# Applying lever theory to foot.

1. Dec 11, 2013

### N Eriksen

So, suppose I want to describe my foot, the ancle joint and my achilles tendon as a lever. I look at my foot as if it was standing on a step of a stair with my heel protruding out into the air, or if i was standing on solid ground, lifting my heels just a millimeter from the ground. Like this:

______L___I
X

Where X is the supporting point, L is my ancle joint and I is the Achilles tendon. I say that the distance from X to L is 2, and the Distance from L to I is 1 (units will not be relevant here).

I suppose that my weight is evenly distributed on both feet. As I am completely stationary, there is no net-torque on my foot.

Now, as I see it, I can describe the above as a first class lever, like a see-saw. L is the fulcrum, and the force on the point X is Fa= ½*bodymass*g. The lever is 2 long. So, the torque on the lever will be τa = bodymass*g on the left side.

On the right side, I have the same torque τab. However, I have a lever only 1 long, so I get the force on the achilles tendon to be Fb = bodymass*g = 2*Fa.

However. I see many pages on the internet looking at the case like this:

The fulcrum is the supporting point, X, the joint, L, is the load point, and the effort is on the point I, the achilles tendon. In this way the system is like a 2nd class lever.

I go on to calculate. The Load on the L is Fa=½*bodymass*g. The lever is 2, so the torque from the load of the body on the joint is τa=bodymass*g. The torque must be zero, so I say, looking only at magnitudes of forces and torques:

τab

2*Fa=3*Fb => Fa=3/2Fb

But HEY! Fa is the same in the two cases.

In the first case I got Fb=2Fa
In the second: Fb=2/3Fa

How can this be? What way should I look at this?

Last edited: Dec 11, 2013
2. Dec 11, 2013

### BeBattey

Your Fa is not the same in both cases. In your case you set Fa to be at x, while in the internet's case their Fa is set at L. Your Fb was the same however, but in your set of equations at the end you're showing the relation of Fb between the X and the L forces.

3. Dec 11, 2013

### N Eriksen

Fa is the same. Its half a bodyweight times g. How could it be anything else?

4. Dec 11, 2013

### BeBattey

Imagine your foot is growing lengthwise between L and X as you're standing there. The force on your foot from X would increase as a function of the distance between L and X. Therein lies the concept that forces are wonky when not applied to the center of mass.

5. Dec 11, 2013

### N Eriksen

I honestly don't see how that makes sense. As the distance from L to X increases, the torque increases. The force stays the same. Otherwise I should be spontaneously gaining mass.

6. Dec 11, 2013

### BeBattey

I should have said Fb, the force you apply with your foot.

I'm not sure how to explain it any better, maybe you should try writing it out as a simple rod attached at certain points with forces applied on it with net torque equal to zero. Don't make any assumptions about what the forces are, just Fa, Fb and Fc.

I think the connection to reality is messing you up. No matter what happens you will fall backwards if you don't shift your center of mass forward, and this negates the assumption that the load is applied vertically. You can't pull yourself up and push yourself down at the same time, your weight and your achilles heel would then be able to cause a net force on you if that were true. Astronauts would be able to float back to the spaceship if they were free-floating in space if that were true.

7. Dec 12, 2013

### N Eriksen

Is someone able to give me a better explanation? Have spent some time on this, and it still does not make sense to me. I really want to understand this fully. My original question was also meant to clarify which approach was correct. I have gotten no further in this. At all.