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Mathematical solution needed for this biomechanics problem

  1. Feb 15, 2015 #1
    Hi all,

    Wasn't sure whether this fits best in the maths or engineering forums, either way I'm hoping that a mathematical solution exists for my problem which I'll now explain.

    My problem is one of biomechanics. One of the primary exercises I use is the barbell squat. Holding a loaded bar on your upper back and squatting down means that the skeleton of your body moves, the joints move and the bones and muscles transmit force through them down to the floor. I am looking for a mathematical model which can take the various measurements of the body and calculate the angles the joints are making with each other, the forces that are being transmitted through each member and the moments that exists around the joints.

    Here is an image showing the position of the skeleton in the squat.

    okctuq.jpg

    The load is positioned on the upper back, and L1 represents the length from the load to the hip joint. L2 is the length of the upper leg. The lower leg of length L3 pivots at the ankle and L4 represents the height of the heel on the shoe, which serves to lift the heel of the foot off the floor slightly. L5 is the length of the foot and the load must always remain on the centreline of the foot, as represented by the red centreline. The load always moves vertically on this centreline. The only inputs to the calculation are the lengths L1 to L5, and the size of the load in kilograms.

    A1 is the angle formed between the spine and the upper leg at the hip. A2 is the angle formed between the upper leg and the lower leg at the knee. A3 is the angle formed between the lower leg and the foot at the ankle. What are these angles for any lengths of L1 to L5?

    The load is stationary in this position, so the system must be in equilibrium. The load must be being transmitted to the ground through the bones. Given the size of the load, the various lengths L1 to L5, and the angles they are creating, what are the forces in each segment and what are the moments being created around each joint?


    Im looking for a set of equations that can describe this model which I can plug in various lengths of L1 to L5 and a size of load variable.

    Many thanks and looking forward to any responses!

    Dan
     
  2. jcsd
  3. Feb 15, 2015 #2
    Do you know any trigonometry? Do you know any engineering mechanics?
     
  4. Feb 15, 2015 #3
    You can solve this problem by steps :

    1. We isolete the spine and we apply the three dynamics equations on it to calculate the forces in A1 as the torque.
    2. Do the same for L2, L3 and L5.
     
  5. Feb 16, 2015 #4
    Hi thanks for the replies.

    The forum mods asked me to mention that this isn't a homework problem - im not a student. Its just something Im trying to figure out for myself.

    Regarding attempts at solving it. I know basic trigonomentry and how to use sine, cosine & tangent. But they aren't right angled triangles so I didn't think these functions worked.

    For example I tried sin(A1)=O/H. But I can't work out O, because the triangle formed by L1 and L2 (and a hypothetical third side, O) isn't a right angled triangle.


    And felmon, thanks, its been so long since Ive done any engineering study I don't know what the three dynamics equations are. And similar to the above, I struggled to break the load down into its vectors because they aren't right angled triangles.
     
    Last edited: Feb 16, 2015
  6. Feb 16, 2015 #5

    OldEngr63

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    Gold Member

    Why do you say this? This seems to imply that the muscles are all slack when the system is under load, and I just don't believe that is true.
     
  7. Feb 16, 2015 #6
    Sorry danlight..., to solve this problem you must review the Newton Mechanic you studied
    Regards
     
  8. Feb 16, 2015 #7
    A good start could be to review the law of sines and cosines. Your drawing shows a side view of a person. I am seeing this as a 3d figure compressed into the y-z plane with the thickness of the person and the barbell projecting into and out of the paper. Is the load perfectly balanced across the person? the components of L1 to L5 can be computed using the law of cosines since you are dealing with triangular shapes that are not right triangles. Once you know the magnitudes and the angles then you can form unit vectors who's length will be 1. If you take the vector representation of L1 as a unit vector for example then multiply it by a new length of the line represented by L1 will give you a new vector with a unique magnitude and same direction.
    The torque can be computed by choosing a axis of rotation and taking the magnitude of the cross product of the two two vectors. depending on the direction of rotation the result will be pointing into or out of the paper
    I hope this helps it has been a while worked mechanics problems
     
  9. Feb 16, 2015 #8
    I was simplifying. The muscles are under tension and the bones under compression in a static position with the load stationary. Forces in the muscles equal and opposite to the forces in the bones, and around the joints, there are moments which are equal and opposite also, again because the load is stationary. I was correct in saying the load is being transmitted through the bones to the ground though, it can only be that way. In this static system, the muscles serve to cancel out the moments around the joints (otherwise the load would move), and the bones act in compression to transmit the force to the ground.

    However what Im interested in is working out what these forces are. Felmon, I'm not capable of working this out for myself, its a complex problem, which is why Im posting here in the first place. Im afraid I need more than just a hint.


    I just looked up the law of cosines. Unfortunately you need to know 2 sides and an angle to be able to compute other sides or angles. In this case we only know 2 sides (L1 and L2), not any of the angles between them. Unless Im misunderstanding this, I'm not sure if it can be used to solve the problem?
     
    Last edited: Feb 16, 2015
  10. Feb 16, 2015 #9
    Maybe this could help you too. would it help if someone setup one of the equations for you with some insight on how they accomplished this?
     
  11. Feb 16, 2015 #10
    Yes it would because I really don't think I can solve it myself.
     
  12. Feb 16, 2015 #11
    I'll do the easy one first:

    M@A3= P×L5×cos(A5)
    where
    P = Load
    A5 = Arcsin(L4/L5)

    assuming we can ignore the weight of the lifter
     
    Last edited: Feb 16, 2015
  13. Feb 17, 2015 #12
    You've picked the only right angled triangle there haven't you lol.

    So you're working out what A5 is (a new label representing the angle between the toe end of the foot and the ground), by taking the inverse sine of opposite (L4) / hypotenuse (L5), which only works for this bit because its a right angled triangle.

    Then to calculate the moment you're taking the load, P, multiplying it by the length L5 and the cosine of A5 which is to somehow get the right vector is that correct?

    Using some real life numbers.

    P=100kg
    L5=0.2m
    L4=0.02m

    M@A3=19.9 kg.m

    So ok, not sure what that number really means in this context. Also I still don't know how to apply this to the other triangles which are not right angled triangles.

    And ive spotted a misleading thing in the picture I posted as well. The foot is actually completely in contact with the ground through the shoe, its not raised off the ground like it looks there. The weightlifting shoe is shaped like a wedge, higher at the heel than the toe end, but the sole is completely in contact with the ground. The length L4 represents the height of the heel on the shoe, which during the lift movement is something which effects the knee position.
     
    Last edited: Feb 17, 2015
  14. Feb 17, 2015 #13
    Maybe you could convert them into right angled triangles.
     
  15. Feb 17, 2015 #14
    You mean by subdividing them? I did think of that, but didn't know where to start with it.
     
  16. Feb 17, 2015 #15
    Well to start with L3, the angle A3 - A5 makes a right angled triangle, dinnit?
     
  17. Feb 18, 2015 #16
    Im not sure what you mean sorry. Are you saying A3 minus A5 there? Or A3 to A5? A5 is the angle you labelled, I assumed it was the angle between the floor and the foot (L5) is that right? I don't see how anything there can be a right angled triangle.
     
  18. Feb 18, 2015 #17
    Yes, I meant A3 minus A5.

    It might help to communicate better if you draw a horizontal line through each joint on your sketch and label the angle each member makes with this horizontal line.
     
  19. Feb 18, 2015 #18
    Like this:

    1zyzhpc.jpg

    Ive labeled the angles between the horizontal lines as B1, B2 and B3.

    Still dont understand what A3-A5 means.
     
  20. Feb 18, 2015 #19
    Yes, that looks good. Hopefully you can see all the right angled triangles now.

    Also, it should be clear that B3 = A5, yes?
     
  21. Feb 18, 2015 #20
    I can see that B3 = A5 yes, but how does that help figure out A3?

    Im still not really sure on the right angled triangles. Are you saying that one of them might be if a line is dropped from the knee joint vertically down to the horizontal line above the foot? In that case L3 is the hypotenuse (side C in pythag), but the length of the hypothetical sides A and B is still unknown, and the angles A3 and A2 are still unknown so I don't see how that helps things.
     
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