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.
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7. Friction, Inclines, Systems
Inclined Planes
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