number two. We're going to determine whether this is static or kinetic friction, and there's two ways to do this. Since some problems you'll actually be able to tell from the text. It will say that the box isn't yet sliding or it is sliding. You'll know whether it's static or kinetic, but in this problem, it's kind of vague. We're supposed to calculate this friction force, but we don't know whether it's still stationary or starts moving. And so what happens is we're gonna have to compare all the non friction forces. Basically, in this problem here, that's just gonna be rmg X. So we're gonna have to compare those non friction forces along the axis of motion to the maximum static friction. We're basically trying to figure out whether our non friction forces like MG X are strong enough to get this object moving right. So we're basically trying to figure out whether we're dealing with F S or F K. And to do that, we're gonna we're just basically going to calculate MG X and F s Max and see which one is bigger. All right, so when we're doing MG X, remember, that's just mg times the sine of data, and we can calculate that we have all those numbers. This is massive 10 9.8, and then we're using the sign of 37 If you go ahead and work this out, you're going to get 59 Newtons for mg X So we know that this is equal to 59. Now what we do is we just compare this to the maximum static friction. Remember, that's that threshold. So remember that the equation for that is Mu s times the normal. We know what the US is. It's just 0.6 that was given to us in the problem here. So what about this? Normal force will remember on inclined planes if you're if you have friction and mg ex and the X axis and then only end in N g y and the Y axis, those forces have to cancel so and is equal to mg y, which is equal to mg times the coastline of data. So we plug in the normal force. We're really just gonna do 10 times 9.8 times not sign of but co sign of 37. So if you go and work this number out where you're gonna get is Newtons. So what happens here is that our Fs Max is 47. That's the maximum static friction but your MG X is stronger than that. So what that means is that if your MG X overcomes the maximum static friction, then therefore the actual friction basically just becomes kinetic friction. So this F actually is f k. And to calculate this, we're just going to use mu k times the normal and we can figure that out as well. This f k here is just going to be, um UK, which is 0.4 and then times the normal force, which again is 10 times 98 times the co sign of 37. So if you go ahead and work this out, you're gonna get 31.3 Newtons. Alright, so basically we know this is going to be 31.3 mg X is strong enough to get the object moving. That friction becomes kinetic and we have that kinetic friction force. All right, so that's the answer to part A. Now we're doing in Part, B is trying to figure out the blocks acceleration. So now in part B, we want to figure out a X. So now if we're trying to figure out a X now, we're going to get into our F equals m A. So we write all the forces in the X axis equals m a X. And once we have our direction of positive, we really just have our mg X. That's positive, minus our friction. Kinetic is equal to m a X, And this is actually really straightforward because we've already calculated all of these values. So we can do here is we can say the MG X is equal to 59. That's what we calculated over here and then going to subtract the friction force, which we just calculated is 31.3 and this is going to be 10 times a X. So we go ahead and solve this. You're gonna get to 100.77 m per second squared. We get a positive number for our acceleration. And that makes sense. It just means that accelerates down the ramp along the direction of positive. All right, so that's it for this one. Guys, let me know if you have any questions. I'd like to get some practice

2

example

## Pushing A Box Up A Ramp

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Hey, guys, let's get started here. So we have a 20 kg block or box that we're trying to push up a moving truck. We're applying it to 110 Newton Force up the incline. So I'm gonna draw a quick little schedule. What's going on here? So I got my block like this. I know this is 20. Basically, I'm just trying to get it up this ramp by pushing it with 110 Newtons. What we want to do is what we're given the coefficients of friction. I want to figure out the magnitude and direction of the acceleration of the box. So let's get started. We know we're going to have to draw a free body diagram and then tilt our coordinate system. So let's go ahead and do that. So basically, I've got my free body diagram that's like this. I've got my mg, which is gonna act downwards, and then I've got my normal force, which acts up like this or sorry, perpendicular to the surface. I've got my applied force that pushes this way, and so we know we're gonna have some friction. But first, before I draw the friction What I want to do is I want to break up this mg. So this is going to be M g y. And we know this is going to be MDX, so we can just get rid of this mg here. So we have this this friction force. But what happens is we have these two forces one of this pushing up the ramp. One of them is pushing down the ramp and we want to figure out what the friction force is going to be. So we're actually going to the second step here. We're gonna go to the second step. We have to determine what kind of friction we're going up against and in what direction is going to push. So here's the deal. If you have ever have the direction of that, friction is not known, right? If this is ever unknown, what you have to do first is you're gonna have to find the net of all the non friction forces along your axis of motion. Basically, what we're gonna do is we're gonna look at the non friction forces F and M G X and figure out which one of them is stronger. Remember that friction always wants to oppose the direction that the object would move if there were no friction, meaning if the if the block was going to go up the ramp, friction would go down. If the block was going to go down the ramp, then friction actually point up. So what we have to do is we have to figure out which one of these stronger which one of these forces are stronger, and then our f is going to be opposite to that. So let's get so basically we do is we look at our F force, which we know is 110 and we're gonna compare this to MG X, which we know is MG times the sine of theta. And so this is basically 15 times 9.8. I'm sorry about 15 is 20. So it's 20 times 98 times the sine of 50. Sorry, that's 15 degrees. So if you go ahead and work this out, you get an MG X that is equal to 50.7. So if you take a look here we have our force that points up the ramp. That's 1 10 mg X is equal to 50.7. So basically we're pushing up the ramp stronger than gravity is pulling it down. What that means is that f is going to be opposite to our that direction. So that means our friction forces actually going to point down the ramp. All right, so now that we know the direction of friction, we still have to figure out what type of friction that is F s or F k. And if we don't know from the problem text, we're going to have to look at the sum of all forces along the axis of motion and figure out whether it's enough to overcome the static maximum threshold. And if it does, then our friction turns to kinetic. So basically, we're going to do here is we have to look at the sum of all forces on the axis of motion. That's basically our F and M g X. So you're some of all forces that are non friction is really just gonna be f minus mg X, and we actually know what those are, right? So this is 1 10 minus 50.7 and you get 59.3. Basically, what witnesses is once you cancel out the 1 10 this way and the 50.7, It's as if you had one force that was going up. That was 59.3. How does this compare to Fs Max? Well, F s Max is going to be mu static times the normal. Which arm you static is 0.3 and then remember, our normal force on inclined planes is going to be M G y, which is mg times the cosine of data. So really, we're just going to do is 20 times 9. times the coastline of 15 degrees. So you end up getting 56 points. Uh, let's see, this is 56.8, I believe. Yeah, so you get 56.8. So what happens is the our forces are net of all the non friction forces is 59.3. So what happens is here are Sigma F is stronger than the F S. Max. So we've overcome the static Max, the static Max threshold. And so therefore, our friction is kinetic friction, which is mu k times the normal. So we have here is our friction are kinetic friction. Here is the force that is pushing down the ramp, Um, the actually getting the box to start moving. So to calculate this kinetic friction force, I'm just gonna use 0.2 times 9.8. Sorry. Uh, times the mass, which is 20 times 98 times the co sign times the coastline of 15 degrees and what you get is you get a kinetic friction force of 37 points. Uh, nine. So that's our kinetic friction. So now what we have to do in order to figure out the acceleration, that's all this whole problem is about is now we have to write our r f equals M A. So we're gonna write out f equals m A along the axis of motion. Now we just pick a direction of positive. So I'm just going to choose the direction up the ramp to be positive, because we know that basically our forces going to go up the ramp, and that's like the one that's stronger, right? So I'm just gonna choose that to be positive. So we're gonna expand our forces. We have f Then we have minus mg X minus. F K is equal to m a So this is what we're looking for here. Okay, so we actually basically know what all of these forces are. We just calculated what this f k is. This kinetic friction is 37.9. We already know what Mg X is because we calculated earlier and we also know what are applied forces. It's the 1 10, So we're just gonna plug everything in. So this is gonna be 1 10 minus 50.7 minus 37.9 is equal to the mass, which is 20 times a. So if you go ahead and plug all this and you're gonna get 21.4 is equal to 28 so a is equal to 1.1 meters per second square. Remember that when we saw for acceleration in these problems, this sign will give us the direction because we got a positive sign here. That just means it points a longer direction of motion. So we look at our answer choices and it has to be answered. Choice C. So that's it for this one. Guys, let me know if you have any questions.

3

example

## Block Launched Up Ramp

7m

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Hey, guys, let's check this one out. So we've got this 2 kg block that is launched up a 40 degree ramp. So I want to just draw this out for just a second Here. We've got this incline like this. I've got a box that's at the bottom. It's given an initial launch speed this V not of 10 m per second. So I know that the angle of this is 40 degrees, and basically what happens is that it's going to come to a stop at some point that is m vertically above the bottom of the ramp. So that means that basically vertically above the bottom, Uh, that means this height of this triangle here is Delta y. I'm gonna call it Delta Y and equals three. So we want to do is we want to calculate the coefficient of kinetic friction. So what I want to do first is I want to draw a free body diagram, so this is gonna be my free body diagram here, So basically, I'm gonna start up with the weight force that's acting straight down. That's mg that I also have the normal force which acts perpendicular to the surface like that. I don't have any applied forces, no tensions. Basically, I've pushed it and it's already starting with an initial 10 m per second. So there's no applied for us or anything like that. But we do know that there is a coefficient of kinetic friction. So what happens is if the block is sliding up the ramp, that means that the friction has to oppose this and it's gonna be f k. All right. And the last thing we have to do is just break up this mg into its components. So I got my M g y here and then I also have a components mg X, so m g X acts down the ramp. All right, so these two forces like this. So we already actually figured out what kind of friction that we're dealing with here, so we actually don't need to even do Step two. We know it's kinetic friction, so we want to figure out this coefficient of kinetic friction here that's going to come from F. K. So let's go ahead and write on our ethical dilemma here. So I've got my F equals m a. And now I just want to choose the direction of positive. So if the block is going up the ramp with a velocity like this, then that's the directional. Choose as positive rights, the direction it's moving. So what this means here is that when I expand my forces, I have these two downwards forces mg X and F K. And so they both pick up negative science. So what I've got here is it got mg exit is negative and then minus f k is equal to mass times acceleration. All right, so I can expand both of these terms out. Remember, MG X is really just mg times the sine of theta. And then f k has an equation as well. It's mu k times the normal. So that's equal to M A. All right, so what I can do here is say this is mg sine theta. And then remember that this mu k times the normal in inclined planes, the normal is just equal to mg times the coastline of data. So I've got minus mu k mg co sign of data is equal to N A. So if I want to figure out what this coefficient of kinetic friction is, I need to know everything else about this equation. I know what mg sine theta is. Those are all just numbers. I know what mg cosign theta is again. Those are all just numbers here and I know what the mass is. The one variable that I don't know is I don't know what the acceleration is. I know that it starts with some velocity. And when it gets to the top of this ramp here, the final velocity is going to equal zero for just a second. And I know that there is going to do this over some displacement, which I'm gonna call Delta X. So how do we solve this acceleration? Well, if we get stuck using forces, we're gonna try to solve it using motion equations. And that's exactly what we do here. If we want to figure out the acceleration so we can plug it back into this formula here and solve for the coefficient of static friction, then we're going to list out our five variables, and we're gonna figure out three out of five and then pick an equation to solve for a So we know this initial velocities. 10 Final velocity is zero and then we don't know anything about A T or Delta X. So, unfortunately, this is two out of five variables, and I've gotten stuck here. I'm gonna have to solve for one of them. So usually when we get stuck, we use f equals M A to figure out the acceleration. But that's actually where we just came from. We just came from F equals M A. So we're really stuck between solving for the time or the displacement were not given any information about the time. But we do know something about the distance vertically above the triangle or the ramp or something like that. So maybe we can figure out what this Delta X is. So basically, if we're trying to figure out this Delta X, which is the hypothesis of this triangle, and we know the Delta y and the angle we can always solve for that other side just using our sign and co sign equations. Basically, what we can do is say that Delta X Times the sine of 40 degrees, is equal to Delta Y. I want to use the sign because the sign of this angle is going to give me the opposite side, which is the one I know. All right, so use the this expression here, and what we do is we can solve for X or Delta X, because this is just equal to Delta Y, which is three rights 3 m above the bottom of the ramp. This is three divided by the sine of 40. And what you get is 4.67 so that there it is. There's my third out of five variables, 4.67. So we can finally set up an equation to solve for this acceleration here. So if we do this, we're gonna use equation number two, which is the one that ignores my time. So we get that the velocity squared is equal to initial velocity squared plus two, eight times Delta X and we want to solve now for this acceleration. So the final velocity is zero. The initial velocity is 10, and then we have this two times the acceleration times 67. So when you move this to the other side to sulfur A, you're gonna get negative. 9. 34 move there. So you're gonna get negative. 9.34 equals 100. And actually, I'm missing an a here, so this is negative. 9.34 A. So when you saw for a you're gonna get negative 10.7. So this is our answer. Uh, sorry. This is our acceleration. Remember, we came here because we wanted this last variable so we can plug it back into our f equals m a. And now we can go ahead and solve from the U. K. So I'm gonna do is I'm just gonna plug in all the values that I know. This is just two times 9.8 times the sine of and this is gonna be minus mu K times two times, 9.8 times the co sign of 40. And this is going to equal the mass, which is two times negative, 10.7. So all of these really just become numbers here, right? So let's go ahead and, um, and solve for each of them. So if you look at, let's see, the first term just becomes negative. 12.6, and then this term here is going to become negative. 15. Exactly, um UK and then this two times negative 10.7 is gonna be negative. 21.4. So if we notice here, all of these numbers are actually negatives. So we can do is basically just turn them all into positives. It's kind of if you like, multiply the equation by negative one. It doesn't really change anything. So we've got here is we've got 15 m. U K plus 12.6 equals 21.4. And then if you subtract this 12.6 over what you end up getting is you get a new getting 15 u k equals 8.8. And so when you solve for this, you're just gonna get 0.59. So if you go to your answer choices and that is going to be answered choice, See, that's your coefficient. So hopefully this makes sense. Guys, let me know if you have any questions and let's move on

4

example

## Force Not Parallel To Ramp

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Hey, guys, Welcome back. Let's check this one out. So we're pushing a mini fridge up this incline and go ahead and sketch this out like this and we have a force. All right, so we've got our mini fridge right here. We have a force that is 120 degrees, but it's not actually perfectly up the incline. We have a force that's basically angled at some angle, which we know is 30 degrees above the axis of the inclines. We know this F here is equal to 1 20. What we want to do is we want to figure out the acceleration of the box. So let's go ahead and get started. We know we're going to have to draw a free body diagram, so let's do that over here. So I've got my box like this, and I've got my mg like that. Now I've got the normal force that points perpendicular to the surface, which means that we split up RMG into M g y N X. So that's our M G force like this. All right. And then we have an applied force as well. We know this applied force is actually gonna point not up the ramp like this. It's actually gonna point at some angle. This is f equals 1 20. And we know that this angle here is actually equal to 30 degrees. So what that means is that just like we did with MG, we had to separate it into its into its components. We have to do the same thing with the F. So we separate this into its f x, X and Y components like this. All right, The last thing we have to do is look for friction. We're told that we have some coefficient of static and kinetic friction. Unfortunately, based on the problem, we actually don't know whether we're pushing this thing up or whether basically, like, we don't know the direction of that friction force. So our friction force here is unknown. And just like we did in a previous problem, whenever we have the direction of friction is unknown. We first have to do is we have to look at all the non friction forces like FX and MGM X and figure out which is stronger. Basically, we have to do is figure out where would this block move If there weren't any friction. So let's go ahead and do that right. So you got our FX, which is really just f times the co sign of 30 degrees. So this is 1 20 times the co sign of 30 and you're going to get 104 So how does this FX compared to mg X So rmg X's mg times the sine of the angle Except we're not gonna use 30. We're going to use 20 because that's the angle of the ramp. So there's two angles here. Don't get confused there. And so basically, this is just gonna be 30 times 9.8 times the sine of 20 degrees. If you go ahead and work this out, you're gonna get 100.5. So what happens is our f X. The X component is one of four. That's the component that's going up the ramp. But our MG X is only 100 if we just kind of round it. So what happens is our FX is actually slightly winning without friction. This fridge would actually go up the ramp. So what does that mean? That means our friction force actually points down the ramp like this So this is our f. So that's the first step. We have that friction now. At least we have the direction. Now we move onto the second step of any problem. Basically, we're going to determine the type of friction, whether we're going up against static or kinetic. And to do that, we have to look at all the non friction forces and see how they compare to the maximum static friction. So what happens is our some of all forces, all the non friction forces is really just gonna be FX minus mg X. And we already calculated those right? We just have 100 100 and four minus 100.5 and you're gonna get 3.5 here. So that is the sum of all forces that are non friction. So how does this compare to F. S? Max will remember, Um, we have to use the equation mu static times, the normal we have metastatic. Now we just have to figure out what this normal forces normally pun intended. Your normal force is going to be m g y, which is mg cosign theta here. But remember that only works if all of your forces are along the incline. But that actually doesn't happen here because this force that we're applying actually isn't perfectly along the incline. It's inclined at some angle. So we have this f y to consider. So we have to do first is we have to go here and look at the sum of all forces in the Y axis, which equals mass times acceleration Now, in the Y axis along the incline that acceleration zero because it doesn't go flying off the ramp or anything like that. So we still have normal plus f y minus mg Y is equal to zero. So basically, what happens is we just have another force to consider. Your normal is equal to mg cosign theta minus your F y. All right. So if we go ahead and expand this what this really means is that we have, uh this is 30 times 9.8 times the co sign of 20 degrees minus. And then remember this f y here is really just gonna be times the sine of the angles. This is going to be a sign of 30 degrees. So basically what happens if this becomes to 76.3. This becomes 60. And so your normal is really just equal to 216.3. So that's the number that we plug into here. So now our F. S Max is equal to use static, which is 0.3, I believe. Yeah. So it's 0.3 times to 16.3. You get an F S. Max, that is 60. Let's see, 64 0.9 Newtons. All right, so this is our fs max. Remember? The whole reason we did this is because we're trying to figure out whether the non friction forces are enough to overcome the maximum static are non friction. Forces only really add up to 3.5 Maximum static friction is almost 65 Nunes. So what happens is because your non friction forces are less than F. S. Max. Then that means that your F your friction force is equal to static friction. And so therefore, the static friction that's acting on the fridge is really just equal to 3. Newtons down the ramp. And because this is static friction, that means that the acceleration of the fridge is equal to zero. Basically, we haven't overcome the maximum static friction. And so therefore, that's your answer. The fridge actually doesn't move the acceleration zero.

5

concept

## Incline with Friction & Critical Angle

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Hey, guys. So now that we've taken a look at objects on rough inclined planes, occasionally in these problems you'll have to calculate something when the object is doing something very specific, like it's starting to slide, are beginning to slide or the block slides at constant speed. And really, these two phrases that are pretty common related to two special angles in these rough inclined plane problems. These angles are called critical angles, and I'm gonna show you how we calculate them in this video. So we're gonna actually come back to these points in just a second here. I want to just jump into the problem because it's very visual to understand. So here we're gonna place a block on an adjustable ramp. So I literally have a block on an adjustable ramp like this. And the whole idea is that we're going to tilt the ramp slowly until the block suddenly starts sliding. So basically what happens is I'm gonna tilt this block right into the point where the block starts moving right there, and we're going to calculate that angle in the second problem. It's going to be very similar, except now we're going to calculate the angle where it already is moving. But now it slides down at constant velocity like that. We want to figure out that angle as well. And really, those are the critical angles. So let's go ahead and check this out here. We're going to just draw some free body diagrams and then use our F equals M A, just like any other forces problem. So if we take a look at the forces that are acting on this block while it's on the ramp, right, so when it's tilted like this, we're gonna have some MG that wants to pull it down the ramp, which means that we have an M G Y components and an MG X components. Why doesn't it start sliding yet even as we start tilting the ramp? That's because there's some friction that's preventing it. So there's some friction that acts up the ramp, and because it's not yet moving, this is going to be static friction, right? As we're tilting the ramp, basically, static friction prevents the object from sliding. Now we know that as we tilt it higher and higher and higher, your MG X gets stronger and stronger right, The force that's pulling it down, the ramp gets stronger, and basically your static friction in order to prevent it from sliding has to also increase. Well, basically, the point right where the block starts to slide. That's where you've overcome the maximum static friction. So really, this mg X at the angle where it starts sliding is actually balancing out the F S. Max. So that's what we're going up against here. So we want to calculate this fate a critical s. And so how do we do that? We're just going to use our F equals, Emma. So we've got f equals m A here and now we just pick a direction of positive, so usually that's just gonna be the direction. But we think the object is going to slide. So down the ramp like this, that's our direction of positive, which means that when we expanded our forces, we have MG X, and then we have F S Max. So what's our acceleration? Well, in these problems, even though we're calculating the moment where it starts to slide, the acceleration is actually equal to zero. Remember, we're trying to calculate the special angle right at the moment before it starts sliding, right? So even though the block starts to slide, we want the angle right before that actually happens, right where MG X is equal to F S. Max. So later on, we're also going to see that when we're sliding in constant speed, acceleration also is going to be zero. So really, both of these are just equilibrium problems. Alright, So basically, what I'm gonna do is I'm going to move my f S Max over to the other side and I'm gonna expand out both of these terms. I know this MG X is mg sine of data and then my f s Max has an equation. Remember, this is just mu static times the normal. So this is going to be a new static times the normal force. And remember that in these problems, your normal force is really just equal to your G y, which is equal to your mg times. The cosine of theta so we can do here is we have mg times. The sine of theta is equal to, um use static times mg co sign of data, which means that the mgs on both sides of the equation actually we'll just cancel out right? You have MG cancels out mg. So I'm basically left with two variables feta and also my mu static. So I want I want to basically figure out an expression for this data here. So I'm gonna move this to the other side because I have my new static here on the rights. So I'm gonna divide out the co sign. I'm gonna move it to the other side so we can basically just collapsed into a single trig function. When you take sine over cosine, remember that becomes a tangent. So tangent data critical s is really just equal to mu static. And the last step we do is just get rid of this tangent by taking the inverse tangent. So tha tha critical s is equal to the inverse tangent of metastatic. And that's the equation. Basically, these are your two critical angle equations for theta critical s. So you have to to critical s equals tangent, inverse of metastatic. If you have a static, you can figure out that a critical. But you also have the other way around. Uh, metastatic is equal to the tangent of theta critical. Basically, it's just two sides of the same equation. If you have one, you can always figure out the other. All right, So basically, if we want to figure out this theater critical for our problem, this is we're just going to take the inverse tangents of the Meuse static that we were given, which is 0.75. So if you do this, you're gonna get is 37 degrees. So this is the angle that you have to tilt the ramp at so that the blocks starts sliding. All right, so now in the second problem, we want to calculate this angle here again, where the block is sliding down at constant speed. So basically, once the object starts moving, we're gonna have to adjust the ramp so that it slides out in constant velocity. If we kept this thing at 37 degrees, it would continue to accelerate down the ramp. So we're gonna have to adjust a little bit, basically like tilted, tilted down a little bit so that the block starts sliding a constant velocity. Alright, so what happens is now we're calculating tha tha critical K and the entire solution is going to be almost the exact same thing that we did on the left side here. I'm gonna copy most of this stuff over because most of it is going to be the same. So we copy this diagram over. The only thing that's different about this problem is that because you are sliding with some velocity already, you're not going up against maximum static friction anymore. You're actually going up against kinetic friction. That's really the only difference. So we're just basically going to replace every single s in the solution with just K's. So here what happens is we have When the block starts sliding, your MG X is equal to your maximum static friction. And we know that again because of equilibrium here, where the block slides a constant speed, we have MG X is equal to kinetic friction instead. All right, so what I'm gonna do is I'm gonna copy this whole entire thing here, and we're basically just going to replace every single s with a K. So I'm gonna go ahead and do that. All right, So this is Kay just using UK, and then finally are two equations. So basically what happens is when you rewrite all of these equations here. Your thought a critical K is going to be the inverse tangents, not of us. But if the U. K and your UK is going to be the tangents, not of fate a critical s but a fatal critical. Okay, so that's the angle where it slides down at constant speed. So now, if we just calculate this Arthur to critical K is going to be the inverse tangents of, and now we're just gonna use arm UK, which is given to us 0.31. And if you go ahead and work this out, you're gonna get 17 degrees. So let's talk about these two angles. We've got 37 17, and it's kind of related to the idea that it's harder to get something moving that it is to keep it moving. So the idea here is that you have to tilt this thing at 37 degrees to overcome the maximum static friction. Once you do that, then you only have to lower it to 17 degrees so that the block continues sliding here at constant velocity. That's why we got these two different angles. All right, so the last point I just want to mention is that notice how both of these equations here, data critical s NK depend only on the coefficient of static and kinetic friction. They don't depend on any other variables. Like, for instance, the mass, which might be counterintuitive. You might think that if the box was heavier, the angle would be different. But actually, it doesn't matter. It doesn't affect the angle at all, because again, this MG cancels out when you do the F equals m a right. So hopefully that makes sense. Guys, let me know if you have any questions.

6

example

## Critical Angle

1m

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Hey, guys, let's check out this problem here is we've got a car that's parked along the street. It's on some angled road. It's 35 degrees and a snowstorm hits the area. And basically, what happens is that the icy road now become slippery enough for the car to begin sliding downhill. So we want to do is we want to figure the coefficient of static friction now between the car and the icy roads. So we know this is a critical angle problem, because what's going to happen is that there are some angles or some, uh, coefficients that are changing for that the car so that the car just begins to slide downhill. That's the critical angle. So we already have the equation form us. Remember, um, us is just going to be the tangent of Arthur to critical s so tangent of Arthur to critical s all right. And so what we have here is we have that angle is 35 degrees, so it's really just as straightforward as plugging this in. So we have tangent of 35 degrees. Make sure your calculator is in degrees mode, and what you should get is you'll get 0.7. So that's the answer. And let me know if you guys have any questions

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Problem

A 3-kg block is at rest on an adjustable ramp. When the ramp is tilted to a 20° angle, the block slides with constant velocity. What is the coefficient of kinetic friction between the ramp and the block?

A

μk = 0.24

B

μk = 0.36

C

μk = 0.52

D

Not enough information given

Additional resources for Inclined Planes with Friction

PRACTICE PROBLEMS AND ACTIVITIES (3)

- A bicycle coasting at 8.0 m/s comes to a 5.0-m-long, 1.0-m-high ramp. What is the bicycle’s speed as it leaves...
- A small block has constant acceleration as it slides down a frictionless incline. The block is released from r...
- A 25.0-kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient ...