1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Biomechanics Biceps Problem - Classical Mechanics (Moment Arms/Forces)

  1. Sep 26, 2014 #1
    Please let me know if I did this wrong or right, and if I did it wrong, please correct me :)

    1. The problem statement, all variables and given/known data

    The biceps brachii, a muscle in the arm, connects the radius, a bone in the forearm, to the scapula in the shoulder (see below). The muscle attaches at two places on the scapula but at only one on the radius. To move or hold the arm in place, the biceps muscle balances the weight of the arm and the force at the elbow joint. The centre of mass of the arm is at 15 cm from the elbow joint. The horizontal force of the elbow joint is 6.5 N when the forearm is held parallel to the ground and the forearm weighs 15.3 N. If the biceps supports the entire weight of the forearm, calculate the necessary force from each branch of the biceps to hold the forearm parallel to the ground and the vertical force at the elbow.

    biceps brachii[2].png

    2. Relevant equations

    net F = 0; net Fx = 0, net Fy = 0
    net torque = 0


    3. The attempt at a solution
    Choose E as point of rotation (i.e., Moment Arm of E, MAE = 0)

    MAarm = (15.3 N)(15 cm) = 229.5 N cm

    Determine angles of insertions of A and B:

    Firstly, I assume that, based on the diagram, points E and B (the point where B attaches to the scapula) line up vertically - is this assumption valid?

    Thus angle of insertion of A = thetaA = tan-1(30/2) = 86.19 deg
    Similarly, thetaB = tan-1(30/5) = 80.54 deg

    Torque balance:

    MAarm = MAA + MAB
    229.5 N cm = FA * (5 cm)sin(86.19 deg) + FB * (5 cm)sin(80.54 deg)

    Rearranging for FA,
    FA = [229.5 N cm - FB * (5 cm)sin(80.54 deg)] / (5 cm)sin(86.19 deg) ... Eq. 1

    Note: The "(5 cm)sin(86.19 deg)" and "(5 cm)sin(80.54 deg)" represent, respectively, the lengths of the lines that originate at E and intersect the FA and FB vectors such that the lines form right angles with said vectors.

    Force balance:

    x-direction:
    FEx = FAx + FBx
    6.5 N = FAcos(86.19 deg) + FBcos(80.54 deg)
    FA = [6.5 N - FBcos(80.54 deg)] / cos(86.19 deg) ... Eq. 2

    Substituting Eq. 1 into Eq. 2 yields:

    [229.5 N cm - FB * (5 cm)sin(80.54 deg)] / (5 cm)sin(86.19 deg) = [6.5 N - FBcos(80.54 deg)] / cos(86.19 deg)
    46 N - 0.989FB = 97.82 N - 2.473FB
    1.484FB = 51.82 N
    FB = 34.92 N

    Substituting this result into either of Eqs. 1 or 2 yields FA = 11.47 N

    y-dir'n force balance:
    15.3 N = FAy + FBy + FEy
    FEy = 15.3 N - (11.47 N)sin(86.19 deg) - (34.92 N)sin(80.54 deg)
    FEy = -30.6 N - am I supposed to get a negative answer here?!
     
  2. jcsd
  3. Sep 26, 2014 #2
    Buuuuump! xD
     
  4. Sep 26, 2014 #3
    It's a fun problem for anyone who enjoys classical mechanics. Can anyone afford the time to work through the problem and see if they get the same numbers as me? Thanks!
     
  5. Sep 26, 2014 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That all looks right. To see whether the vertical force should be up or down, consider moments about the point where the muscle attaches to the radius.
     
  6. Sep 26, 2014 #5
    Thanks for responding! Was my use of trigonometry correct, particularly when calculating the moment arms? I neglected the 6 cm width of the forearm, not sure if it was okay to do so. That's where I had a bit of uncertainty. Also, in retrospect, FEy should be negative; the elbow ALWAYS exerts a net downward force in this configuration.
     
  7. Sep 26, 2014 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The 6cm width is not in itself relevant, but your working is not quite accurate because the point of attachment of the muscle to the radius is slightly above the joint. The height difference is unclear - it looks to be less than 3cm.
     
  8. Sep 27, 2014 #7
    Thank you, that is precisely where my uncertainty lies in my attempt to solve the problem. Do you know of a more accurate way to solve the problem than assuming a height differential of, say, 2.5 cm? That is, I would instead use "(5.6 cm)sin(86.19 deg)" and "(5.6 cm)sin(80.54 deg)", respectively, when calculating MAA and MAB, where the "5.6 cm" is obtained via

    c2 = a2 + b2
    c2 = 52 + 2.52
    c = 5.6 cm

    The other uncertainty is the assumption that points B and E are in vertical alignment. What do you think of this assumption? Is it necessary to solve the problem? Valid?

    Many thanks for helping me out!
     
    Last edited: Sep 27, 2014
  9. Sep 27, 2014 #8
    Bump! :p
     
  10. Sep 27, 2014 #9
    By assuming a height differential of 2.5 cm between E and the point of insertion of the biceps into the radius (i.e., a distance of 5.6 cm between E and said point of insertion), I get FA = 3.26 N, FB = 38.24 N, and FEy = 25.67 N [down]. Can anyone verify this? Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Biomechanics Biceps Problem - Classical Mechanics (Moment Arms/Forces)
  1. Biomechanic problem (Replies: 1)

  2. Moment of arm (Replies: 1)

  3. Moment Arms (Replies: 1)

Loading...