7. Friction, Inclines, Systems

Inclined Planes

1

concept

## Intro to Inclined Planes

6m

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Hey, guys. So in this video, we're gonna start talking about incline plane or ramp problems. These are super important problems. You're gonna see a lot of in physics. So it's very important that you learn this correctly. I'm gonna show you easy way how to solve all these problems. Let's go ahead and get to it. So we're actually gonna come back to this in just a second here. I want to start off the problem. We have this 5 kg block that's on a frictionless incline and it's angled at 37 degrees. So we know this data relative to the horizontal is actually 37. We want to do first in this first problem is we want to draw the free body diagram and that's actually the first step to basically solving any forces problem. So let's get to it. So we know that the free body diagram we look for the weight first and the weight is always going to act straight down. So this is gonna be our weight force, which is M g. Then we look for any applied or tension forces which we have none of. And then we're also told that this is frictionless, but we do have the surfaces in contact, which means there is going to be a normal force. But it's not going to point straight up like this, because remember, normals always point perpendicular to the surface perpendicular to this incline here. So your normal actually points in this direction, and this is our normal. That's the free body diagram. All right, so that's done. Now we want to do is we want to calculate the acceleration down the incline. So here's what happens. What does that mean? What does the acceleration down the incline mean? Well, if you just imagine that you take this block and you put this on the ramp and then you let it go, it's not going to accelerate to the right like this. It's actually going to accelerate, sort of down parallel to the incline. So it's really important here is that rather than using our old coordinate system where we had Y and X like this, we usually did this because we had forces and accelerations in the horizontal and the vertical. But now we're going to have an acceleration that points down along the inclines slope. So we're going to do is we're going to basically tilt our coordinate system. That's the second step. So in these kinds of inclined plane problems, you're gonna tilt your XY plane and you do this to basically line up the new X axis to be parallel to the inclines slopes. You want to do this parallel to the incline slope? All right, so that means there were no longer to use these these cordon systems anymore. We're actually going to use the tilted one. So this is gonna be my new plus y and my new plus X. I'm gonna call this plus X New plus y new. All right, So now what happens is we have our normal force. It's gonna point in our new wide direction, which is good. But now this mg kind of lies sort of halfway in between the negative y axis and the positive y axis. And so whenever this happens, you have to decompose this mg. So after you tilt your coordinate system, mg will always have to be decomposed. All right. So basically, we're just gonna decompose this into its X and Y components. So this is gonna be my mg now in the X direction, and this is gonna be my mg in the Y direction you can think about. This is basically like there's a component of the weight that is trying to push it down the ramp. That's the MG X and the component of weight that's basically pushing it into the ramp, which is RMG y all right. So it's what we have to decompose this, and we're going to decompose this just like we would any other forces by using our normal co sign and sign equations. But what's really important about this is that your components of mg are actually going to be opposite from the usual component equations for forces, meaning we have a force that's at an angle theta X. Then we calculate F X by just using f cosign theta, and we use F y. So basically, why goes with sine theta? What's really important about MG in inclined plane problems is that it's actually going to be flipped backwards. The X components, the MG X is actually going to go not with co sign, but with Sign and the M G Y is actually going to go with co sign. So basically, RMG X is going to be mg times the sine of theta X and M G Y is going to be mg times the co sign of data X. This is always going to be true for your incline plane problems. And it's basically just because this angle is actually the same as this smaller angle over here, which is relative to the Y axis, and usually that's bad. But that's basically why that works. All right, so now that we've done that, now we want to write the acceleration. We want to calculate it. So we're gonna have to use f equals M. A. We know in this in this part B we're trying to solve for the acceleration only in the X axis, right? Basically the acceleration down the ramp like this. So we're gonna have to use our f equals m a in the x and y axis. I'm gonna start off with the X axis first. Some of the forces is in the X equals m a X, and I really only have one X force. Now, my mg X, that is all the forces that act on the X axis. So I've got mg X equals M a X Remember mg X. You're just gonna replace it with the mg sine theta. So this is going to be my mg times sine theta X equals m a X. Now, remember, we're trying to solve the acceleration so we can actually cancel out the masses that appear on both sides. And what you get is that your acceleration is g times sine data X. So you're acceleration is 9.8 times the sine of 37 and you're gonna get 5.9 m per second squared. So that is the acceleration that is down the incline. All right. And so now let's move on to the last part here. We're just going to write an expression for the normal force. So remember that the normal force pops up in the Y axis in your new Y axis now. So now we're just going to write f equals m A for our wife forces. Right. So we have f y equals m a y. So let's think about this for a second here we had an acceleration that was down the incline. That was a X, But do we have one in the Y axis? Well, any acceleration of the Y axis would mean it would fly upwards off the ramp or would somehow be going into the ramp, which doesn't really make a whole lot of sense. So what happens here is that the acceleration is actually equal to zero in these problems, So that brings us to really important points. The acceleration on inclined planes only happens along the X axis always. And that's because your acceleration, the Y axis is zero, which basically means that all the UAE forces will cancel. And so what we've seen here is that there's no other forces that are acting on an object. The acceleration in the X axis only depends on data. It's just g times the sine of data X. And again, this happens that there's no other forces. All right, so we go ahead and solve for R f Y. We have the normal force. This is gonna point along our direction of positive, and then this is going to be M G y, which is gonna point negative. This is gonna be zero. So it means that we have n is equal to remember M. G. Y has an expression. We're just going to replace it with mg times the coastline data So we have em is equal to mg Cosign theta x. All right, so here is the expression now for the normal force. That's it for this one. Guys, let me know if you have any questions.

2

example

## Runaway Ramp

4m

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check out this problem here. We've got large trucks on runaway ramps and what happens is we're told that this truck is moving at 18 m per second. We're told that this ramp has a 20% grade. We want to figure out how long the ramp should be to bring the truck to a stop. So I'm going to draw this out real quick. Basically got this incline like this. It's kind of like a runaway ramp is, um So I got this truck that's at the bottom here. The initial velocity is 18 m per second and it's moving to the rights. What happens is the truck is going to go up the ramp and it's eventually going to come to a stop. So this V final is going to be zero. What we're trying to do is we're trying to figure out how long basically a Delta X that this ramp needs to be in order to come to a stop, right? So if we're looking for Delta X notice, the other variables that we have in this problem we have an initial velocity V, not we also have a final velocity, so we're having a lot of these motion or charismatics type variables so we can try to solve this using our cinematics equations from trying to find Delta X. I'm going to list out all of my five variables v, not the final acceleration in time. If I can find three out of five that I can find an expression for the Delta X and the reason we can do this is remember that the acceleration on an inclined plane is constant. It's always going to be the same value down the ramp. All right, so we know that this initial velocity is going to be 18 and we know the final velocity is going to be zero. We don't know the acceleration or the time, and that's kind of bad, because that just means that we have only two out of five variables and we need three. So if I want to find Delta X, which is my target variable, then I mean you have I'm gonna have to find either acceleration or time. Now, remember, on inclined planes, we actually have an expression that we can solve for the acceleration. If there's no other forces that are acting on our object other than gravity. Remember that the acceleration on an inclined plane is just equal to G times the sine of theta X. So if I can figure out this acceleration using this function or this this equation here, I'm just going to plug it back in, and then I can finally find Delta X. The problem, though, is that I know G right, This is just 9.8. But how do I take the sign of this? 20%. What does that mean? So what happens here is when you are given sometimes these theta angles as percentages, What you're gonna have to do is you're gonna first have to convert it to a decimal, and then you can convert it to degrees by using this very simple equation here. If you want this theta angle in terms of degrees, then you just take the inverse tangent of the percentage that was given to you divided by 100. So, for example, what we're gonna do here is R Theta X is gonna equal the tangent inverse of 20% divided by 100 so really fatal X in terms of degrees, it's just gonna be the inverse tangent of 0.2. And if you work this out, where you're gonna get is 11.3 degrees. So this is the number I'm going to use in this equation here to find the acceleration. Somebody's you 9.8 times the sine of 11.3. Where you're going to get here is you're gonna get 1.92 m per second squared as your acceleration. Now, remember, that acceleration usually gives you the correct sign and the and the direction and indicates the direction here. So the big question is, while this truck is going up, the ramp is the acceleration, positive or negative is up the ramp or down. And if you think about this, if the truck is going 18 and then it goes to zero, that actually means that the acceleration is down the ramp. This is a equals 1.92. So we do here is this actually has to pick up a minus sign. So really, our acceleration is going to stop our initial velocity, so it has to be negative, Right? So this is gonna be negative 1.92. So now we have 3 to 5 variables, and we can pick the equation that ignores time. So time is going to be our ignored variable. Here, That's going to be equation number two. So, really, this is the final velocity equals initial velocity plus two, eight times Delta X. There is our target variable. Now we just plug in everything else, right? So this is gonna be zero squared equals the initial velocity 18 plus two times a, which is 1.92 and then this is going to be Delta X. So when you move this to the other side, what you're gonna get, you're gonna get 3.84 times Delta X is equal to, and this is going to be 324 when you take 18 squared. And so finally, we're going to get is Delta X is equal to 84 groups. This is gonna be 84.3 m. And if you look at your answer choices Wow, this is gonna be 84 m, all right? And that's the answer. So that's how long this runaway ramp needs to be. In order for the truck to come to a stop

3

Problem

ProblemA 4.0-kg box sits on a frictionless inclined plane that makes a 21° angle with the horizontal. It is held in place by a cord parallel to the plane. Calculate the tension in the cord.

A

14 N

B

37 N

C

33 N

D

21 N

4

example

## Connected Block on a Ramp

5m

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Hey, guys, let's get started here. We've got these two connected objects that are on a frictionless incline. So this actually is going to be a connected systems of objects on an inclined planes. Really? Just gonna combine a lot of steps that we know about connected systems and also inclined plane. So let's check this out. Right. Who? We've got these two objects. Let's go ahead and start with the free body diagram for object. A. So this is the one that's on the inclined plane. So remember, we're going to have the weight force that acts straight down, So this is going to be m a G. Now we have a normal force. That normal force doesn't point upwards. It acts perpendicular to the surface. So the normal force for a actually is going to look like this. This is gonna be my normal. There's also a cable that's attached this box here. So we know we're gonna have some tension for us at points this way. But that's it. There's no friction. There's no applied forces, anything like that. Let's look at object B. This is a little bit simpler. So we have these just, uh, two forces. We really just have the weight force, which is MBG that act straight down. And then we have the tension force and these tensions are actually gonna be the same because they're part of the same cable that goes over the pulley. So that's the first step. However, one of these objects is on an inclined plane. So we have to do first is we actually have to tilt the coordinate system. So we want the new Why access to be basically in the direction of that normal force, and then we have a choice. We can either choose down the incline to be positive or up the incline to be positive. So if you think about this, what's going to happen is that when you release this system, object B is going to go down like this, right? It's gonna accelerate downwards. And so because of that, this object is going to go up the ramp. So I'm gonna choose this direction to be my positive X. All right, so that's what I've got here, which means that I'm gonna have to decompose this mg into its components. So I've got my m g y. That's going to be down here. This is M A G Y. And this component here is M A G X, and that completes our free body diagram. Now, we're just gonna choose the direction of positive. But we already just talked about that. Basically, anything that goes up and down like this, like over and down, is going to be the direction of positive. All right, so that's that. Now we want to write the F equals M A. Starting with the simplest object. The simplest object is the one with the fewest forces in that object B. So we have here is F equals M B times A. Except we're not looking at the X Forces were looking at the white forces right now, remember, for object, be anything that points downwards is going to be positive. So when it comes to our f y, we expand, this are MBG is positive and our tension is negative and this equals m B times A. So we can figure this out real quickly. We know that the mass is equal to five for this B block, so this mg is actually just gonna be 49 once you multiply five by 9.8. So you get this 49 minus T equals five A. I want to figure out the acceleration, but I can't because I have this tension force. It's unknown, so I'm just going to go to the other object now. Object? A So this is the sum of all forces in the X axis. I'm now looking up the incline like this, and this is gonna be mass eight times acceleration. Remember the object a is going to accelerate this way. Right? Okay, So all your forces in the X axis are gonna be tension and tension is along the direction of positive. So this is our T then we have minus M. A G X is equal to L a times a. Now remember, this mg X component of gravity is not mg. It's actually mg times the sine of theta. So this is equal to mass times acceleration, so we can go ahead and figure this out. Mass is two g is 9.8 and now we're going to use the sine of the angle that was given to us. This is 53 degrees right here. So we have the sign of 53 this is equal to mass, which is two times acceleration. So I'm just gonna try to scoop this over. All right, So we get here once you plug this all in is you're gonna get 15.7, so you get 15.7 here is equal to two times a, so we have the acceleration. And again, we have this tension force that's unknown here. So I'm gonna have these two equations with two unknowns. I'm going to label them one and two. And here what I'm gonna do is now, I'm gonna have to solve this by using equation Substitution. We do. That equals May. Now, we're gonna solve this by using equation addition. So this is my first equation here. Remember, this is gonna be tension minus 15.7 equals to a now, for the second equation would have lined up so that the non target variable goes away. So this is going to be 49 minus T and then I'll line up the equation. Signs like this equals five a day. So this is my equation Number two. You're gonna add them straight down like this so the tensions will cancel. You're going to get 49 minus 15.7 is equal to seven A. So when you go and actually saw for this, you're gonna get it as an acceleration that is four points, 8 m per second squared. All right, so that's it for this one, guys. And that basically means that our equation or our answer is answer choice A. That's the magnitude of the acceleration.

Additional resources for Inclined Planes

PRACTICE PROBLEMS AND ACTIVITIES (3)

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- A 4000 kg truck is parked on a 15 degrees slope. How big is the friction force on the truck? The coefficient o...
- A rifle with a barrel length of 60 cm fires a 10 g bullet with a horizontal speed of 400 m/s. The bullet strik...