Introduction to mechanical advantage | Work and energy | Physics | Khan Academy
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Introduction to mechanical advantage | Work and energy | Physics | Khan Academy


Welcome back. We’ll now use a little bit of
what we’ve learned about work and energy and the conservation
of energy and apply it to simple machines. And we’ll learn a little bit
about mechanical advantage. So I’ve drawn a simple
lever here. And you’ve probably
been exposed to simple levers before. They’re really just kind
of like a seesaw. This place where the
lever pivots. This is called a fulcrum. Just really the pivot point. And you can kind of view this
as either a seesaw or a big plank of wood on top of a
triangle, which essentially is what I’ve drawn. So in this example, I have
the big plank of wood. At one end I have this 10
newton weight, and I’ve written 10 in there. And what we’re going to figure
out is one, how much force– well, we could figure out
a couple of things. How much force do I have
to apply here to just keep this level? Because this weight’s going
to be pushing downwards. So it would naturally
want this whole lever to rotate clockwise. So what I want to figure out is,
how much force do I have to apply to either keep the
lever level or to actually rotate this lever
counterclockwise? And when I rotate the lever counterclockwise, what’s happening? I’m pushing down on this
left-hand side, and I’m lifting this 10 newton block. So let’s do a little thought
experiment and see what happens after I rotate this
lever a little bit. So let’s say, what I’ve drawn
here in mauve, that’s our starting position. And in yellow, I’m going to draw
the finishing position. So the finishing position
is going to look something like this. I’ll try my best to draw it. The finishing position is
something like this. And also, one thing I want to
figure out, that I wanted to write, is let’s say that the
distance, that this distance right here, from where I’m
applying the force to the fulcrum, let’s say that
that distance is 2. And from the fulcrum to the
weight that I’m lifting, that distance is 1. Let’s just say that, just for
the sake of argument. Let’s say it’s 2 meters and 1
meter, although it could be 2 kilometers and 1 kilometer,
we’ll soon see. And what I did is I pressed down
with some force, and I rotated it through
an angle theta. So that’s theta and this
is also theta. So my question to you, and
we’ll have to take out a little bit of our trigonometry
skills, is how much did this object move up? So essentially, what
was this distance? What’s its distance in the
vertical direction? How much did it go up? And also, for what distance did
I have to apply the force downwards here– so that’s this
distance– in order for this weight to move up this
distance over here? So let’s figure out
either one. So this distance is what? Well, we have theta. This is the opposite. This is a 90 degree
angle, because we started off at level. So this is opposite. And this is what? This is the adjacent angle. So what do we have there? Opposite over adjacent. Soh Cah Toa. Opposite over adjacent. Opposite over adjacent. That’s Toa, or tangent. So in this situation, we know
that the tangent of theta is equal to– let’s call
this the distance that we move the weight. soon. So that equals opposite over
adjacent, the distance that we moved the weight over 1. And then if we go on
to this side, we can do the same thing. Tangent is opposite
over adjacent. So let’s call this the distance
of the force. So here the opposite of the
distance of the force and the adjacent is this 2 meters. Because this is the hypotenuse
right here. So we also have the tangent of
theta– now you’re using this triangle– is equal to
the opposite side. The distance of the force
over 2 meters. So this is interesting. They’re both equal to
tangent of theta. We don’t even have to
figure out what the tangent of theta is. We know that this quantity is
equal to this quantity. And we can write it here. We could write the distance of
the force, that’s the distance that we had to push down on
the side of the lever downwards, over 2, is equal to
the distance of the weight. The distance the weight traveled
upwards is equal to the distance, the weight,
divided by 1. Or we could say– this
1 we can ignore. Something divided
by 1 is just 1. Or we could say that the
distance of the force is equal to 2 times the distance
of the weight. And this is interesting, because
now we can apply what we just learned here to figure
out what the force was. And how do I do that? Well, when I’m applying a
force here, over some distance, I’m putting energy
into the system. I’m doing work. Work is just a transfer of
energy into this machine. And when I do that, that
machine is actually transferring that energy
to this block. It’s actually doing work on the
block by lifting it up. So we know the law of
conservation of energy, and we’re assuming that this is a
frictionless system, and that nothing is being lost to
heat or whatever else. So the work in has to be
equal to the work out. And so what’s the work in? Well, it’s the force that I’m
applying downward times the distance of the force. So this is the work in. Force times the distance
of the force. I’m going to switch colors
just to keep things interesting. And that has to be the same
thing as the work out. Well, what’s the work out? It’s the force of the weight
pulling downwards. So we have to– it’s essentially
the lifting force of the lever. It has to counteract the force
of the weight pulling downwards actually. Sorry I mis-said it
a little bit. But this lever is essentially
going to be pushing up on this weight. The weight ends up here. So it pushes up with the
force equal to the weight of the object. So that’s the weight of the
object, which is — I said it’s a 10 newton object — So
it’s equal to 10 newtons. That’s the force. The upward force here. And it does that for
a distance of what? We figured out this object, this
weight, moves up with a distance d sub w. And we know what the distance
of the force is in terms of the distance of w. So we could rewrite this as
force times, substitute here, 2 d w is equal to 10 d w. Divide both sides by 2 you d w
and you get force is equal to 10 d w 2 two d w, which is
equaled to, d w’s cancel out, and you’re just left with 5. So this is interesting. And I think you’ll see where
this is going, and we did it little complicated this time. But hopefully you’ll realize
a general theme. This was a 10 newton weight. And I only had to press down
with 5 newtons in order to lift it up. But at the same time, I pressed
down with 5 newtons, but I had to push down
for twice as long. So my force was half as much,
but my distance that I had to push was twice as much. And here the force is twice as
much but the distance it traveled is half as much. So what essentially just
happened here is, I multiplied my force. And because I multiplied
my force, I essentially lost some distance. But I multiplied my
force, because I inputted a 5 newton force. And I got a 10 newton force out,
although the 10 newton force traveled for
less distance. Because the work was constant. And this is called mechanical
advantage. If I have an input force of 5,
and I get an output force of 10, the mechanical
advantage is 2. So mechanical advantage is equal
to output force over input force, and that should
hopefully make a little intuitive sense to you. And another thing that maybe
you’re starting to realize now, is that proportion of the
mechanical advantage was actually the ratio of this
length to this length. And we figured that out by
taking the tangent and doing these ratios. But in general, it makes sense,
because this force times this distance has
to be equal to this force times this distance. And we know that the distance
this goes up is proportional to the length of from the
fulcrum to the weight. And we know on this side the
distance that you’re pushing down, is proportional to the
length from where you’re applying the weight
to the fulcrum. And now I’ll introduce you
to a concept of moments. In just a moment. So in general, if I have, and
this is really all you have to learn, that last thought
exercise was just to show it to you. If I have a fulcrum here, and
if we call this distance d 1 and we called this
distance d 2. And if I want to apply
an upward force here, let’s call this f 1. And I have a downward force,
f 2, in this machine. f 2 times d 2 is equal
to d 1 times f 1. And this is really all
you need to know. And this just all falls
out of the work in is equal to the work out. Now, this quantity isn’t
exactly the work in. The work in was this force–
sorry, F2– is this force times this distance. But this distance is
proportional to this distance, and that’s what you
need to realize. And this quantity right here is
actually called the moment. In the next video, which I’ll
start very soon because this video is about to end. I’m running out of time. I will use these quantities to
solve a bunch of mechanical advantage problems. See

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55 thoughts on “Introduction to mechanical advantage | Work and energy | Physics | Khan Academy

  1. Mr. Khan, even if you're lifting the object upwards, how can u have a 90 degrees triangle. Because i mean, the lenth of the lever didnt extend or anything, it's the same lever. So, even if the lever turns upwards, it has to move in a circular motion.
    Another way u can look at it is, it's like a clock. 1 m = hour hand 2 m = minute hand,
    it's just turning around like a circle. having 90 degrees thr is actually impossible.

  2. @marcellohro that would be the inductive method…you gather the data, insert it in a graph, and you use simple 9th grade mathematics to solve the problem.

  3. Too complex – Very complicated all round. Use similar triangles instead of all that trig., same angle twice as large. Very complicated all round. Also your argument is circular. You assume the conservation of energy is the same at the conservation of work done and that is what you prove.

  4. I think the instruction is wrong. There are no way that degree is 90 degree. The length of is 1m and 2m and the ending length is still 1m and 2m so there are no way that angle is 90 and therefore it cannot use tan(theta) to calculate.

  5. you messed up on your angles, the hypotenuses would be 1m and 2m. they stay the same length but shift an angle theta

  6. Hello Salman first of all i want to thank you for all your efforts.Salman when you draw this picture andwhen you applied Tangent on both of the triangles you said that both of the angles were equal to theta well i want to ask that if we separate these theta as theta1 and theta2.will they will be also equal to each other after considering them as theta1 and 2 🙂

  7. Guys I suggest you watch trignometry videos before watching this in order to understand how all this is possible and watch some kinematics video in order to develop your logic more.

  8. Hi Sal, a question for me here is that it would seem that 5N would only BALANCE the 10N in equilibrium and even assuming a frictionaless system it would take some extra force to actually move the weight. Furthermore since the system is rotating the "opposite" side would be curved and the weight tipped to the left, though over such a short distance I might agree it doesn't matter very much.

  9. this is what young future scientists really need an intuition not cramming formulas. thanks Mr sal

  10. I think you have a mistake. The hyp is 2m and 1m in the triangles because when you apllyed the force the length of this platue/stick is not changing

  11. Is it true to say that
    The greater the mechanical advantage the more efficient the machine is
    And
    The greater the velocity ratio the less efficient the machine is

  12. This appears to have a reasonably significant flaw. In the diagram, the lever appears to just become longer, which makes no sense. It would make more sense that it goes in a circular path but by similar circles the ratio of the distances is still 2.

  13. wow ! im like reading the comments and everbody is saying stuff like " but dosnt 3=5%50/200=300+50= 3$00(**80)" and im just sitting here like whaaaaa????

  14. My new PHYSICS SOLVING APP.More then 150+ formulas,Solves for any variable you want,Covers up all physics.download now.https://play.google.com/store/apps/details?id=com.physics.lenovo.myapplication

  15. I saw this video, I’m interested in, can we make a more efficient drive train connector, other than a Clutch or a torque converter, that can be instantaneous. This will improve acceleration. When accelerating, there is always delay because torque multiplication is always too slow, so automatics are ok, but the torque converter needs to do a better job at 2 things; speed and multiplying torque

  16. Forget all the formulae that we use for energy/work/moments of force' that explain levers. Now describe to me physically how a larger force can materialise from a smaller force via the medium of the lever? Further , explain to me why there is a dependency on the position of the fulcrum? Is it some dependency/relationship relating to 'elastic restoring forces' between the atoms/molecules of the medium? That somehow the position of the fulcrum relative to the forces applied on either side of it will cause a difference in the elastic restoring forces in the medium (and a mechanical advantage)? If you search the internet for an answer you will get opinions but no-one seems to have a confirmed explanation.

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