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 going to need to know how to solve problems where an object is said to be in equilibrium. So we're going to be talking about what that equilibrium term means in this video, and we're going to do some examples. Let's check it out. So guys, the big idea here is if all the forces that are acting on an object cancel out. If all the forces cancel, then 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 0, and so therefore your acceleration is equal to 0. Let's do a couple of examples so I can show you how this works here. So we've got these 2 equal forces that are acting on this box or pulling, and we know that this box is going to be moving at a constant 5 meters per second. So we know that v equals 5. It doesn't matter if it's moving left or right. So I'm just going to point to 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 box's acceleration. So we want to calculate acceleration. We've got forces involved. We know we're going to have to use some F equals m*a. But first, we're going to have to draw the free body diagram. So the first thing we would do is we would check for the weight force. But remember that this problem has, we're going to assume that the box has no weight. We're going to skip that one. We've already got the applied forces here, and there's no cables, so there's no tensions. And then there are no 2 surfaces in contact, so there are no normals or frictions or anything like that. So our free body diagram is done, and now we just 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 2 applied forces. So we're going to 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 direct direction, I've got 10. And then that means that this one is negative 10 because it points in the opposite direction. Then I equals m*a. So we know that the 10 and the negative 10 are going to cancel out to 0. So this means 0 equals m*a. And what that means here is that the acceleration is equal to 0, and that's what we said equilibrium was. The forces canceled out, so that means that the acceleration is equal to 0. So let me go ahead and actually make a conceptual point here that's really important. So one of the 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 0. We actually saw from the problem that the v, that the velocity of the box is 5 meters per second. Equilibrium means that the object isn't accelerating, which means from F equals m*a, then we know a is equal to 0. So this box is just going to keep moving at 5 meters per second. Alright? So it's an important conceptual point that you will need to know to, you know, make sure you remember that. Alright. Let's move on to the next one. So here we've got a 2-kilogram book, and it's going to rest on the table, and it's going to stay at rest. And we're going to assume that the book does have or the books do have weight, and we want to calculate the forces that are acting on this book. So, again, just like we did in the last problem, we'd 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 m*g. And then we don't have any applied forces or tensions or anything like that, but we do have 2 surfaces in contact, unlike how we did in the first example. So we do have a normal force that points perpendicular to the surface, which is basically just the table that it's resting on. So this is our normal force. We don't know what that is. We 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 m*a. So we've got F equals m*a here. We know that the normal force is going to be up, so I'm going to choose that to be positive. So I've got normal force, and then my m*g points down. So I've got minus m*g, and this is equal to m*a. Well, unlike what we did in the first example, what we have to do in 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 going to be at rest and stays at rest. So what that means is that the acceleration is equal to 0. 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 0, which means that we have n equal minusmg equals 0. And so when you move it to the other side, we have that n is equal to m*g. So that means that both of the forces are going to be equal and opposite. We have 2 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 see 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 going to start off from F equals m*a, and there's basically going to be 2 different kinds of problems. In some problems, you're going to know by all the forces that they're going to cancel, like we did in this first problem here. We had 2 equal forces. So you know on the left side of F equals m*a that all the forces are going to cancel. That means F is, the sum of forces is 0, and that means that the acceleration is equal to 0. 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 going to have to basically extract or learn from the problem that the acceleration is equal to 0. And so if the acceleration is equal to 0, then that means that you know the forces are going to cancel. It's basically 2 different sides of the same coin here. So if you know 1, you know the other. If you know the forces cancel, you know a is 0. If you know a is 0, then you know the forces cancel. So, hopefully, that makes sense, guys. But that's it for this one. Let's move on.

# Vertical Equilibrium & The Normal Force - Online Tutor, Practice Problems & Exam Prep

### Equilibrium

#### Video transcript

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

### Multiple Cables on a Loudspeaker

#### Video transcript

Hey, guys. Let's check out this problem here. We've got this loudspeaker that's held in place by 4 vertical cables. We know the tension in each cable is 30. And so basically, I'm going to draw this out real quick. Right? So we have this loudspeaker like this. I don't know what the mass is, but I do know it's held in place not by 1 cable but actually by 4. And I know the tension in each one of these cables is 30. What I want to do in 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. Alright? So what I want to do is I want to draw a free body diagram because I've got some tensions and I’ve got, you know, things like that. So I want to draw a free body diagram first. So I got my dot like this. Remember, I have the weight force that acts down. This is going to be my m g. This is my target variable here. And then if there were only one cable that was attached to this, I would draw one arrow like this. Right? There are actually 4. So what I need to do is I need to draw 4 identical arrows like this and these are all going to be the tensions. Tensions. These are all supposed to be the same size. So each one of these tensions here is 30. So if I only had 1, it would be t equals 30, but I basically have to account for all 4 of these tensions. And I don't have to write it all 4 times, right, because I know each one of them is 30. Alright? Now I 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 going to expand out our f's. There's a couple ways to do this. Right? We have actually 4 tensions that point upwards. So one way you could do this is you could have tension plus tension plus blah blah blah, but that kind of gets annoying. So instead, what you can do is say, well, if all these tensions are the same, then which is just 4 times t. Right? Four times the same tension. They're all the same value. And then we've got mg downwards. So this is minus mg equals m a. Alright. 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. We 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 0, 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 0. There's going to be no change in velocity. Alright? So what we could do is we can imply or infer from the problem, you're supposed to figure it out, that this acceleration is equal to 0, and so this loudspeaker is at equilibrium. So that means we can just set 4t minus mg equal to 0, and now we can set up our equation and solve for m. So basically, when we move mg to the other side, we're going to get 4 t equals mg, and now we just divide this g over to the other side. So really our m becomes 4×tg, which is 4×309.8. If you work this out you're going to get 12.12 kilograms. And so you look through your answer choices, and it's answer choice b. Alright? So each one of these tensions here is 30 newtons, but all of them together add up to 120. This kind of makes some sense. If there was only one cable, it has it’ll have to support the entire weight. If there are 2 cables, they could kind of split it evenly. There are 3 cables, they could split the weight, distribute among the 3 cables. If there's 4 and so on and so forth, you get the picture. So that's why each one of these cables here doesn't have to support that much weight. Alright? So that's it for that one.

### The Normal Force

#### Video transcript

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 going to 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 2 surfaces in contact, if one surface is pushed up against another in any direction, then that surface is going to push back with a force called the normal. Now there are different symbols, letters that we use for the normal force. One of the ones you might see is the letter N. You might see a little n. You might even see, an FN with your professors. There are lots of different ways to write it. But here at Clutch, we're going to use a capital N. That's just what I'm used to. So what this normal force means is it's 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=mg pushing down, that is pushing this block against the surface so the surface pushes back, and that's going to be upwards with a force called the normal. What I usually just 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 could actually push a block against a 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 and pushes back like that. And if you have a block against a ramp and your normal points that way. Now the most and your normal points that way. Now the most important thing about this normal force that you absolutely need to know is that unlike W=mg, the weight force, there's no magical equation to solve for the normal. You always calculate that normal force by using F=ma. So what we're going to do here is we're going to use a familiar example that we've seen before and we're going to do some variations so I can show you how this works. So we've got this 2.04 kilogram book that is going to rest on the table, and we want to calculate the normal force. So we've already seen what the free bond diagram looks like 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 2.04. And if you do 9.8, this W is going to equal 20 newtons. And it's actually going to be the same throughout all of our problems here. So we have W equals 20, W equals W equals 20. Alright, W equals 20. Right? So it's the same weight 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. And so we want to calculate what that is. So now that we've got our free body diagram, we just have to use F=ma. Right? So you've got F equals ma here. And so we got all the forces, which is the normal plus our weight force, which is downwards, our mg. This is equal to mass times acceleration. Well, just like we did in the previous video, what 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 0 and therefore, the forces have to cancel. So that means that your normal force minus your mg is equal to 0. So what that means here is that your normal is equal to mg, and that's just 20 newtons. So we have n=20 here. Now we're going to talk about this last bullet point in just a few minutes here, but I just want to 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. We're going to 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 we know this is going to be 10 downwards. Now here we've got 2 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=ma. So F=ma. Now here, we've got a normal force. And then these two forces point downwards, so they're going to have to pick up a negative sign. So this is going to be negative F+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 0 and all the forces will cancel just they do before. So now we just have another force to consider. So here, once you move everything over to the other side, you're going to have m is equal to mg plus F because remember this is going to be 0. So what happens is now we've got our mg which is 20 and our normal force is 10. So now our normal force is going to, or sorry, our applied force is 10. So our normal force is going to be 30. Alright. 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've got this 15. You're trying to pull the book up with 15, but the weight force is stronger. It's still 20. So this the weight force wins. There's still going to be 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 going to calculate that using F=ma. So here, we've got our normal force but now our applied force also points upwards so it stays positive and our mg is going to be negative and this is equal to m a. Now just like the previous two examples, this box is this book is still going to be at equilibrium. Right? All the forces still have to balance out because it's still you basically, your force isn't strong enough to lift the book off the table. So it's still going to be some normal force and all the forces are going to have to cancel. So this a is equal to 0. So now what we've got here is we've got n, and once you move mg to the other side, you're going to have to subtract F. So this is going to be 20 minus 15, and so you're going to end up with a normal force of 5. So we know here that this n=5. And finally, for the last problem here, we're going to double our force and we're going to pull this book up with 30 newtons instead of 15. And so now, what we want to do is calculate the acceleration. So here, what we've got is we're going to double that applied force and our F is going to be 30 instead of 20. So now what happens to this normal force? Well, unlike this previous example that we did here, our 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 your pulling force wins. So what that means here is that this book is actually going to 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 0. It doesn't exist anymore. So to calculate this acceleration, we just use F=ma. So here we have an applied force of 30 and now we 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 0 because we know again that this box is going to go flying upwards. So here we've got our 20 and here we got our, negative sorry, 30 and our negative 20, and this is equal to 2.04 times a. So you work this out, and what you're going to get is that a is equal to 4.9 meters 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 forces, n is equal to mg. And the second one, we push the book down. So if you're pushing the book down, which means your applied forces along with mg, then your normal force has to basically balance out both forces. And we saw that n is greater than mg. It went from 20 30. Here we pulled the book with the book up, but not enough to lift it. What that just means is that our applied force of 15 was less than our mg of 20. And in this case, what we saw is that our normal force was only 5 newtons. So in this case, it turns out to be less than mg. And then finally, if you pull it with enough force to lift it, which basically means your 30 was greater than the 20, then that means that your normal force is equal to 0. There's no more surface push anymore. Alright. So do you have to memorize all these situations? No. But you should always remember that you're already are you gonna calculate n by using F equals ma. These are just some of the results that you might see. So hopefully, that made sense. That's it for this one, guys. Let's move on.

## Do you want more practice?

More sets### Your Physics tutor

- When jumping straight up from a crouched position, an average person can reach a maximum height of about 60 cm...
- (II) A 20.0-kg box rests on a table.<IMAGE>(b) A 10.0‑kg box is placed on top of the 20.0-kg box, as sho...
- A steel rod of radius R = 15 cm and length ℓ₀ stands upright on a firm surface. A 78-kg man climbs atop the ro...
- A 1130-kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E5.8). The cable ma...