University of North Carolina at Chapel Hill
Normal Forces - Example 2


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Video Transcript

welcome to our second example video. Looking at normal forces in this video, we're going to consider the problem of you're standing in a room and for some reason you're very tall and you are holding a book up against the ceiling. And we want to ask ourselves, What are the forces acting on this book? Okay, so when we draw the free body diagram of the book, we know we have a force up due to you. We'll call that Force One, and then it's gonna experience a force down due to gravity, and it's also going to experience a force down due to the normal force of the ceiling. So let's go ahead and write that out. Then we have F one is positive and up minus F G is down, so it's negative minus normal force is down, and that's gonna be equal to the mass of the book times its acceleration, which should be zero if it's just sitting against the wall, meaning that Force One is equal to F G plus f n, or perhaps more insightful, Lee than normal force is equal to the force that you are applying minus the force due to gravity, so nothing here should be earth shattering or super revelatory. But what it is showing you is that we have a normal force pointing down in the same direction as F. G. So normal forces don't always just keep objects from falling through the floor. Occasionally, they can stop objects from going through the ceiling. It's all about where they're being, what they're being pushed against and how firm the object is there being pushed against. In this case, we find that the normal force will be equal to how hard you're pushing, subtracted by how hard the weight of the object is being is being pulled down the amount of force that gravity is applying to it. On the other hand, what if we decided we wanted to pull the book down and then push it against the wall? When we draw this free body diagram, we still have F g down because it's always there and now we have a Force F one this way, and we have a normal force due to the wall this way. But in order to keep the box from sliding down, what we really need is a force this way, pointed up to counter Act F G because when we write these out in our X direction and y directions separately in our X direction will have fn minus F one equals, we hope zero Newton's because we don't want it to accelerate in the X direction in the Y direction, we're going to have a question mark minus F g hopefully again equal to zero Newton's, which means whatever this mystery forces it has thio completely counter act. The weight of the book has to be equal to the weight of the book. Now it turns out as well find in a later unit that f question mark is going to be the force due to friction between the book and the wall. And this is gonna be a very interesting problem that we consider here where the question would be. How hard do you have to push the book against the wall in order to keep it from sliding down? And we'll be able to investigate that as we learn more about friction in a future video

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