How does a lever arm increase force?

A lever and a fulcrum is able to to multiply the input force to the output force and trade moving distance for increased force. The same happens in automobile transmissions, taking advantage of high motor RPMs.

What exactly is the physical mechanism behind this? I am very aware you can mathematically prove it, for instance via an energy approach where energy in roughly equals energy out. Still, is it possible to explain the phenomena down to atomic level? That is, how is it in qualitative terms, that the force on the other side of the lever is amplified?

• What's wrong with a geometrical explanation? Why should atoms be relevant? May 15 at 21:02
• @PM2Ring I wonder if there exists a more qualitative explanation, something out of geometry, energy laws and such.
– Erik
May 15 at 21:04
• The book by Spivak, ELEMENTARY MECHANICS FROM A MATHEMATICIAN'S VIEWPOINT is possibly what you want, if you're willing to spend a lot of time. Lecture 5 is all about rigid bodies: the telling quote is "...our analysis shows, especially when we think of the lever as bending slightly, that it is the internal forces of the lever that make the weights balance; in short, all the "extra force" that one obtains by pushing at a large distance from the fulcrum is supplied by the lever itself, in its effort to preserve rigidity..." May 16 at 8:12
• I personally find that thinking about pulleys or bike gears is better for making sense of this intuitively. Bike gears are kind of working the opposite way, though. They typically have a mechanical advantage that is less that one. May 16 at 17:26
• Simply put, the force is increased by reducing the distance travelled. May 16 at 21:02

The "extra" force is generated by the fulcrum. Consider the following setup:

We load a seesaw with a weight of 1 and 3. The fulcrum has to supply a force of 4 in the opposite direction, otherwise the net force is not zero and the seesaw would accelerate through the fulcrum. In the setup above, the seesaw is perfectly balanced. We know this because the net torque around the fulcrum is zero: $$\tau=\sum_ir_i\times F_i$$. It's as if the lever multiplies the force, but the force is supplied by the fulcrum. Due to the fact that torque scales with distance, we have that in equilibrium the largest force is the closest to the point of rotation.

You may still not be convinced, because the seesaw is not doing useful work. Consider the same bar of 4 units long, but now we want to pry open a crate using our lever. At one end we are pushing down and the other two ends are stuck. Consider the scenario where you are pushing down, but the crate/lever is not moving yet. Once again, the total force and total torque must be zero. So the force in the middle must be larger than the rightmost force to ensure a net force of zero. You can add an arbitrarily large force couple to the forces that touch the crate and still have a net force of zero, so this leaves one degree of freedom. We can use the fact that the torque must be zero again to show that the center/right forces are larger than the input force.

So why can you create large forces with a lever? I would say two things:

1. When two objects are pressed together, they create a normal force at the surface. This normal force can arbitrarily large, provided the objects don't break.
2. Torque scales with lever arm.

• I have the feeling that OP really means what you mention in a half sentence Due to the fact that torque scales with distance and Torque scales with lever arm. - why (on a physical/atomic, "practical" level) is this the case, what mechanism (on the atomic scale) creates this phenomenon? (And no, I have no idea how to answer this in a different way than you did, but still...)
– AnoE
May 16 at 7:14
• @AnoE Exactly! I study engineering and have done moment equations equal to zero for a long time, but I still wonder - where the physical nature of this comes into play. If you somehow could study the microscopic stress carried through the lever, why does it feel a need to increase to main the equilibrium? Energy and such obviously tell that it "must" be this way, but still, qualitatively, why does it happen?
– Erik
May 16 at 8:08
• @Erik I don't quite get what you mean here. It happens because when you press on the end of the lever, the only other alternative(assuming the bar doesn't break or bend) is that it passes through the fulcrum, and that would be ridiculous. May 16 at 12:39
• @Erik I think you may be asking about the nature of the normal force, which always pushes back exactly as hard as the input force. It's usually modeled as the van der Waals force, which becomes very large as distances become very small. When you push atoms together, they push back, the alternative is matter passing through other matter or something breaking. May 16 at 13:00
• Why does the lever distribute the force from the fulcrum to it's ends unevenly? I think this is the core of the question. May 16 at 14:38

You can think through the mechanics of this phenomenon with a bendy lever.

Suppose you press down at one end that is 3 m from the pivot and the other end is 1 m from the pivot in the other direction. The first thing that happens is that the point you press goes down, which pulls the part next to it down too. By how much? Let's view the lever as having 4 equal parts, 3 on your side. Your pressing causes the 1st part to bend, which causes the 2nd part to bend, and so the 3rd part bends too, and of course finally the 4th part. If the lever's pivot and opposite end is fixed, this will reach an equilibrium in which they all bend the same amount, since any part that is more bent will cause its neighbours to bend more instead.

Here is a highly exaggerated diagram of the bendable lever in an equilibrium state:

Now every part is the same, and each bends by an amount roughly proportional to the bending stress on it, which is just how strongly one end is twisted relative to the other. At equilibrium, the bending stress is identical in every part. Thus the end of the lever that you press down moves 3 times as much as the other end of the lever. The force exerted at a point by a part is proportional to the difficulty in moving that point, so the end of the lever that you press down is 3 times easier to move than the other end.

Therefore the force you exert on your end is 1/3 of the force that the other end exerts.

Note that you can also figure out the force at the pivot relative to your end by exactly the same reasoning, simply treating the other end as the immovable pivot instead, and so the force you exert on your end is 1/4 of the force exerted at the pivot.

• I believe I understand the intuition of this answer, but a diagram would help immensely (even a crude mspaint diagram). Also, I think a discussion the limit as the stiffness of the lever aproaches infinity would make clear how the answer relates to a rigid lever. May 16 at 13:41
• @Vaelus: Feel free to put in a diagram. I am quite busy so I didn't make one... And I'm not sure that we should talk about rigid levers as the consequence of taking limits, since that's never what happens in reality. Think about Newton's cradle with 5 balls and how you would explain the result of releasing 1 from the left and 2 from the right at the same time. Rigid mechanics cannot pin down the result, and elasticity is crucial. May 16 at 14:13
• This made a lot of sense to me so I made a diagram based on what the other answer posted. I tried to show that the weight of $1$ makes the first segment of the long bar bend a little bit, that first segment makes the second segment bend the same bit, and that makes the third segment bend the same bit. By contrast, the weight of $3$ makes its single segment bend the same angle as the other three combined. Solid bars don't want to bend that much because of atomic bonds so it ends up being more force on that end. i.stack.imgur.com/FmH1K.png or i.stack.imgur.com/pXBgW.png May 16 at 17:12
• The first image has the left-hand segments each rotated at 2° and the right-hand segment at 6°. The second image uses 3° and 9° for a more bent look. May 16 at 17:15
• @EngineerToast: Your diagram is incorrect. The explanation I gave shows that the bending is uniform across the lever (for small weights). In your diagram, the lever would immediately flex itself to balance out the excessive bending stress that you put at the pivot. The equal bending becomes untrue only if the forces on the lever do more than bend it (i.e. if it gets overall stretched or compressed too). I have decided to include a diagram in my post to illustrate the correct picture clearly. May 17 at 7:28

You can look at this from the perspective of energy conservation.

The force applied to the long side of the lever is doing work, and because of energy conservation the short side of the lever is doing the same amount of work on its load.

$$W_{\text{in}} = F_{\text{in}} \cdot X = W_{\text{out}} = F_{\text{out}} \cdot Y$$

The distance that the long and short sides can move is constrained by the mechanics of the system. Therefore, the output force is the only variable that can account for the necessity of energy conservation.

In other words, X > Y requires that F(out) > F(in). As we know, the force multiplier is proportional to the ratio of the lengths of the levers.

$$F_{\text{out}} = \frac{X}{Y} \cdot F_{\text{in}}$$

Essentially, energy conserved everywhere in the system. At the contact points on the lever, and everywhere in the lever. Applying force to one side of the lever transmits force through the atomic bonds in the lever, and out the other side. From the classical perspective, it's Newton's Third Law in action.

As a counter example, if the lever was elastic, then the lever could flex and the simple proof above would not be true. Energy would be absorbed by the lever and the fulcrum and converted to heat or deformation.

So, it's the integrity of the rigid lever that requires this to be true, otherwise the lever would not be a rigid lever at all.

• The last paragraph is true but irrelevant to the actual question here, because the question is actually about the real lever, and a real lever must be flexible in order to transmit the forces. A 100% rigid lever will simply break when a force is applied on one end! See my answer for why the lever mechanics hold for a flexible lever, and why the ratio of forces is approximately as expected. May 17 at 7:36