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Moments (part 2)

2 more moment problems. Created by Sal Khan.

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  • starky ultimate style avatar for user Andrew
    Why aren't the clockwise forces marked with a negative sign when Sal does the algebra? As always, thanks a bunch.
    (7 votes)
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    • blobby green style avatar for user kong.m.siu
      The reason why there is no negative sign in the equation is because He is setting it as in equilibrium, the rightwise is equal to the counter-clockwise. This is like up is equal to down, it is implied that the forces are already acting in opposite directions so the negative sign is not needed. Another way to look at this would be The net movement of the table equals zero; 0 = clockwise force minus counter-clockwise force. Add counter-clockwise force on each side and you get. clockwise force = counter-clockwise force.
      (24 votes)
  • leaf green style avatar for user Nick Schmit
    I'm not sure whether or not I may have missed a title, but is there a video that addresses the parallel axis theorem? I have a textbook that briefly addresses the theorem, but it gives a poor example and moves right along.
    (11 votes)
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    • duskpin ultimate style avatar for user shingamba
      I don't think there is one specifically for that topic, but the parallel axis theorem as I know states that the moment of inertia of a body about any axis is equal to the sum of [the moment of inertia of the body about an axis (parallel to the axis taken) at its center of mass] and [the product of the mass of the body and the square of the distance between the two parallel axes].
      That is : I = Icm + (m).(d^2)
      {where I is the moment of inertia about the axis taken; Icm is the moment of inertia about the center of mass and parallel to the axis taken; And m is the mass of the body, d is the distance between the two parallel lines}

      I hope the statement wasn't so confusing...
      (12 votes)
  • old spice man green style avatar for user soumita1212
    please can give me answer of this ques , 10N weight can be lifted in one end of a thin rod by applying a 5N force on the
    other end. If the length of the rod is 1.5 meter, where is the fulcrum situated? and please tell me how to measure
    (6 votes)
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  • blobby green style avatar for user fatimadorgham142
    I didn't get how the forces of 20 N and and the force exerted by the book are clockwise ?
    (6 votes)
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  • piceratops seed style avatar for user jespat_sousa
    when calculating the counterclockwise forces, why isn't the weight of the right leg counted? only the force up is counted not the one going down? why is that ?
    (4 votes)
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  • piceratops tree style avatar for user Panha
    I'm 12 years old and I understand this. My friends says that it is impossible and my dad says if I learn to much I might get really stressed. I don't think learning stresses me out much. Should I continue? I really like learning physics.
    (2 votes)
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    • leaf blue style avatar for user Hakan
      If you are comfartable and you enjoy it. Then why not! However listen to your parents too. I mean try to find an optimized way. Do extra but don't burn your head. It's good that you learn more but everything has a time. Just don't push so hard.
      (3 votes)
  • leaf green style avatar for user Jeramy Huerto
    Wouldn't the axis of rotation be at the base of the left leg? If you removed the right leg, the table would rotate on the base of the left leg. If the axis was at the top, that left leg would slip off the floor and rotate clockwise. That would be weird.
    (2 votes)
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    • piceratops ultimate style avatar for user Laukik Rege
      Yes you are right...the axis should be at the base of the left leg...but when calculating the torque it wont matter as perpendicular distance of the 100 N force from the axis will be same in both cases(you know, since force is a vector you can translate the force vector or extrapolate its line of action downwards without changing its value) ....

      Hope this helps :)
      (3 votes)
  • spunky sam blue style avatar for user katy
    what is the difference between moment of inertia and moment to force?
    (1 vote)
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    • leafers ultimate style avatar for user Diogo Ribeiro
      It's kinda hard for me to explain, so I'll be quoting wikipedia:

      "Moment of inertia is the mass property of a rigid body that determines the torque needed for a desired angular acceleration"

      "Moment of force (or Torque), is the tendency of a force to rotate an object"

      If you translate these concepts to motion of particles, torque would be the force and moment of inertia would be the mass.
      (2 votes)
  • blobby green style avatar for user INDU SACHDEVA
    How to calculate torques when the plank or supporting board also has significant weight and changes the cheese of the system
    (1 vote)
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    • aqualine ultimate style avatar for user vinnv226
      You add in an extra torque for the weight of the plank. The force will have magnitude mg, where m is the mass of the plank and g is 9.8 m/s^2. The position of the force will be at the plank's center of gravity (on Earth's surface, this is at its center of mass also) and it will point straight down towards Earth's center.
      (2 votes)
  • blobby green style avatar for user Vandana Kumari
    i didn't understood that second problem.. Why did sal add 85N force ?
    (1 vote)
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Video transcript

I'm going to do a couple more moment and force problems, especially because I think I might have bungled the terminology in the previous video because I kept confusing clockwise with counterclockwise. This time I'll try to be more consistent. Let me draw my lever again. My seesaw. So that's my seesaw, and that is my axis of rotation, or my fulcrum, or my pivot point, whatever you want to call it. And let me throw a bunch of forces on there. So let's say that I have a 10-Newton force and it is at a distance of 10, so distance is equal to 10. The moment arm distance is 10. Let's say that I have a 50-Newton force and its moment arm distance is equal to 8. Let's say that I have a 5-Newton force, and its moment arm distance is 4. The distance is equal to 4. That's enough for that side. And let's say I have a I'm going to switch colors. Actually, no, I'm going to keep it all the same color and then we'll use colors to differentiate between clockwise and counterclockwise so I don't bungle everything up again. So let's say I have a 10-Newton force here. And, of course, these vectors aren't proportional to actually what I drew. 50 Newtons would be huge if these were the actual vectors. And let's say that that moment arm distance is 3. Let me do a couple more. And let's say I have a moment arm distance of 8. I have a clockwise force of 20 Newtons, And let's say at a distance of 10 again, so distance is equal to 10, I have my mystery force. It's going to act in a counterclockwise direction and I want to know what it needs to be. So whenever you do any of these moment of force problems, and you say, well, what does the force need to be in order for this see saw to not rotate? You just say, well, all the clockwise moments have to equal all of the counterclockwise moments, So clockwise moments equal counterclockwise. I'll do them in different colors. So what are all the clockwise moments? Well, clockwise is this direction, right? That's the way a clock goes. So this is clockwise, that is clockwise. I want to go in this direction. And so this is clockwise. What are all the clockwise moments? It's 10 Newtons times its moment arm distance 10. So 10 times 10 plus 5 Newtons times this moment arm distance 4, plus 5 times 4, plus 20 Newtons times its moment arm distance of 8, plus 20 times 8, and that's going to equal the counterclockwise moments, and so the leftover ones are counterclockwise. So we have 50 Newtons acting downward here, and that's counterclockwise, and it's at a distance of 8 from the moment arm, so 50 times 8. Let's see, we don't have any other counterclockwise on that side. This is counterclockwise, right? We have 10 Newtons acting in the counterclockwise direction, and its moment arm distance is 3, plus 10 times 3, and we're assuming our mystery force, which is at a distance of 10, is also counterclockwise, plus force times 10. And now we simplify. And I'll just go to a neutral color because this is just math now. 100 plus 20 plus 160 is equal to-- what's 50 times 8? That's 400 plus 30 plus 10F. What is this? 2, 50 times 8. Right, that's 400. OK, this is 120 plus a 160 is 280. 280 is equal to 430-- this is a good example-- plus 10F, I just realized. Subtract 430 from both sides. So what's 430 minus 280? It's 150. So it's minus 150 is equal to 10F. So F is equal to minus 15 Newtons in the counterclockwise direction. So F is minus 15 Newtons in the counterclockwise direction, or it means that it is 15 Newtons. We assumed that it was in the counterclockwise direction, but when we did the math, we got a minus number. [SNEEZE] Excuse me. I apologize if I blew out your speakers with that sneeze. But anyway, we assume it was going in the counterclockwise direction, but when we did the math, we got a negative number, so that means it's actually operating in the clockwise direction at 15 Newtons at a distance of 10 from the moment arm. Hopefully, that one was less confusing than the last one. So let me do another problem, and these actually used to confuse me when I first learned about moments, but in some ways, they're the most useful ones. So let's say that I have some type of table. I'll draw it in wood. It's a wood table. That's my table. And I have a leg here, I have a leg here. Let's say that the center of mass of the top of the table is here. It's at the center. And let's say that it has a weight. It has a weight going down. What's a reasonable weight? Let's say 20 Newtons. It has a weight of 20 Newtons. Let's say that I place some textbooks on top of this table, or box, just to make the drawing simpler. Let's say I place a box there. Let's say the box weighs 10 kilograms, which would be about 100 Newtons. So let's say it weighs about 100 Newtons. So what I want to figure out, what I need to figure out, is how much weight is being put onto each of the legs of the table? And this might not have even been obviously a moment problem, but you'll see in a second it really is. So how do we know that? Well, both of these legs are supporting the table, right? Whatever the table is exerting downwards, the leg is exerting upwards, so that's the amount of force that each of the legs are holding. So what we do is we pick-- so let's just pick this leg, just because I'm picking it arbitrarily. Let's pick this leg, and let's pick an arbitrary axis of rotation. Well, let's pick this is as our axis of rotation. Why do I pick that as the axis of rotation? Because think of it this way. If this leg started pushing more than it needed to, the whole table would rotate in the counterclockwise direction. Or the other way, if this leg started to weaken and started to buckle and couldn't hold its force, the table would rotate down this way, and it would rotate around the other leg, assuming that the other leg doesn't fail. We're assuming that this leg is just going to do its job and it's not going to move one way or the other. But this leg, that's why we're thinking about it that way. If it was too weak, the whole table would rotate in the clockwise direction, and if it was somehow exerting extra force, which we know a leg can't, but let's say if it was a spring or something like that, then the whole table would rotate in the counterclockwise direction. So once we set that up, we can actually set this up as a moment problem. So what is the force of the leg? So the whole table is exerting some type of-- if this leg wasn't here, the whole table would have a net clockwise moment, right? The whole table would tilt down and fall down like that. So the leg must be exerting a counterclockwise moment in order to keep it stationary. So the leg must be exerting a force upward right here. The force of the leg, right? We know that. We know that from basic physics. There's some force coming down here and the leg is doing an equal opposite force upwards. So what is that force of that leg? And one thing I should have told you is all of the distances. Let's say that this distance between this leg and the book is 1 meter-- or the box. Let's say that this distance between the leg and the center of mass is 2 meters, and so this is also 2 meters. OK, so we can now set this up as a moment problem. So remember, all of the clockwise moments have to equal all of the counterclockwise moments. So what are all of the clockwise moments? What are all of the things that want to make the table rotate this way or this way? Well, the leg is the only thing keeping it from doing that. So everything else is essentially a clockwise moment. So we have this 100 Newtons, and it is 1 meter away. Its moment arm distance is 1. So these are all the clockwise moments, 100 times 1, right? It's 100 Newtons acting downwards in the clockwise direction, clockwise moment, and it's 1 meter away, plus we have the center of mass at the top of the table, which is 20 Newtons, plus 20 Newtons, and that is 2 meters away from our designated axis, so 20 times 2. And you might say, well, isn't this leg exerting some force? Well, sure it is, but its distance from our designated axis is zero, so its moment of force is zero. Even if it is exerting a million pounds or a million Newtons, its moment of force, or its torque, would be zero because its moment arm distance is zero, so we can ignore it, which makes things simple. So those were the only clockwise moments. And what's the counterclockwise moment? Well, that's going to be the force exerted by this leg. That's what's keeping the whole thing from rotating. So it's the force of the leg times its distance from our axis. Well, this is a total of 4 meters, which we've said here, times 4 meters. And so we can just solve. We get 100 plus 40, so we get 140 is equal to the force of the leg times 4. So what's 140-- 4 goes into 140 35 times? My math is not so good. Is that right? 4 times 30 is 120. 120 plus 20. So the force of the leg is 35 Newtons upwards. And since this isn't moving, we know that the downward force right here must be 35 Newtons. And so there's a couple of ways we can think about it. If this leg is supporting 35 Newtons and we have a total weight here of 120 Newtons, our total weight, the weight at the top of the table plus the bookshelf, that's 120 Newtons. So the balance of this must be supported by something or someone. So the balance of this is going to be supported by this leg. So it's 120 minus 35 is what? [PHONE RINGS] Oh, my phone is ringing. 120 minus 35 is what? 120 minus 30 is 90. And then 90 minus 5 is 85 Newtons. It's so disconcerting when my phone rings. I have trouble focusing. Anyway, it's probably because my phone sounds like a freight train. Anyway, so there you go. This type of problem is actually key to, as you can imagine, bridge builders, or furniture manufacturers, or civil engineers who are bridge builders, or architects, because you actually have to figure out, well, if I design something a certain way, I have to figure out how much weight each of the supporting structures will have to support. And as you can imagine, why is this one supporting more weight? Why is this leg supporting more weight than that leg? Well, because this book, which is 100 Newtons, which is a significant amount of the total weight, is much closer to this leg than it is to this leg. If we put it to the center, they would balance, and then if we push it further to the right, then this leg would start bearing more of the weight. Anyway, hopefully you found that interesting, and hopefully, I didn't confuse you. And I will see you in future videos.