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Mechanical advantage (part 2)-ի սղագրությունը

  • 0:00Welcome back.
  • 0:01When I left off, I was hurrying a little bit.
  • 0:03But we'd hopefully come to the conclusion that if I have a
  • 0:06simple lever, like I have here, and I know the distances
  • 0:09from where I'm applying the force, to the
  • 0:12fulcrum, to the pivot.
  • 0:13And I know the distance from the pivot to where the machine
  • 0:16is essentially applying the force, the machine being the
  • 0:18lever in this situation, I know the relationship between
  • 0:21the two forces I'm applying.
  • 0:23The input force-- so actually I shouldn't call this a force
  • 0:26too, I should call this an input force-- anyway, the
  • 0:28input force times the distance from the input force to the
  • 0:31fulcrum is equal to the output force times the distance from
  • 0:36the output force to the fulcrum.
  • 0:39And that all fell out of what we did in the last video.
  • 0:41The conservation of energy and that the work in has to equal
  • 0:45the work out.
  • 0:45And all work is, is a transfer of energy, so the transfer of
  • 0:48energy in has to be the transfer of energy out,
  • 0:50assuming we have no friction and none of the energy is
  • 0:53lost.
  • 0:53And how is this useful?
  • 0:55Well we could do a bunch of problems with this.
  • 1:00Let's say that I have a 100 newton object
  • 1:07right here, 100 newtons.
  • 1:10And let's say that I know, no matter what I do, my maximum
  • 1:15strength that I could push-- well let me draw this a little
  • 1:18different-- let's say it's like this, cause of my goal is
  • 1:19to lift the 100 newton object.
  • 1:20So the 100 newton object is right here.
  • 1:23That's a 100 newtons.
  • 1:24And let's say I know that the maximum downward force that
  • 1:27I'm capable of applying is only 10 newtons, right?
  • 1:31So I want my force to be multiplied by 10
  • 1:34to lift this force.
  • 1:36So let's figure out what would happen.
  • 1:37My input force is 10.
  • 1:39And I want to figure out the distance.
  • 1:42So let's say my input force is 10.
  • 1:44And let's call this the input distance.
  • 1:47And I want the output force to be 100, right?
  • 1:50And let's call this the output distance.
  • 1:54So if I have a fulcrum here, this is the input distance and
  • 2:00this is the output distance.
  • 2:01Let me switch colors.
  • 2:02This is getting monotonous.
  • 2:03This is the output distance, from here to here.
  • 2:05And let's figure out what the ratio has to be, for the ratio
  • 2:09of the input distance to the output distance.
  • 2:11Well, if we just divide both sides by 10, we get the
  • 2:14distance input.
  • 2:15It has to be 10 times the distance output, right?
  • 2:18100 divided by 10.
  • 2:20So if the distance from the fulcrum to the weight is, I
  • 2:24don't know, 5 meters, then the distance from where I'm
  • 2:27applying the force to the fulcrum has
  • 2:30to be 10 times that.
  • 2:31It has to be 50 meters.
  • 2:34So no matter what, the ratio of this length to this length
  • 2:37has to be 10.
  • 2:38And now what would happen?
  • 2:40If I design this machine this way, I will be able to apply
  • 2:4310 newtons here, which is my maximum strength, 10 newtons
  • 2:45downwards, and I will lift a 100 newton object.
  • 2:49And now what's the trade off though?
  • 2:50Nothing just pops out of thin air.
  • 2:52The trade off is, is that I am going to have to push down for
  • 2:56a much longer distance, for actually 10 times the distance
  • 2:59as this object is going to move up.
  • 3:02And once again I know that because the work in has to
  • 3:05equal the work out.
  • 3:06I can't through some magical machine-- and if you were able
  • 3:10to invent one, you shouldn't watch this video and you
  • 3:12should go build it and become a trillionaire-- but a machine
  • 3:15can never generate work out of thin air.
  • 3:18Or it can never generate energy out of thin air.
  • 3:19That energy has to come from some place.
  • 3:21Most machines actually you lose energy to friction or
  • 3:23whatever else.
  • 3:25But in this situation, if I'm putting in 10 newtons of force
  • 3:30times some distance, whatever that quantity is of work, the
  • 3:33work cannot change.
  • 3:34The total work.
  • 3:35It can go down if there is some friction in the system.
  • 3:38So let's do another problem.
  • 3:45And really they're all kind of the same formula.
  • 3:47And then I'll move into a few other types of simple systems.
  • 3:55I should use the line tool.
  • 3:59We'll make this up on the fly.
  • 4:03And you could always create problems where you can
  • 4:05compound it further and et cetera, et cetera, using some
  • 4:07of the other concepts we've learned.
  • 4:08But I won't worry about that right now.
  • 4:16So let's say that I'm going to push up here.
  • 4:24Well no let me see what I want to do.
  • 4:26I want to push down here with a force of-- let's say that
  • 4:35this distance right here is 35 meters, this distance is 5
  • 4:44meters-- and let's say I'm going to push down with the
  • 4:46force of 7 newtons, and what I want to figure out is how
  • 4:49heavy of an object can I lift here.
  • 4:52How heavy of an object.
  • 4:53Well, all we have to do is use the same formula.
  • 4:55But the moments-- and I know I used that word once before, so
  • 4:58you might not know what it is-- but the moments on both
  • 5:00sides of the fulcrum have to be the same.
  • 5:02Or the input moment has to be the output moment.
  • 5:05So what's the moment again?
  • 5:06Well, the moment is just the force times the distance from
  • 5:10the force to the fulcrum.
  • 5:12So the input moment is 7 newtons times 35 meters.
  • 5:19And realize that that does not work, because the distance
  • 5:21this force is traveling is not 35 meters.
  • 5:24The distance this force is traveling is
  • 5:26something like, here.
  • 5:28But this 35 meters is going to be proportional to the
  • 5:31distance that this is traveling when you compare it
  • 5:33to this other side.
  • 5:34So this quantity, 7 newtons times 35
  • 5:36meters, is the moment.
  • 5:38And that is going to be equal to the moment on this side,
  • 5:41the output moment.
  • 5:43So that is equal to 5 meters times the force that I'm
  • 5:47lifting, or the lifting force of the machine, times let's
  • 5:51say the force out.
  • 5:53So we can figure out the force out by just dividing both
  • 5:56sides by 5.
  • 5:58So let's see, 35 divided by 5 is 7, so you get 7 times 7
  • 6:03equals the force out, or 49 newtons.
  • 6:07And you can see that, because you can see that the length of
  • 6:10this side of the lever is 7 times the length of this side
  • 6:13of the lever.
  • 6:14So when you input a force of 7, you output a force of 7
  • 6:18times that.
  • 6:19And of course, in order to move the block 1 meter up in
  • 6:23this direction, you're going to have to
  • 6:24push down for 7 meters.
  • 6:27And that's where we know that the input work is equal to the
  • 6:31output work.
  • 6:32Well anyway hopefully I didn't confuse you and you have a
  • 6:35reasonable sense of how levers work.
  • 6:38In the next couple of videos, I'll introduce you to other
  • 6:41machines, simple machines like a wedge-- I've always had
  • 6:43trouble calling a wedge a machine, but it
  • 6:45is one-- and pulleys.
  • 6:47I'll see you in the next video.