Equation for force on the knee and arms in this exercise

[Metapress]

There is a body weight exercise I'm doing for the knee where I hold it for time isometrically. This is an ASCII picture:

Edited for clarity
   3
   |
   | B (vertical line from knee to shoulder joint at 3)
   |
 S |
---2
1

#1 is the feet, #2 is at the knees and #3 is where the shoulders are while arms are straight out holding onto the wall around a doorframe, so no extra depth calculations have to be made about how far the arms are ahead of the body. S is the horizontal lines representing the length of the shin and B is the vertical line representing the length of the body from knee to shoulder. W is not shown but is the weight of everything from the knees up. I know I will be able to look up average body segment length and weight info to get exact numbers when I need them.

As a basic lever equation, since the shins are parallel to the floor, the total force that the knees and arms are doing together is F=S*W. Right? I'm sure equations could be done finding how the weight of the shins add in, and the weight of the feet are irrelevant since the ankles are pressing to the ground.

But how do I find the difference between the force endured by the knees compared to the shoulders? The length between them is B and it is a vertical 90 degree angle where the arms are holding on. How would I know how much force is done by which part? I'm sure there are equations that can help me. Any tips? Thank you!

 

[stapel]

This is an ASCII picture:

3
|B
|
S |
--2
1

The forum script strips out extra spaces. To preserve spacing, use "code" tags. For instance:

This is ASCII:
   3
   |B
   |
 S |
---2
1

 

[Metapress]

Thank you Stapel! I revised my original post to reflect your suggestion.

 

[Metapress]

I decided to bump this thread since I did not get any answers so far. Does anyone think I posted this in the wrong forum? I know it is a physics questions and I *think* it goes in calculus, but perhaps I should post the question elsewhere? Or if someone knows the answer that's great too! Thanks.

 

[DrPhil]

There is a body weight exercise I'm doing for the knee where I hold it for time isometrically. This is an ASCII picture:

Edited for clarity
   3
   |
   | B (vertical line from knee to shoulder joint at 3)
   |
 S |
---2
1

#1 is the feet, #2 is at the knees and #3 is where the shoulders are while arms are straight out holding onto the wall around a doorframe, so no extra depth calculations have to be made about how far the arms are ahead of the body. S is the horizontal lines representing the length of the shin and B is the vertical line representing the length of the body from knee to shoulder. W is not shown but is the weight of everything from the knees up. I know I will be able to look up average body segment length and weight info to get exact numbers when I need them.

As a basic lever equation, since the shins are parallel to the floor, the total force that the knees and arms are doing together is F=S*W. Right? I'm sure equations could be done finding how the weight of the shins add in, and the weight of the feet are irrelevant since the ankles are pressing to the ground.

But how do I find the difference between the force endured by the knees compared to the shoulders? The length between them is B and it is a vertical 90 degree angle where the arms are holding on. How would I know how much force is done by which part? I'm sure there are equations that can help me. Any tips? Thank you!

Sorry we did not answer before. The picture is still not completely clear to me. Are the toes on the floor but the knees NOT resting on the floor?

Note that "Weight" is another word for "Force of gravity." The quantity S*W is a Torque acting on the shim bone .. that is, the force of gravity is trying to rotate the shin clockwise, so that the knees drop. If the knees are on the floor, then the force of the knees against the floor is W, and the length S does not matter.

I also don't understand what is happening with the arms and shoulder. As drawn, difference of force between knee and shoulder is just W. If the arms are extended against the door frame, a fraction of W can be supported by the frictional force between the hands and the frame. In principle that could be any fraction of W, which would decrease the load at the knees by an equal amount.

 

[Metapress]

Thank you Dr. Phil for replying to my question!

Sorry we did not answer before. The picture is still not completely clear to me. Are the toes on the floor but the knees NOT resting on the floor?

Yes. The shins are parallel to the floor. The purpose is to put a lot of torque on the knees for a quadriceps/knee strength exercise.

Note that "Weight" is another word for "Force of gravity." The quantity S*W is a Torque acting on the shim bone .. that is, the force of gravity is trying to rotate the shin clockwise, so that the knees drop.

In this case I think S*W is the torque that is borne by the knees, but it does not account for the opposing force from the shoulders at height B that keeps the body vertical from knee to shoulder. So I need to find what force the knees are truly dealing with.

I also don't understand what is happening with the arms and shoulder. As drawn, difference of force between knee and shoulder is just W. If the arms are extended against the door frame, a fraction of W can be supported by the frictional force between the hands and the frame. In principle that could be any fraction of W, which would decrease the load at the knees by an equal amount.

Any frictional forces in this only show up by accident. The intended purpose of the arms and shoulder are only to keep the body upright by pushing straight ahead. There is a lot of torque that would naturally throw the body forward and the arms are only pushing so that the knees stay off the floor so their muscles, the quadriceps, can exert force.

The answer I am looking for is the equation to find the torque borne by the knees as expressed in B, the vertical line from the knee to the shoulder. For example, if the length B changed (a higher hold from the hands would put more or less torque on the knee) I'd like to know what the equation would be to find out how much force the quadriceps are putting at the knee. If B doesn't matter (since it is an equal opposing force to keep the body upright) than I guess the knees' torque would be S*W plus a bit accounting for the weight of the shins as a lever arm. If length B truly matters than the equation might get much more complicated.

I edited this message after posting to clean it up a bit. Does this make more sense?

Thank you again for responding to this question. I really do appreciate it!

 

[DrPhil]

There is a body weight exercise I'm doing for the knee where I hold it for time isometrically. This is an ASCII picture:

Edited for clarity
   3
   |
   | B (vertical line from knee to shoulder joint at 3)
   |
 S |
---2
1

#1 is the feet, #2 is at the knees and #3 is where the shoulders are while arms are straight out holding onto the wall around a doorframe, so no extra depth calculations have to be made about how far the arms are ahead of the body. S is the horizontal lines representing the length of the shin and B is the vertical line representing the length of the body from knee to shoulder. W is not shown but is the weight of everything from the knees up. I know I will be able to look up average body segment length and weight info to get exact numbers when I need them.

As a basic lever equation, since the shins are parallel to the floor, the total force that the knees and arms are doing together is F=S*W. Right? I'm sure equations could be done finding how the weight of the shins add in, and the weight of the feet are irrelevant since the ankles are pressing to the ground.

But how do I find the difference between the force endured by the knees compared to the shoulders? The length between them is B and it is a vertical 90 degree angle where the arms are holding on. How would I know how much force is done by which part? I'm sure there are equations that can help me. Any tips? Thank you!

My old knees hurt just thinking about that pose!!

ok - I think we agree that the torque at the knee is W*S .. but I think W includes the weight of the head - everything but the shins and feet.

At the shoulder, there is a couple with horizontal forces: the wall pushing on the arms pushing on the shoulder, and the floor pushing horizontally on the toes (friction required) pushing on the knee. The lever arm for that torque is B. Mechanically, for static equilibrium, the sum of torques must be zero, so S*W (clockwise) must equal B*F (counterclockwise), and
,,,,,,F = (S/B)*W
is the force of the arms and toes.

 

[Metapress]

My old knees hurt just thinking about that pose!!

Definitely don't do it! I should have prefaced this first that it would be an advanced exercise, perhaps like starting a leg extension machine with an unknown (i.e.: potentially dangerous) amount of weight and if the hip is not immediately over the knee and the torque is not EXACTLY along the hinge of the knee that could also be dangerous.

I should have posted that as a warning in the first part of the thread that I was looking for new and advanced body weight leg exercises when I came up with this idea.

ok - I think we agree that the torque at the knee is W*S .. but I think W includes the weight of the head - everything but the shins and feet.

At the shoulder, there is a couple with horizontal forces: the wall pushing on the arms pushing on the shoulder, and the floor pushing horizontally on the toes (friction required) pushing on the knee. The lever arm for that torque is B. Mechanically, for static equilibrium, the sum of torques must be zero, so S*W (clockwise) must equal B*F (counterclockwise), and
,,,,,,F = (S/B)*W
is the force of the arms and toes.

Yes! This is it. I found a lot of torque equations on the web but what added with static equilibrium is the piece I was missing completely. This is exactly what I was looking for. Thank you Dr. Phil!