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## Physics library

### Course: Physics library > Unit 5

Lesson 3: Mechanical advantage# Mechanical advantage (part 2)

More on mechanical advantage, levers and moments. Created by Sal Khan.

## Want to join the conversation?

- hey guys,just out of curiosity,can we apply this mechanical advantage stuff to multiply the forces in machines like gears and pulleys etc??(13 votes)
- That's exactly what these machines are designed to do:

Make one's life easyer. i.e. you can do things with less force.

But remember: you are paying for this advantage: you have to apply the force over a larger distance, because the work has to stay the same. Look at the video at2:50(15 votes)

- I paused the video and since 7*35/5=7*7=49, can he lift a 49 N weight if he is exerting 7 N of force and is 35 meters away from the fulcrum while the weight will be placed 5 meters from the fulcrum?(5 votes)
- technically, he can't, because it would only be in equilibrium with 49N, but any force above that would be used to lift the weight. However, this still is constrained by conservation of energy, as he must move his side of the lever down about 5 times as much than the distance he lifts the weight.(9 votes)

- What happens if you add to much force?

For example, from4:28on, what would happen to a 42N weight?(4 votes)- The weight would accelerate upwards until gravity brings it back down. It would "jump", as you can try with any lever in real life.(6 votes)

- Is this concept having any relation with Newton's third law of motion?(3 votes)
- Not in the way you are thinking. Yes, when you push down on the lever, the other end goes up, but Newtons 3rd law is stating that when you push down on the lever, you are being pushed up with the same force you apply.(6 votes)

- 4:30anyone else think that the force be drawn perpendicular to the lever and not directly down? It may be an ok as an approximation, but my impression was Moment = (distance * force perpendicular)(3 votes)
- Just wondering...does mechanical advantage affect energy?(1 vote)

- Does anybody know where I can find mechanical energy? I need to know how to determine mechanical energy at a certain point above earth's surface, and it's a different formula than the traditional Wm=Wp+Wk, as it deals with the gravitational field around earth. I thought that this may be where I would find answers, but it's not. Please help.(2 votes)
- How do you incorporate the mass of the lever in the calculation since it too plays a role. My mind tells me that it plays a larger role as the fulcrum is moved further from the center. Thanks(2 votes)
- You would need to know the dimensions of the lever and it's density. Then you could calculate the centre of mass of each half, assume that the entire mass was being acted on through this point and include it in the calculations that way. Of course, if it is a very thick lever it gets even more complex since the movement of the lever would also change the centre of mass of each part of the lever to a greater degree depending on the thickness. For these reasons we often assume the lever to be without mass, or perfectly symmetrical and very thin unless it is a real world problem that needs a more realistic solution.(2 votes)

- Does the basic F*d=F*d equation work for other classes of levers? That is, when the fulcrum is not between the force and the load?(2 votes)
- Okay, i don't know if I am fully qualified to answer this question, but I believe what you are referring is a system similar to a wheelbarrow. I believe it will act in a very similar way. The way to visualize this is by looking at the distance the weight on the wheel barrow is moving up and compare to the distance the handle you are holding on is going up. So a weight of 3N one meter away from the wheel, being lifted .1m and handles that are 3m long would require an upwards force greater than 1N to lift it, and 1N to maintain equilibrium to a height of .3m. You increased the distance the force had to travel by 3 times so the force is cut by one third and the amount of work is conserved. The key is F*D=F*D work in = work out. Someone please corroborate this or correct me if I am wrong.(1 vote)

- In general ho w would you find the mechanical advantage if you knew the weight of the load, and the lengths from both ends to the fulcrum?(2 votes)
- At3:14Sal says that energy cannot be made out of thin air.

But what about if the machine is 101% efficient , say the finsurd's perpectum machine.

Perpetual motion?(1 vote)- Energy cannot be created out of thin air, so you know that any machine that claims to be 101% efficient is a fraud or a trick.(3 votes)

## Video transcript

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