Inclined Planes - Online Tutor, Practice Problems & Exam Prep

Intro to Inclined Planes

Guys, so in this video, we're going to start talking about inclined plane or ramp problems. These are super important problems you're going to see a lot of in physics. So it's very important that you learn this correctly. I'm going to show you an easy way how to solve all these problems. Let's go ahead and get to it. We're actually going to come back to this in just a second here. I want to start off the problem. We have this 5-kilogram block that's on a frictionless incline and it's angled at 37 degrees. So we know that this theta relative to the horizontal is actually 37. What 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. We know that, in the free body diagram, we look for the weight first. And the weight is always going to act straight down. So this is going to be our weight force which is mg. 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. This is our normal. That's the free body diagram. Alright? That's done. Now, what we want to do is we want to calculate the acceleration down the incline. So here's what happens. What does the acceleration down the incline mean? Well, if you just imagine that you take this block and you put this on a 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 what'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 inclined slope. So what 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 going to tilt your x y plane. And you do this to basically line up the new x-axis to be parallel to the incline's slope. So you want to do this parallel to the incline slope. Alright? So that means that we're no longer going to use these coordinate systems anymore. We're actually going to use the tilted one. So this is going to be my new plus y and my new plus x. I'm going to call this plus x new plus y new. Alright. So now what happens is we have our normal force. It's going to point in our new y 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. Alright? So basically, we're just going to decompose this into its x and y components. So this is going to be my mg now in the x direction and this is going to be my mg in the y direction. You can think about this as basically like there's a component of the weight that is trying to push it down the ramp, that's the mgx, and the component of weight that's basically pushing it into the ramp which is our mgy. Alright? So it's, which we'd have to decompose this and we're going to decompose this just like we would any other forces using our normals cosine and sine equations. But what's really important about this is that your components of mg are actually going to be opposites from the usual component equations for forces. Meaning, if we have a force that's at an angle theta x, then we calculate fx by just using fcosine theta. And we use fy, so basically y goes with sine theta. What's really important about mg in inclined plane problems is that it's actually going to be flipped backward. Components, the mgx, are actually going to go not with cosine but with sine. And the mgy is actually going to go with cosine. So basically, our mgx is mgsinθ and mgy is mgcosθ. This is always going to be true for your inclined 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 it works. Alright? So now that we've done that, now we want to write the acceleration. We want to calculate it. So we're going to have to use f equals ma. We know 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 going to have to use our f equals ma in the x and y-axis. I'm going to start off with the x-axis first. The sum of our forces is in the x, equals max. And I really only have one x force now, my mgx. That is all the forces that act in the x-axis. So I've got mgx equals max. Remember mgx, you're just going to replace it with the mg sine theta. So this is going to be my mg times sine theta x equals max. 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 sin theta x, so your acceleration is 9.8 times the sin of 37 and you're going to get 5.9 meters per second squared. So that is the acceleration that is down the incline. Alright? 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 ma for our y forces. Right? So we have fy equals may. So let's think about this for a second. Here we had an acceleration that was down the incline that was ax. But do we have one in the y-axis? Well, any acceleration in the y-axis would mean they would fly upwards off the ramp or it'd 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 0 in these problems. So that brings us to a really important point. The acceleration on inclined planes only happens along the x-axis always. And that's because your acceleration in the y-axis is 0 which basically means that all the y forces will cancel. And so what we've seen here is that if there's no other forces that are acting on an object, the acceleration in the x-axis only depends on theta. It's just g times the sin of theta x. And again, this happens if there's no other forces, alright? So if we go ahead and solve for our Fy, we have the normal force that's going to point along our direction of positive, and then this is going to be mgy, which is going to point negative. This is going to be 0. So that means that we have m is equal to, and remember mgy has an expression, we're just going to replace it with mg times the cosine theta. So we have m is equal to mg cosine theta x. Alright. Here is the expression now for the normal force. That's it for this one, guys. Let me know if you have any questions.

A 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

Runaway Ramp

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 meters 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, I've got this incline like this. That's kind of like what a runaway ramp is. So I've got this truck that's at the bottom here. The initial velocity is 18 meters per second, and it's moving to the right. So what happens is the truck is going to go up the ramp, and it's eventually going to come to a stop. So this final velocity is going to be 0. 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, final velocity, acceleration, and time. If I can find 3 out of 5, then I can find an expression for delta x. And the reason we can do this is that the acceleration on an inclined plane is constant. It's always going to be the same value down the ramp. So we know that this initial velocity is going to be 18, and we know the final velocity is going to be 0. We don't know the acceleration or the time, and that's kind of bad because that just means that we have only 2 out of 5 variables, and we need 3. So if I want to find delta x, which is my target variable, then I'm going to 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 are 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 ⁢ sin ⁢ ( θ )

If I can figure out this acceleration using this expression 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 sine of this 20%? What does that mean? So what happens here is when you're given sometimes these theta angles as percentages, what you're going to have to do is you're going to 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 going to do here is our theta x is going to equal the tangent inverse of 20 percent divided by 100. So really, theta x in terms of degrees is just going to be the inverse tangent of 0.2. And if you work this out, what you're going to get is 11.3 degrees. So this is the number I'm going to use in this equation here to find the acceleration. So I'm going to do 9.8 times the sine of 11.3. What you're going to get here is you're going to get 1.92 meters per second squared as your acceleration. Now remember that acceleration usually gives you the correct sign and the direction. It indicates the direction here. So the big question is, while this truck is going up the ramp, is the acceleration positive or negative? Is it up the ramp or down? And if you think about this, if the truck is going 18 and then it goes to 0, that actually means that the acceleration is down the ramp. This is equal to 1.92. So what 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 going to be negative 1.92. So now we have 3 out of 5 variables, and we can pick the equation that ignores time. S_time is going to be our ignored variable here. That's going to be equation number 2. So, really, this is the final velocity equals initial velocity plus 2a times delta x. There is our target variable. Now we just plug in everything else. Right? So this is going to be 0 squared equals the initial velocity 18 plus 2 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 going to get is you're going to get 3 0.84 times delta x is equal to, and this is going to be 324 when you take 18 squared. And so finally, what we're going to get is delta x is equal to 84.3 meters, and if you look at your answer choices, wow, this is going to be 84 meters. Alright. 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.

Connected Block on a Ramp

Guys, let's get started here. We've got these 2 connected objects that are on a frictionless incline. So this actually is going to be a connected system of objects on inclined planes. We're really just going to combine a lot of steps that we know about connected systems and also inclined planes. So let's check this out. Right? So we've got these 2 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 mag. 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 going to be my normal. There's also a cable that's attached to this box here, so we know we're going to have some tension force that points this way. But that's it. There's no friction. There's no applied forces or anything like that. Let's look at object B. This is a little bit simpler. So we have these just two forces. We really just have the weight force, which is mbg that acts straight down, and then we have the tension force. And these tensions are actually going to 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 what we have to do first is we actually have to tilt the coordinate system. So we want the new y-axis 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 going to accelerate downwards. And so because of that, this object is going to go up the ramp. So I'm going to choose this direction to be my positive x. Alright? So that's what I've got here, which means that I'm going to have to decompose this mg, into its components. So I've got my mgy that's going to be down here. This is magy. And then this component here is magx. And that completes our free body diagram. Now we're just going to choose the direction of positive, but we already just talked about that. Basically, anything that goes up and down like this or like over and down is going to be the direction of positive. Alright? So that's that. Now we want to write the F=ma starting with the simplest object. The simplest object is the one with the fewest forces and that's object B. So we have here is F=mba except we're not looking at the x forces, we're looking at the y forces. Right? Now remember, for object B, anything that points downwards is going to be positive. So when it comes to our Fy, when we expand this, our mbg is positive and our tension is negative. And this equals mba. So we can figure this out real quickly. We know that the mass is equal to 5 for this B block. So this mg is actually just going to be 49 once you multiply 5 by 9.8. So you get this, 49 minus t equals 5a. 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 going to be mass A times acceleration. Remember the object A is going to accelerate this way, right? Okay, so all your forces in the x-axis are going to be tension, and tension is along the direction of positive. So this is our T, then we have minus ma times gx which is equal to ma times a. Now remember this mgx 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 2, 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 sin53°, and this is equal to mass which is 2 times acceleration. So I'm just going to try to scoop this over. Alright. So we get here once you plug this all in, is you're going to get 15.7 So you get 15.7 here is equal to 2 times a. So we have the acceleration and again, we have this tension force that's unknown here. So I'm going to have these 2 equations of 2 unknowns. I'm going to label them 1 and 2. And here, what I'm going to do is now I'm going to have to solve this by using equation substitution. So we did F=ma. Now we're going to solve this by using equation addition. So this is my first equation here. Remember, this is going to be tension minus 15.7 equals 2a. Now for the second equation, I want to line it up so that the nontarget variable goes away. So this is going to be 49 minus T, and then I'll line up the equation signs like this equals 5a. So this is my equation number 2. You're going to add them straight down like this so that tensions will cancel. You're going to get 49 minus 15.7 is equal to 7a. So when you go and actually solve for this, what you're going to get in is an acceleration that is 4.8 meters per second squared. Alright. 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.