Anderson Video - Acceleration on an Incline

Professor Anderson
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>> Hello class, Professor Anderson here. Let's ask a question about this g sine theta. What we said was if you have a box on a frictionless incline, then the acceleration of that box down the incline is G sine theta. Okay? How do we know that? Well, you can reason it out from looking at the limits of theta, but how can we actually prove this? The way we prove it is using vectors. If I think about my box, that box is going to undergo an acceleration G due to gravity. G is always straight down but any vector, I can always break into components. So I can write it as a vector in that direction plus a vector in that direction, where this is a right angle. And now if I want this side of the triangle, all I need to know is that angle right there. We don't know exactly what that angle is yet, but let's just call it phi. What is the rest of my triangle look like? Here's what the rest of the triangle looks like. That is angle Theta. This is of course, a right angle and this let's call, a some other angle Beta. And now with just a little bit of geometry here, we can probably figure out how these things match up. Alright? So we know that the sum of angles in a triangle equals 180 degrees. So let's take a look at this triangle right here. I have theta plus beta plus 90 degrees equals 180 degrees. Theta plus beta plus 90 degrees equals 180 degrees. If I move that 90 over to the other side, I get theta plus beta equals 90 degrees. Alright, that's good, but now let's look at these two angles right here. The way we drew our vectors beta plus phi is also equal to 90 degrees. Right? That is a right angle there. And so now look what I have. I have phi plus beta equals 90 degrees. I have theta plus beta equals 90 degrees, and so the only conclusion that you can make is phi is of course, equal to theta. So when you draw the box and you say there's some component of gravity in that direction, some component of gravity in that direction, you know that this one has to be G sine of theta and this one has to be G cosine of theta, because if I go back to this triangle, this is sine and this is cosine. Okay? So that's a little proof of where G Sine Theta comes from. Hopefully, that's clear. If not, come see me in office hours. Cheers.
>> Hello class, Professor Anderson here. Let's ask a question about this g sine theta. What we said was if you have a box on a frictionless incline, then the acceleration of that box down the incline is G sine theta. Okay? How do we know that? Well, you can reason it out from looking at the limits of theta, but how can we actually prove this? The way we prove it is using vectors. If I think about my box, that box is going to undergo an acceleration G due to gravity. G is always straight down but any vector, I can always break into components. So I can write it as a vector in that direction plus a vector in that direction, where this is a right angle. And now if I want this side of the triangle, all I need to know is that angle right there. We don't know exactly what that angle is yet, but let's just call it phi. What is the rest of my triangle look like? Here's what the rest of the triangle looks like. That is angle Theta. This is of course, a right angle and this let's call, a some other angle Beta. And now with just a little bit of geometry here, we can probably figure out how these things match up. Alright? So we know that the sum of angles in a triangle equals 180 degrees. So let's take a look at this triangle right here. I have theta plus beta plus 90 degrees equals 180 degrees. Theta plus beta plus 90 degrees equals 180 degrees. If I move that 90 over to the other side, I get theta plus beta equals 90 degrees. Alright, that's good, but now let's look at these two angles right here. The way we drew our vectors beta plus phi is also equal to 90 degrees. Right? That is a right angle there. And so now look what I have. I have phi plus beta equals 90 degrees. I have theta plus beta equals 90 degrees, and so the only conclusion that you can make is phi is of course, equal to theta. So when you draw the box and you say there's some component of gravity in that direction, some component of gravity in that direction, you know that this one has to be G sine of theta and this one has to be G cosine of theta, because if I go back to this triangle, this is sine and this is cosine. Okay? So that's a little proof of where G Sine Theta comes from. Hopefully, that's clear. If not, come see me in office hours. Cheers.