Inclined Planes with Friction - Online Tutor, Practice Problems & Exam Prep

In problems involving objects on rough inclined planes, understanding the forces at play is crucial. The friction force can be static or kinetic, determined by comparing the gravitational component down the ramp, $\mathrm{mg}\sin \left(\theta \right)$, with the maximum static friction, $\mu$_s⁢mg⁢cos(θ). Critical angles for static and kinetic friction can be calculated using $\theta =\tan \mu \left(\theta \right)$, highlighting the relationship between friction coefficients and motion. Understanding these concepts is essential for analyzing motion on inclined planes.

Intro to Incline Plane with Friction

Guys, in previous videos, we've seen how to solve inclined plane problems and we've seen friction problems. So we're really just going to combine these two concepts together in this video because sometimes you're going to run into 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 kilogram block that's on a ramp at 37 degrees. We have our coefficients of static and kinetic friction. And in part a, what we want to do is calculate the friction force that's acting on the block when it's released.

So in part a, 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 going to need a free body diagram. So 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 applied forces or tensions, but we are going to have a normal force that points perpendicular to the ramp's 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 free body diagrams for inclined planes, we're going to tilt our coordinate system so that the x axis points down the ramp like this. Basically, this is our plus x and plus y. When we do that, remember our mg basically just gets split up into its components. So we have mgy and then we have mgx. Alright. So those are our mg components here. So now what we have to do is we have to determine what kind of friction we're using, or we have in this problem. Whether it's static or kinetic friction. So that brings us to step number 2. We're going to determine whether this is static or kinetic friction, and there are 2 ways to do this. So in some problems, you'll actually be able to tell from the text. It'll 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 going to have to compare all the non-friction forces. Basically, in this problem here, that's just going to be our mgx. So we're going to 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 mgx are strong enough to get this object moving. Right? So we're basically trying to figure out whether we're dealing with fs or fk. And to do that, we're just basically going to calculate mgx and fs max and see which one is bigger.

Alright, so when we're doing mgx, remember it's just mg times the sine of theta, and we can calculate that. We have all those numbers. This is a mass of 10, 9.8, and then we're using the sine of 37. If you go ahead and work this out, you're going to get 59 newtons for mgx. So we know this is equal to 59. Now what we do is we just compare this to the maximum static friction. Remember that's that threshold. Remember that the equation for that is μs times the normal. We know what μs is. It's just 0.6 that was given to us in the problem here. What about this normal force? Well remember on inclined planes, if you have friction and mgx in the x-axis and then only n and mgy in the y-axis, those forces have to cancel. So n is equal to mgy which is equal to mg times the cosine of theta. So when we plug in the normal force we are really just going to do 10 times 9.8 times not sine of 37 but cosine of 37. So if you go ahead and work this number out, what you're going to get is 77 newtons. So what happens here is that our fs max is 47. That's the maximum static friction, but your mgx is stronger than that. So what that means is that if your mgx overcomes the maximum static friction then therefore the actual friction basically just becomes kinetic friction. So this f actually is fk and to calculate this we're just going to use μk times the normal and we can figure that out as well. This Fk here is just going to be μk which is 0.4 and then times the normal force which again is 10 times 9.8 times the cosine of 37. So if you go ahead and work this out, you're going to get 30.8 newtons.

Alright, so basically, we know this is going to be 30.8. MGX is strong enough to get the object moving, that friction becomes kinetic, and we have that kinetic friction force. Alright? So that's the answer to part a. Now what we're doing in part b is trying to figure the block's acceleration. So now in part b, we want to figure out ax. So now if we're trying to figure out ax, now we're going to get into our F equals ma. So we write all the forces in the x-axis equals max, and once we have our direction of positive, we really just have our mgx that's positive, minus our friction kinetic is equal to max. And this is actually really straightforward because we've already calculated all of these values. So what we can do here is we can say that mgx is equal to 59. That's what we calculated over here. Then you're going to subtract the friction force which we just calculated as 30.8. This is going to be 10 times ax. So if you go ahead and solve this, you're going to get 2.82 meters per second squared. You get a positive number for our acceleration, and that makes sense. It just means that it accelerates down the ramp along our direction of positive. Alright. So that's it for this one, guys. Let me know if you have any questions and I'd like to get some practice.

Pushing A Box Up A Ramp

Guys, let's get started here. So we have a 20 kilogram block or box that we're trying to push up a moving truck. We're applying a 110 Newton force up the incline. So, I'm going to draw a quick little sketch of what's going on here. I've got my block like this. Basically, I'm just trying to get it up this ramp by pushing it with 110 newtons. What we want to do is when 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. I've got my free body diagram that's like this. I've got my mg which is going to 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 going to have some friction. But first, before I draw the friction, what I want to do is I want to break up this mg. m g y . And we know this is going to be m g x . So we can just get rid of this mg here.

So, we have this friction force. But what happens is we have these two forces. One of this is pushing up the ramp. One of them is pushing down the ramp. We want to figure out what the friction force is going to be. So we're actually going to stick to the second step here. We have to determine what kind of friction we're going up against and in what direction that is going to push. If you ever have the direction of that friction is not known, what you have to do first is you're going to have to find the net of all the non-friction forces along your axis of motion. Basically, what we're going to do is we're going to 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 block was going to go up the ramp, friction would go down. If the block was going to go down the ramp, then friction would actually point up. So what we have to do is we have to figure out which one of these forces is stronger and then our f is going to be opposite to that.

So basically, what we do is we look at our f force which we know is 110 and we're going to compare this to m g x which we know is sin ⁡ ( θ ) . And so this is basically 20 times 9.8 times the sine of 15 degrees. So if you go ahead and work this out, you get an m g 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 110. m g 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 direction. That means our friction force is actually going to point down the ramp.

Alright, so now that we know the direction of friction, we still have to figure out what type of friction that is, fs or fk. 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. So basically, what we have 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, your sum of all forces that are non-friction is really just going to be f minus m g x . And we actually know what those are. Right? So this is 110 minus 50.7, and you get 59.3. Basically, what this is, is once you cancel out the 110 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 f s max ? Well, f s max is gonna be μ s static times the normal which our μ s static is 0.3. And then remember our normal force in inclined planes is going to be m g y which is mg times the cosine of theta. So really you just do is 20 times 9.8 times the cosine of 15 degrees. So you end up getting 56.8, I believe. Yes. So you get 56.8. What happens is the forces are net of all the non-friction forces is 59.3. So what happens here our Σ f is stronger than the f s max , so we've overcome the static maximum threshold. And so, therefore, our friction is kinetic friction which is μ k times the normal. So what we have here is our friction, our kinetic force, I'm just going to use 0.2 times 9.8. Oops. Sorry. Times the mass, which is 20 times 9.8 times the cosine times the cosine of 15 degrees, and what you get is you get a kinetic friction force of 37.9. So that's our kinetic friction.

So now what we have to do in order to figure out the acceleration that's what this whole problem is about is now we have to write our f equals ma. So we're going to write out f equals ma 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 force is going to go up the ramp and that's, like, the one that's stronger. Right? So I'm just going to choose that to be positive. So we're going to expand out our forces. We have F then we have minus m g x minus fk is equal to ma. So this is what we're looking for here. So we actually basically know what all of these forces are. We just calculated what this fk is, what this kinetic friction is. It's 37.9. We already know what m g x is because we calculated it earlier, and we also know what our applied force is it's the 110. So we're just going to plug everything in. So this is going to be 110 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 in, you're going to get 21.4 is equal to 20a, and so a is equal to 1.1 meters per second squared.

Now, remember that when we solve for acceleration in these problems, this sign will give us the direction. It's because we got a positive sign here, that just means it points along the direction of motion. So we look at our answer choices, and it has to be answer choice C. So that's it for this one guys. Let me know if you have any questions.

Block Launched Up Ramp

Guys, let's check this one out. We've got a 2 kilogram block that is launched up a 40 degree ramp. So I want to just draw this out for 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, v₀, of 10 meters 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 3 meters vertically above the bottom of the ramp. So that means that basically vertically above the bottom, that means this height of this triangle here is Δy. I'm going to call it Δy and it equals 3. So what we want to do is 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 going to be my free-body diagram here. So basically, I'm going to start out with the weight force that's acting straight down. That's mg. Then I also have the normal force, which acts perpendicular to the surface like that. I don't have any applied forces, no tensions. I've pushed it, and it's already starting with an initial 10 meters per second, so there's no applied force or anything like that. But we do know 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 going to be f_k. Alright. And the last thing we have to do is just break down this mg into its components. So I've got my mg_y here, and then I also have a component mg_x. So mg_x acts down the ramp. Alright? So there are these two forces like this. 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 2. 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 out our F=ma here. So I've got my F= <math><mrow><mi>m</mi><mi>a</mi></mrow></math>. And now I just want to choose a direction of positive. So if the block is going up the ramp with a velocity like this, then that's the direction I'll choose as positive, right? It's the direction it's moving. So what this means here is that when I expand my forces, I have these two downward forces, mg_x and f_k, and so they both pick up negative signs. So what I've got here is I've got mg_x which is negative and then minus f_k is equal to mass times acceleration. Alright, so I can expand both of these terms out. Remember, mg_x is really just mg sin(θ), and then f_k has an equation as well, it's <math><mrow><mi>mu</mi><mi>k</mi><mspace linebreak="newline" /><mrow><mi>m</mi><mi>g</mi><mi>cos</mi><mo>(</mo><mi>theta</mi><mo>)</mo></mrow></mrow></math>. So that's equal to ma. So what I can do here is say this is mg sin(θ) and then remember that this <math><mrow><mi>mu</mi><mi>k</mi><mo>times</mo><mi>the</mi><mi>normal</mi></mrow></math>, in inclined planes, the normal is just equal to mg cos(θ). So I've got minus mu_k mg cos(θ) is equal to ma. 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 sin(θ) is; those are all just numbers. I know what mg cos(θ) 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 0 for just a second. And I know that it's going to do this over some displacement, which I'm going to call Δx. So how do we solve this acceleration? Well, if we get stuck using forces, we're going to 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 kinetic friction. Then we're going to list out our 5 variables, and we're going to figure out 3 out of 5 and then pick an equation to solve for a. So we know the initial velocity is 10, we know the final velocity is 0, and then we don't know anything about a, t, or Δx. So unfortunately, this is 2 out of 5 variables, and I've gotten stuck here. I'm going to have to solve for one of them. So usually when we get stuck, we would use F=ma to figure out the acceleration, but that's actually where we just came from. We just came from F=ma. So we're really stuck between solving for the time or the displacement. We're 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 Δx is. So bascially, if we're trying to figure out this Δx, which is the hypotenuse of this triangle, and we know the Δy and the angle, we can always solve for that other sign just using our sine and cosine equations. Basically, what we can do is say that Δx times the sine of 40 degrees is equal to Δy. I want to use the sine because the sine of this angle is going to give me the opposite side, which is the one I know. Alright, so we use this expression here, and what we can do is we can solve #x# or Δx because this is just equal to Δy, which is 3. Right? It's 3 meters above the bottom of the ramp. So this is 3 divided by the sine of 40 and what you get is 4.67. So there it is, there's my 3rd out of 5 variables, 4.67. So we can finally set up an equation to solve for this acceleration here. So if we do this, we're going to use equation #2, which is the one that ignores my time. So we get that the velocity squared is equal to initial velocity squared plus 2 a times Δx. We want to solve now for this acceleration. So the final velocity is 0, the initial velocity is 10, and then we have this 2 times the acceleration times 4.67. So when you move this to the other side to solve for a, you're going to get negative 9.34 a equals 100. And actually, I'm missing an a here. So this is negative 9.4 or a. So when you solve for a, you're going to get negative 10.7. So this is our answer of, sorry. This is our acceleration. And remember, we came here because we wanted this last variable so we can plug it back into our F=ma. And now we can go ahead and solve for mu_k. So what I'm going to do is I'm just going to plug in all the values that I know. This is just 2 times 9.8 times the sine of 40, and this is going to be minus mu_k times 2 times 9.8 times the cosine of 40, and this is going to equal the mass, which is 2 times negative 10.7. So all of these really just become numbers here, right? So let's go ahead 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, mu_k. And then this 2 times negative 10.7 is going to be negative 21.4. So if we notice here all of these numbers are actually negatives. So what we can do is basically just turn them all into positives. It's kind of as if you like multiply the equation by negative one. It doesn't change anything. So what we've got here is we've got 15 mu_k plus 12.6 equals 21.4. And then if you subtract this 12.6 over, what you end up getting is you end up getting 15 mu_k equals 8.8. And so when you solve for this, you're just going to get 0.59. So you go to your choices, and that is going to be answer choice c. That's your coefficient. So, hopefully, this made sense, guys. Let me know if you have any questions, and let's move on.

Force Not Parallel To Ramp

Guys, welcome back. Let's check this one out. So we're pushing a mini fridge up this incline. I'm going to go ahead and sketch this out like this, and we have a force. Alright. 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 incline. So we know this force here is equal to 120. What we want to do is 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 our mg into mgy and x. So that's our mg force like this. Alright. And then we have an applied force as well. We know this applied force is going to point not up the ramp like this. It's actually going to point at some angle. This is f equals 120, 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 components. We have to do the same thing with the f. So we separate this into its x, x, and y components like this. Alright. The last thing we have to do is look for friction. We're told that we have some coefficients of static and kinetic friction. Unfortunately, based on the problem, we don't know whether we're pushing this thing up or, you know, which direction that friction force is. So our friction force here is unknown. And just like we did in a previous problem, whenever we have the direction of friction as unknown, we first have to look at all the non-friction forces like fx and mgx and figure out which is stronger. Basically, we have to figure out where this block would move if there weren't any friction. So let's go ahead and do that. Right? So you've got our fx which is really just f times the cosine of 30 degrees. So this is 120 times the cosine of 30 and you're going to get 104. So how does this fx compare to mgx? So our mgx is mg times the sine of the angle. Except we're not going to use 30; we're going to use 20 because that's the angle of the ramp. So there are 2 angles here. Don't get confused there. And so, basically, this is just going to be 30 times 9.8 times the sine of 20 degrees. If you go ahead and work this out you're going to get 100.5. So what happens is our fx, the x component, is 104 that's the component that's going up the ramp, but our mgx 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 fr. So that's the first step. We have that friction now. We have at least we have the direction. Now we move on to the second step of any problem: 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 sum of all forces, all the non-friction forces, is really just going to be fx minus mgx, and we already calculated those right. We just have a 100. That's where a 104 minus a 100.5, and you're going to get 3.5 here. So that is the sum of all forces that are non-friction. So how does this compare to fs max? Well, remember, we have to use the equation times the normal. We have mu static now; we just have to figure out what this normal force is. Normally, pun intended, your normal force is going to be mgy, which is mg cosine theta here. But remember that only works if all of your forces are along the incline. But that doesn't happen here because this force that we're applying isn't perfectly along the incline. It's inclined at some angle. So we have this fy to consider. So 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 is 0 because it doesn't go flying off the ramp or anything like that. So we still have normal plus fy minus mgy is equal to 0. So, basically, what happens is we just have another force to consider. Your normal is equal to mgcosine theta minus your fy. Alright. So if we go ahead and expand this, what this really means is that we have, this is 30 times 9.8 times the cosine of 20 degrees minus and then remember this fy here is really just going to be 120 times the sine of the angles, this is going to be a sine of 30 degrees. So, basically, what happens is this becomes 276.3. 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 fs max is equal to mu static, which is 0.3, I believe. Yeah. So it's 0.3 times 216.3. We get an fs max that is 64.9 Newtons. Alright. 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. Our non-friction forces only really add up to 3.5. Maximum static friction is almost 65 Newtons. So what happens is because your non-friction forces are less than fs max, then that means that 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 3.5 Newtons down the ramp. And because this is static friction, that means that the acceleration of the fridge is equal to 0. Basically, we haven't overcome the maximum static friction. So therefore, that's your answer. The fridge actually doesn't move; the acceleration is 0.

Incline with Friction & Critical Angle

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 or beginning to slide or the block slides at constant speed. And really these two phrases that are pretty common are related to 2 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 gonna 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 gonna calculate that angle. In the second problem, it's going to be very similar, except now, we're gonna calculate the angle where it already is moving, but now it slides down at constant velocity like that's and 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 gonna just draw off 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 mgy components and an mgx 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, the static friction prevents the object from sliding. Now we know that as we tilt it higher and higher and higher, your MGX 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. Friction. So really, this mgx 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 theta critical s. And so how do we do that? We're just gonna use our f equals m a. 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 we think the object is gonna slide, so down the ramp like this. That's our direction of positive, which means that when we expand out our forces, we have mgx 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 0. Remember, we're trying to calculate this 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 mgx is equal to f s max.

So later on, we're also gonna see that when we're sliding in constant speed, the acceleration also is going to be 0. 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 mgx is mgsine of theta, and then my f s max has an equation. Remember, this is just mu static times the normal. So this is gonna be mu static times the normal force. And remember that in these problems, your normal force is really just equal to your mgy, which is equal to your mg times the cosine of theta. So what we can do here is we have m g sin θ = mu _s m g cos θ , which means that the mgs on both sides of the equation actually will just cancel out.

Right? If m g cancels out m g. So I'm basically left with 2 variables, theta and also my mu static. So I want I want to basically figure out an expression for this theta here. So I'm gonna move this to the other side because I have my mu static here on the right. So I'm gonna divide out the cosine, and I'm gonna move it to the other side so we can basically just collapse it to a single trig function. When you take sine over cosine, remember that becomes a tangent. So Tan θ = mu _s . And the last step we do is just get rid of this tangent by taking the inverse tangent. So θ = tan - 1 mu s , and that's the equation. Basically, these are your 2, critical angle equations for theta critical s. So you have theta critical s equals tangent inverse of mu static. If you have mu static, you can figure out theta critical, but you also have the other way around. mu s = tan θ . Basically, it's just two sides of the same equation. If you have one, you can always figure out the other. Alright. So basically, if we want to figure out this theta critical for our problem, this is we're just gonna take the inverse tangents of the mu static that we were given, which is 0.75. So if you do this, what you're gonna get is 37 degrees. So this is the angle that you have to tilt the ramp at so that the block starts sliding. Alright. 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 it slides down a constant velocity. If we kept this thing at 37 degrees, it would to accelerate down the ramp. So we're gonna have to adjust a little bit b

Critical Angle

Guys, let's check out this problem here. We've got a car that's parked along the street, positioned on an angled road at 35 degrees, and a snowstorm hits the area. The icy road now becomes slippery enough for the car to begin sliding downhill. We want to figure out the coefficient of static friction between the car and the icy road. This is a critical angle problem because there are some angles or some coefficients that change for the car so that it just begins to slide downhill. We already have the equation for μ_s, which is just going to be the tangent of our θ_critical. So, with the given angle at 35 degrees, it's really just as straightforward as plugging this in: tan⁡35 degrees. Make sure your calculator is in degrees mode, and what you should get is approximately 0.7. That's the answer, and let me know if you guys have any other questions.

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