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6. Intro to Forces (Dynamics)

1

concept

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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?

A

0 N

B

3 N

C

29 N

D

32 N

3

example

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

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Hey, guys. So early on in this chapter, we took a look at all the forces you might possibly see in your problems, and one of those forces was called the Normal Force. So we're gonna go through a little bit more detail about this normal force and I'll show you how it works. Let's check it out. The whole thing here, guys, is that whenever you have two surfaces in contact, if one surface is pushed up against another in any direction, then that surface is gonna push back with the force called the normal. Now there's different symbols letters that we use for the normal force. One of the ones you might see it as a letter end. You might see a little end. You might even see uh, f n with your professors. Uh, there's lots of different ways to write it, But here, a clutch. We're going to use a capital end. That's just what I'm used to. So what this normal force means is actually kind of like a fancy engineering term, but it really just means perpendicular, so that just means 90 degrees to the surface. So what we saw in a previous video is that if you have the weight force, that's w equals mg. Pushing down that is, pushing this block against the surface to the surface pushes back, and that's going to be upwards with a force called the normal. But obviously like to do is use my finger and thumb to figure out the direction you point your thumb along the surface like this, and then your finger points in the direction of the normal. Now it doesn't necessarily always have to be up. You can actually push a block against the wall like this with an applied force, you have surfaces in contact and the surface pushes back with the normal force. So you take your thumb like this and your finger points in the direction of the normal pushes back like that. And if you have a block against a ramp or an incline or something, you have the weight force that pushes down. It's w equals mg, and then your normal force is you're just gonna tilt your thumb like this in your normal points that way. Now, the most important thing about this normal force, you absolutely need to know, is that unlike W equals MG the weight force. There's no magical equation to solve For the normal, you always calculate that normal force by using f equals m A. So what we're gonna do here is we're going to use a familiar example that we've seen before. We're gonna do some variations so I can show you how this works. So we've got this 2.4 kg book that is going to rest on the table and we want to calculate the normal force. So we've already seen with the free body diagram looks for that we've got the W which is equal to mg, which, by the way, you can calculate because you know, the mass, which is 2.4 And if you do 9.8, this w is going to equal 20 Newtons and it's actually gonna be the same throughout all of our problems here. So we have w equals 20 w equals and then w equals 20 right? So it's the same way because the same mass throughout. Now, in this case, we've got a normal force and it's going to point up. We've already seen that. So we want to calculate what that is So now that we've got our free body diagram, we just have to use f equals m a. Right. So you've got f equals m a here. And so we got our other forces, which is the normal plus our wait for which is downwards RMG. This is equal to mass times acceleration. We'll just like we did in the previous video we have to do is extract from the problem that this book is resting on the table. So what that means here is that this box is at equilibrium or this book is at equilibrium. The acceleration is equal to zero and therefore the forces have to cancel. So that means that your normal force minus your M G is equal to zero. So what that means here is that your normal is equal to mg and that's just Newtons. So you have n equals 20 here. Now, we're gonna talk about this last bullet point in just a few minutes here, but I just want to go ahead, go ahead and move on with the next problem here. What we got is the weights. But now we're going to push the book with an additional 10 Newtons are gonna push the book downwards. What that means is that in our free body diagram we have to write another force which is going to be F and you know this is going to be 10 downwards. Now, here we've got two pushes downwards, which means the surface has to push upwards just like it did before. So we have a normal force like this. We want to calculate that. So we have to use f equals M A. So f equals m a. Now here we got a normal force and then these two forces point downwards. So they're gonna have to pick up a negative science. This is gonna be negative F plus negative mg. And this is equal to m A now, just like we did in the last problem. This book is still at equilibrium, right? It doesn't go flying off the table. It doesn't go crashing through the surface. So we have to do is extract from the problem that the book is still going to be at rest on the table. So a is equal to zero and all the forces will cancel just like they did before. So now we just have another force to consider. So here, once you move everything over to the other side, you're gonna have N is equal to MG plus f because remember, this is going to be zero. So what happens is now we've got our MG, which is 20 and our normal forces 10 So near now are normal. Force is going to start. Our apply forces tend to our normal force is going to be 30. All right, so that's our normal force right here and for this next one same thing. But now we're going to pull the book up with 15 Newtons instead of pushing it down. So now, instead of our applied force being downwards, it points upwards. So here we've got an F, and this is equal to 15. Now, if you take a look here, what happens is we got this 15. You're trying to pull the book up with 15, but the way force is stronger, it's still 20. So that's the way force winds. There's still gonna be some some weight downwards. And so the surface is still going to have to push back upwards with the normal force. So that's our normal. And we're gonna calculate that using f equals m a. So here we've got our normal force, but now are applied. Force also points upwards, so it stays positive and RMG is going to be negative. And this is equal to a now, just like the previous two examples this box is still this book is still going to be an equilibrium, right? All the forces still have to balance out because it's still your Basically, your force isn't strong enough to lift the book off the table, so it's still gonna be some normal force and all the forces are going to have to cancel. So this is a is equal to zero. So now we've got here is we've got N in which you move m g to the other side. You're gonna have to subtract f So this is going to be 20 minus 15, So you're gonna end up with a normal force of five. So we know here that this end is equal to five. And finally, for the last problem here, we're going to double our force, and we're gonna pull this book up with 30 Newtons instead of 15. And so now what we want to do is calculate the acceleration. So here we've got is we're going to double that applied for us and our f is going to be instead of 20. So now what happens to this normal force? Well, unlike this previous example that we did here are force now is strong enough to overcome the force of gravity. The weight force of 20. If you're pulling with 30 but gravity is only 20 then you're pulling force winds. So what that means here is that this book is actually gonna go flying upwards with some acceleration. So what happens to the normal? Well, unlike before now, there's no surface push. So that means that the normal in this case is equal to zero doesn't exist anymore. So to calculate this acceleration, we just use f equals m A. So here we have an applied force of 30 and now we've got mg downwards, and this is equal to mass times acceleration. And unlike the last three examples here, we can't just go ahead and assume this is zero. Because we know again that this box is gonna go flying upwards. So here we've got our 20 and here we've got our negative Sorry, 30 and are negative. 20. And this is equal to 2.4 times a. So you work this out and what you're gonna get is that a is equal to 49 m per second squared. So what happens is we know that this box or book is going to fly up with 4.9. So let me just recap all of these different scenarios here. So when you have the book that is just resting on the table, that means that there's no other applied forces. And in this case, what we saw is that your normal force was equal to mg. So in these cases, if there's no other applied courses and is equal to mg and the second one, we push the book down. So if you're pushing the book down, which means you're applied forces along with MG, then your normal force has to basically balance out both forces. And we saw that and is greater than mg. It went from 2030 here. We pulled the book with the book up, but not enough to lift it. What that just means is that are applied force of 15 was less than RMG of 20. And in this case, what we saw is that our normal force was only five Newtons. So in this case, it turned out to be less than mg. And then finally, if you pull it with enough force to lift it, which basically means you're 30 was greater than the 20 then That means that your normal force is equal to zero. There's no more surface push anymore. All right, So do you have to memorize all these situations? No. But you should always remember that you are you Are you going to calculate and by using F equals m A. These are just some of the results that you might see, So hopefully that makes sense. That's about this one, guys, let's move on.

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