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.