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Introduction to mechanical advantage

Introduction to simple machines, mechanical advantage and moments. Created by Sal Khan.

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  • leaf green style avatar for user Chris
    Well, how can the object move if the summ of the forces that act on them is equal to 0? That makes no sense, but in all these videos they "move", somehow....what did I miss?
    (49 votes)
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    • blobby green style avatar for user rpmcruz
      Sure. Sal explains as much in the beginning of the video. He is only calculating a baseline force: a force necessary to keep any motion going, but not enough to move the system if resting. You therefor can either assume there was an impulse applied (remember, even the weight of a feather would move suffice), or just assume the object was already in motion.
      (11 votes)
  • blobby green style avatar for user LunaEques
    Why are the triangles around assumed to be right triangles. It seems to me as if the hypotenuse and a leg would both be equal to one, so why do we use the trig functions, as the triangle must be isosoles and does not always have a 45 degree angle.
    (15 votes)
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  • aqualine ultimate style avatar for user Kipras
    Isn't this a geometry mistake? I mean, when the swing moves down, its first position and the final position do not make a right triangle but rather an isosceles one. You could not use trigonometric equations there. Or am I somehow wrong?
    (11 votes)
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    • leaf green style avatar for user janismac
      you are right, for larger values of the angle theta this would not work. but in this method we assume that theta is very (in fact infinitely) small. Thus the error disappears. This method is called "virtual work", idk if sal mentions it. you could do this with the correct geometry and sould get the same result for F.
      (11 votes)
  • blobby green style avatar for user cpt.nemo07
    At and beyond.

    The hypotenuses of the right-angled triangles indicate that length of each side of the swing has increased after rotating through theta. Clearly this is not possible in real life, and so how is it possible to solve this problem using these triangles?

    Also, if you are to trace the path of a point at either end of the swing, isn't it a circular path, and not straight perpendicular lines?
    (12 votes)
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    • leafers tree style avatar for user Brandon Battey
      Strange responses..
      The diagram and explanation imply that it rotates, even though it 'looks' longer.

      Also, yes it is a circular path. The fact that it is circular or not doesn't change the fact that by change in height he means change in Y, not total distance traveled over the curve. We could factor in X here just fine by using some trig, but the ME ratio would again come out to be 2, the force moves twice as much to the right, as the block moves to the left.
      (17 votes)
  • blobby green style avatar for user mycoeco
    I am having a hard time explaining to a friend that mechanical advantage doesn't mean you are doing less work. Just spreading the work (energy used) over a longer distance. In your video above you do not emphasize the law of conservation of energy.
    Do you have a video that emphasizes that.
    (5 votes)
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    • blobby green style avatar for user mycoeco
      OK, I have watched the video. Say I am riding my bike (lots of mechanical advantage) from point A to point B at 10 mph. I walk the same distance at 3 mph. I should be using the same amount of energy plus the energy it takes to pedal the mass of the bike. If there is less friction with the bike I would think it would still take more energy. ?
      (6 votes)
  • leaf orange style avatar for user Arun Nagpal
    So, I see that at , Sal uses Newtons both as a unit of force and a unit of weight. Will this work for other units? e.g. "pounds of force" or "tonnes of force"
    If so, can this theoretically be done with any weight unit?
    (5 votes)
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  • blobby green style avatar for user niko.pasanen
    At How can the lever move up if the object's weight is 10N(downwards) and the force is 10N (upwards) ? Isn't the system then in equilibrium and nothing moves?
    (3 votes)
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  • blobby green style avatar for user S A
    Could one say that we are determining a ratio of work equivalency?
    ex. 10N*1m = 1N*10m
    We arrange Forces and distances to match: Work in = Work Out
    ... setting up newton *meter or joule congruence..
    (4 votes)
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  • leaf green style avatar for user Stanislav Grigorievich Ossovsky
    Guys please help I am lost. Perhaps it is sound silly, but I don't understand why the weight STARTS to move. When it is moving and the work is done it is all clear - the energy and work conservation law. But why does the lever multiply the force when the system is stationary? I mean why does the machine START to move? Sorry I don't know how to explain better...
    (4 votes)
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    • spunky sam blue style avatar for user SuperCipher
      I believe that you're asking why and how the lever multiplies the force even when the system is stationary. It is because of the following principle of levers: for two forces F1 and F2 on opposite sides of the lever, where the distances from each force to the fulcrum of the lever are D1 and D2 respectively, the lever balances when the product of F1 and D1 is equal to the product of F2 and D2. Once the lever is balanced, it is in a stable state, and remains in such a state until it is disturbed.
      (1 vote)
  • hopper jumping style avatar for user Sorgan71
    so if im correct if the object is in the same place no matter of how it got there the work is the same
    (2 votes)
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Video transcript

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