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6. Intro to Forces (Dynamics)

# Vertical Equilibrium & The Normal Force

1
concept

## Equilibrium 5m
Play a video:
Hey, guys. So by now, what we've seen is that when you have a net force that pushes an object, it's going to accelerate in that direction. Well, you're gonna need to know how to solve problems where an object is said to be in equilibrium. So we're gonna be talking about what the equilibrium term means in this video. We're gonna do some examples. Let's check it out. The guys. The big idea here is if you have all the forces that are acting on an object, cancel out all the forces cancelled. This object is at equilibrium. So what that means from your F equals M A is that the sum of all forces is equal to zero, and so therefore, your acceleration is equal to zero. Let's do a couple of examples so I can show you how this works here. So we've got these two equal forces that are acting on this box are pulling, and we know that this box is gonna be moving at a constant 5 m per second. So we know that vehicles five, it doesn't matter if it's moving left to right, So I'm just gonna point the right there, so we're going to assume that the box has no weight. And what we want to do is we want to calculate the boxes acceleration. So we want to calculate acceleration. We've got forces involved. We know we're gonna have to use some f equals m A. But first, we're gonna have to draw the free body diagram. So the first thing we would do is we will check for the weight force. But remember that this problem has, uh we're gonna assume that the boxes, no weights. We're gonna skip that one. We've already got. The applied Force is here and there's no cable, so there's no tensions. And then there's no two surfaces in contact, so there's not gonna be normals or frictions or anything like that. So our free body diagram is done, and now you have to write f equals m A. So we've got f equals m A. Here. Now what happens is we have all the forces involved. We know the two applied forces, so we're gonna start from the left side of F equals m A. We have all the forces, and I'm just going to make this the one the positive directions. I've got 10. And then that means that this one is negative. 10 because it points in the opposite direction, then equals Emma. So we know that the 10 and the negative 10 I'm gonna cancel it to zero. So this means zero equals m A. And what that means here is that the acceleration is equal to zero. And that's what we said it, Librium, was the forces cancelled out. So that means that the acceleration is equal to zero. So let me go ahead and actually make a conceptual point here. That's really important. So one of things we've seen here is that equilibrium doesn't mean that an object isn't moving. It doesn't mean that V is equal to zero. We actually solve the problem that the V that the velocity of the boxes 5 m per second equilibrium means that the object isn't accelerating, which means from Africa. Then you know a is equal to zero. So this box is just gonna keep moving at 5 m per second. All right, so it's an important conceptual point that you will need to know to make sure you remember that. All right, let's move on to the next one. So here we've got a 2 kg book and it's gonna rest on the table and it's going to stay at rest. And we're going to assume that the box does have or the books does have weights, and we want to calculate the forces that are acting on this book. So again, just like we did in the last problem, we have to draw a free body diagram. But here we actually do have to account for the weight force. So the weight force points down. This is W equals mg. And then we don't have any applied forces or tensions or anything like that. But we do have two surfaces in contact, unlike how we did in the first example. So we do have a normal force. It points perpendicular to the surface, which is basically just the table that it's worsening resting on. So this is our normal force. We don't know what that is. We don't actually want to calculate the forces that are acting on the book. So now we've got our free body diagram. We just go into F equals Emma. So we got f equals m A. Here. We know that the normal force is going to be up, so I'm gonna choose that to be positive. So I've got normal force and then my mg points down. So I've got minus mg, and this is equal to a well unlike what we did in the first example we have to do with this problem is we have to start on the right side of F equals m A. Because one thing we know about this problem is we're told that the book is gonna be at rest and stays at rest. So what that means is that the acceleration is equal to zero. It's going to stay at rest and it doesn't accelerate or anything like that. So here we know that a is equal to zero, which means that we have end equal minus mg equals zero. And so when you move it to the other side, we have that m s and is equal to mg. So that means that both of the forces are going to be equal and opposite. We have two times 9.8 and we get 19.6 Newtons. So here we have the magnitude of those two forces. We know that N is going to be 19.6 up and m g is going to be 19.6 down. But just like the first example, what we've seen here is that the forces are going to cancel. You have the same equal forces just in opposite directions. So that actually brings me back to the point. The whole point of equilibrium is that you're gonna start off from ethical dilemma. And there's basically going to be two different kinds of problems in some problems. You're gonna know by all the forces that they're going to cancel like we did in this first problem here, we had to equal forces. So, you know, on the left side of F equals M A that all the forces are going to cancel. That means, uh, some of the forces zero. And that means that the acceleration is equal to zero now, in other problems like we did in this problem over here is we actually have that. This book is at rest. So in other problems, you're gonna have to basically extract or learn from the problem that the acceleration is equal to zero, and so the acceleration is equal to zero. Then that means that you know the forces are going to cancel. It's basically two different sides of the same coin here. So if you know one, you know the other. If you know the forces canceled a zero a zero, then you know the forces canceled. So hopefully that makes sense, guys. But that's it for this one. Let's move on.
2
Problem

A 3-kg box of junk is being lowered on a string at a constant speed. What is the tension in the string?

3
example

## Multiple Cables on a Loudspeaker 3m
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Hey, guys, let's check out this problem here. We got this loudspeaker that's held in place by four vertical cables. We know the tension in each cables. 30. So, basically, I'm gonna draw this out real quick. Right? So we have this loudspeaker like this, I don't know what the masses, but I do know it's held in place not by one cable, but actually by four. And I know the tension. Each one of these cables is 30. What I want to do this problem is I want to figure out the mass of this loudspeaker. So I actually want to figure out what is this M here? All right, So what I wanna do is I want to draw a free body diagram because they've had some tensions that I've got, you know, things like that. So I want to draw a free body diagram first. So I got I got like this, remember, I have the weight force that acts down. This is gonna be my mg. This is my target variable here. And then if there are only one cable that was attached to this, I would draw one era like this, right? There's actually four So what I need to do is I need to draw four identical arrows like this, and these are all going to be the tensions. These are all supposed to be the same size. So each one of these tensions here is 30. So if I only had one would be t equals 30. But I basically have to account for all four of these tensions. And I don't have to write it all four times, right, cause I know each one of them is 30. All right, now I've got no applied forces normals or friction or anything like that. I'm not directly pushing or pulling it, and it's also not in contact with anything, right? So those are the only two forces. So now I want to draw. I want to write out my f equals M a. Right. That's how we how we solve problems. So we have our f equals m a here. So the sum of all forces equals mass times acceleration. We just choose our upward direction to be positive. Now we're just gonna expand our EFS. There's a couple ways to do this, right. We have actually four tensions that point upwards, so one way you can do this, you could have tension plus tension plus bah blah. But that kind of gets annoying. So instead what you can do you say? Well, if all these tensions are the same then we're just four times t right, four times the same tension. They're all the same value and then we got mg downwards. So this is minus mg equals m A. All right, so we're trying to find this mass here. I know the tension. So I also know the G. The only other variable I don't know is the acceleration. What is the acceleration of this loudspeaker that's being held by these cables? So here's what we do. You have to go back to the problem and see if we can figure it out. What this problem tells us, though, is that this loudspeaker is held in place by these cables. So what that means is that the velocity is equal to zero and it's held in place. It's not going anywhere, and it's also going to stay that way, which means that the acceleration is equal to zero. There's gonna be no change in velocity. Alright, so we can do is we're gonna imply and infer from the problem. You're supposed to figure it out that this acceleration is equal to zero. And so this loudspeakers at equilibrium. So that means we can just set four t minus mg equal to zero. And now you can set up our equation and solve for M. So basically, when we move m g to the other side, we're gonna get 40 equals mg. And now we just divide this G over to the other side. So really are m becomes four times the tension divided by G, which is four times 30 divided by 98 You work this out, you're gonna get 12.12 kg. So you look for your answer choices and that's answer choice. Be alright. So each one of these tensions here is 30 Newton's. But all of them together add up to 1 20. This kind of makes some sense. If there was only one cable, it has to have to support the entire weight. If there are two cables, they could kind of split it evenly. There are three cables that could split the way distribute among the three cables. If there's four and so on and so forth. You get the picture. So that's why each one of these tables these cables here doesn't have to support damage. Wait. All right, That's it. That one.
4
concept

## The Normal Force 8m
Play a video: 