This is really bothering me for a long time, because the math is easy to do, but it's still unintuitive for me.

I understand the "law of the lever" and I can do the math and use the torques, or conservation of energy. or whatever... And I can see that a lever can amplify a force you apply to it if you apply a force on the longer side of the beam.

If I were to look at the molecular lever and see what actually happens when I push on the lever, and I give acceleration to the molecules, how does it actually happen that more force is transmitted to the other side?

Thank you all

p.s I'm looking just for an explanation in terms of forces and acceleration, it's clear to me how to do this in terms of energy or torques

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    $\begingroup$ Here is a mental model that might help. Think of a series of balls connected in a straight line by very rigid springs floating in space with no external forces acting on it. When it is in its 'rest state' there are no stresses on it. Now give the ball at one end of the assembly a push at right angles from the line. When you do that the spring connecting it to the next ball bends a bit - transmitting the force to the next ball as it tries to straighten out the line. From there to the next ball, and so on. $\endgroup$ – Snowhare Mar 29 '12 at 1:26
  • $\begingroup$ There is not "more force" transmitted to the other side... regardless of looking at molecules, it is the torque that causes acceleration, not force. $\endgroup$ – Chris Gerig Mar 29 '12 at 5:55
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    $\begingroup$ @Chris: No. Emphatically no. It is always force which causes acceleration, and there is more force, just acting over a shorter distance (thereby conserving energy ala $W = \int \vec{F} \cdot d\vec{s}$). It is, however, the torque equation which shows you what the coefficient is. $\endgroup$ – dmckee Mar 29 '12 at 15:41
  • $\begingroup$ Ah...I see. You question comes down to "What is the origin of the forces that let the bar (or indeed any solid) maintain it's shape?", which means that @BenjaminFranz's comment is the core of a good answer. $\endgroup$ – dmckee Mar 29 '12 at 17:18
  • $\begingroup$ @BenjaminFranz so it's the electromagnetic forces between molecules that actually generate the extra force? i.e these molecular bonds do not allow the bar to bend and thus create an extra force? $\endgroup$ – fiftyeight Mar 29 '12 at 17:24

I agree with Benjamin Franz that the ball-and-spring model of a solid is helpful and that when a solid exerts a contact force the bonds between the atoms are distorted in that region. If you take a beam, clamp down its ends, and then apply a force to it off-center, the bonds on the short side are distorted more than the bonds on the long side. Therefore, more force is exerted on the clamp that is closer to the applied force. The diagram below illustrates this: enter image description here

  • $\begingroup$ Although the springs from the longer side are less distorted there is a higher quantity of them, so why wouldn't the net distortion equal or be greater than the distortion from the short side? What's the reason? $\endgroup$ – 21Brunoh Nov 5 '14 at 10:59
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    $\begingroup$ This answer does not explain the phenomenon. For an explanation see my answer of the following question $\endgroup$ – rmhleo Jun 6 '16 at 14:05
  • $\begingroup$ I don't think this answer answers the question sufficiently. $\endgroup$ – ja72 Jul 12 '17 at 17:49
  • $\begingroup$ @21Brunoh from the point of view of the mass on the right, the only "springs" it would care about are those attached directly to the "balls" its in contact with. Say that the mass on the right has a mass of 2M, and the mass on the left has a mass of M. The rod is floating in space. I think (I'm almost sure) that at small enough distances, the force between bonded atoms in a rigid body is proportional to the distance between them. We want to apply the force (put the triangle) in such a place that the deformation of the rod (between any two particles) on the right side is twice as much... $\endgroup$ – Joshua Ronis Nov 6 at 3:30
  • $\begingroup$ ....as the deformation of the rod on the left side. That is, we want the to apply the force in such a place that the vertical distance between any two adjacent (horizontally) atoms on the right side increases (for a brief moment, before the entire thing starts accelerating, since it's in space) by twice the amount that the bonds between adjacent atoms on the on the left side increase by, so that twice the force is exerted (vertically) on each particle on the right side than on each particle on the left side. (Note that since the x distances between particles are so small, increasing the... $\endgroup$ – Joshua Ronis Nov 6 at 3:33

There are two fairly straight forward ways to understand this:

  • As a problem in "statics" involving forces and torques on the lever.
  • In terms of conservation on energy between the work done by the person operating the lever and on the load lifted.


We will, for simplicity, consider the situation where the lever is essentially horizontal (showing that the results hold at other angles is left as an exercise), and will treat the lever as a straight bar of length $l = l_1 + l_2$. Three forces act of the bar, the applied force $F_a$ acts downward atdistance 0, the fulcrum force $F_f$ acts upward at distance $l_1$, and the load $F_l$ acts downward at distance l.

Note that so far I have not said anything about the ratio $l_1/l_2$.


We require that $\sum F_i = 0$ and $\sum \tau_i = 0$ (the sum of the forces and the sum of the torques acting on the bar are zero). I'll measure the torques around the fulcrum.

$$ -F_a + F_f - F_l = 0 $$ $$ F_a \cdot l_1 + F_f \cdot 0 -F_l \cdot l_2 = 0 $$

Immediately we can see that the system is underconstrained and we have one free parameter; that the weight of the load, so we'll express $F_a$ and $F_f$ in terms of $F_l$.

From the torque equation we get $F_a = \frac{l_2}{l_1} F_l$, and plugging that into the forces equation we get $F_f = (1 + \frac{l_2}{l_1}) F_l$.

Energy concerns

The best case is that the machine wastes no energy; we assume this case.

While the bar moves through a small angle $\alpha$ near the horizontal the applied force moves through a distance $-\alpha \cdot l_1$, and the loaded end through a distance $\alpha \cdot l_2$, computing the work done my each end we get

$$ W_a = -F_a \alpha l_1 $$ $$ W_l = F_l \alpha l_2 $$

By assumption these must add to zero, so

$$ F_a = \frac{l_2}{l_1} F_l $$

as before.


If the load is on the short end then $l_2 < l_1$ and $\frac{l_2}{l_1} < 1$ and you require less force to lift the load, but the load moves a shorter distance.

If the load is on the long end then $l_2 > l_1$ and $\frac{l_2}{l_1} > 1$ and you require more force to lift the load, but the load moves a longer distance.

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    $\begingroup$ This is a good answer, especially the statics part, but the thing that's mostly bothering me is what actually creates the extra force that raises the object and "amplifies" the force I'm doing, Is it the forces between molecules that keep the shape of the bar that actually amplify the force, what helped me in your answer is putting the fulcrum itself into the picture, because it creates some constraint for the system that got me thinking about the molecules that keep the shape of the bar. $\endgroup$ – fiftyeight Mar 29 '12 at 17:23
  • $\begingroup$ it is the forces between molecules, which are limited only by the breaking strength of the material (which isn't infinite--- you can't move the whole Earth). The transmission of forces conserves the energy, not the force. Force is not a conserved quantity. $\endgroup$ – Ron Maimon Mar 29 '12 at 17:39
  • $\begingroup$ @fiftyeight: Forces need not be conserved. No one needs to create an extra force. Similar things happen in hydraulics--you can amplify a force in one piston by connecting it to a smaller piston. $\endgroup$ – Manishearth Mar 30 '12 at 3:12
  • $\begingroup$ The presence of the earth is significant. @Ron One can move the whole earth by jumping -- just not very much. $\endgroup$ – Peter Morgan Mar 30 '12 at 14:10
  • $\begingroup$ @Peter: The presence of something to push against is significant--that is why you have to include the Fulcrum force for a complete analysis. IN any case, we have both answered the wrong question. $\endgroup$ – dmckee Mar 30 '12 at 15:34

It is all relative to the pivot point in the lever, and to energy expended, not the force applied. If the pivot point is one quarter of the levers length, from the bottom of the lever, and you apply a force F, to the top of lever, to move the top through adistance D, the result will be that the bottom of the lever will move through a third of the distance of the top. (IE 3/4 of length divided by 1/4 of length about pivot point). The energy expended at the top of the lever is FxD. Since energy in, equals energy out, and the bottom of the lever moves only 1/3 of D, then the force that is exerted at the bottom of the lever is 3D. (IE 3 times the force applied at the top of the lever) but it has been exerted over a shorter distance. Hope that this is what you are looking for, and hope I have made it clear. It is 60 years since I was taught this.

  • $\begingroup$ See my answer, lever can be understood either in terms of energy or in terms of forces. $\endgroup$ – dmckee Mar 29 '12 at 17:06

I take dmckee's answer to be flawed because it doesn't mention the earth.

At the coarsest level, the earth accelerates down while the large object accelerates up. At the level of Newtonian mechanics, every action has an equal and opposite reaction.

In more detail, the center of mass of the earth, the fulcrum, the lever, the person who pushes, and the large object, taken together as a single composite system, stays motionless (or, rather, at constant velocity), but the positions and velocities of the five internal components relative to each other are changed by the actions of the contact forces (which we can take ultimately to be non-contact gravitational, electromagnetic and nuclear forces, and an understanding of the constitution of matter does ultimately require QM) that act between them. At this level of modeling, the earth's acceleration (in the model) will be slightly different (and the same as the acceleration of the fulcrum), because part of the person who pushes is also accelerating downwards, and the acceleration of the various parts of the lever would have to be taken into account.

At increasing levels of detail, each of the five components is also composite. I can bend my arm to exert a downward force because I can adjust the internal geometry of my arm relative to another part using chemical energy (which again we can take to be ultimately electromagnetic and nuclear energy, and QM).

Although you tagged this QM, it can be understood moderately well in terms of classical mechanics and EM. The constitution of matter was a concern for late 19C Natural Philosophers, but everything was enough under control that they barely noticed that they were sweeping troubles under the carpet until Planck.

  • $\begingroup$ thank you, the QM mechanics tag is a mistake, I didn't mean to put it, I was mainly interested in the EM and Classical Mechanics parts. $\endgroup$ – fiftyeight Mar 30 '12 at 17:39

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