University of North Carolina at Chapel Hill
Normal Forces - Example 3


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Video Transcript

welcome to our third example Video for normal forces in this video will consider the problem again of the inclined plane. But with one extra wrinkle, we will have two boxes stacked on top of one another. So in this problem, in order to really make it complex later on will also add in friction and consider the effects of friction. But in this problem, will could stick with just the two boxes and no friction. So we have box one box too, and an angle Fada. Now, when we go to draw our free body diagrams, they should look something like this. We have are inclined plane shape and we have the force of gravity on box one. Then we have normal force again. So this previously we had called this the force of the inclined plane. Now we're calling it the normal force. So this is a normal force on box one. And then when we draw for box, too, we have a box to box one box two, we have force of gravity down, and then we have a normal force pointed up, and this is normal force on box to notice here that the normal force on box too is applied by box one. So that means box one is pushing on. To which means Box two has to push back on box one. Which means we need to add another vector here, which will I'm just gonna label it as fn too, because I know that's going to be the same Magnitude is FN two because it is the reactionary force for box one pushing on box to we get box to pushing back on X one box one. So this will make it a little more complicated. So let's ah, let's take a shot at this. Yeah, we have a situation then where we want to write our equations of motion over here. We're going to get kind of similar that we've had before. If we tilt our axes that we have the X axis parallel to the inclined plane and the Y axis perpendicular to the inclined plane, then we'll find that the normal forces entirely in the Y direction and we will have a small component. That'll be mg cosine, theta of gravity. That is also in the UAE. Remember, we're drawing are triangle here and that will be the angle Fada so we have M G or M two g or, in other words, F g Tua's. I've labeled it down here m two g times cosine, theta. Hopefully that's equal to zero Newton's and then in the X direction we have f to G for, in other words, m two g sine theta which will be equal to Mass two times acceleration notice. Here that acceleration comes out as g sine theta as we would hope to see, and on the left hand side here for bucks one in the Y direction we have FN one. Then we have FN two and then we also have m one g cosine theta and we hope that comes out to be zero. Newton's notice over here that we have fn two is going to be equal to m two g cosine theta so we could plug that in and then we would be able to say that FN one has to be equal to m one plus m two times g cosine theta in the X direction. Then we still only have m one g cosine theta is equal to mass one times acceleration. So again we find that acceleration is equal to G o my my mistake. They're not mung cosign, but M one g sine theta is equal to mass one times acceleration, since that's what it will be again if you're confused about what I'm doing there, remember, we're having a right triangle that looks like this where we have F G and Seda. And then we have our Y direction and X direction oriented like this. Which means that our X component will be f g science data and r y component will be f g cosine theta. So we've said all these up when we include friction in this friction will also show up acting in the X direction, which will make the X direction a little bit more complicated and make things a little more interesting when we move into that unit.

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