1. Limited time only! Sign up for a free 30min personal 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!

Torque in bicep, please help (Urgent)

  1. Aug 16, 2012 #1
    1. The problem statement, all variables and given/known data

    a man holds a 5.00kg weight in his hand. His arm is represented here by a pair of thin rods attached with a hinge, with a cord for the bicep muscle. The forearm makes a 20 degree angle with the horizontal, and the bicep (which is attached 4.0 cm away from the elbow joint) makes a 55 degree angle from the horizontal. assumme that the forearm has a mass of 1.2 Kg and is 36cm long, and has its centre of mass midway through its length. find the tension in the bicep muscle and the components of force in the elbow joint!

    DIAGRAM:
    http://farm9.staticflickr.com/8445/7797836656_015d170c93.jpg

    2. Relevant equations
    Torque =F*d


    3. The attempt at a solution

    Okay so ill show you two ways i've don't it because I'm confused about how its actually done

    1st way it think it may be wrong

    Sin(55)(4)Ft=(sin70*9.8*1.2)(18)+(sin70*9.8*5*36)

    solveing for Ft = 556N

    Attempt two

    sin(55+20)*4(ft)=(sin70*9.8*1.2)(18)+(sin70*9.8*5*36)

    Ft= 480N

    and the vertical component =

    9.8*1.2+5*9.8-480*sin55=-332N or 332N to the vertical

    horizontal component = sin55*480=393N

    Am i going wrong anywhere, which method is correct PLEASE HELP!
     
  2. jcsd
  3. Aug 17, 2012 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    In your first way, you are incorrectly calculating the torque that the bicep produces about the elbow joint.

    The second way looks fine for torque.

    Here's the graphic for anyone viewing this:
    7797836656_015d170c93.jpg
     
  4. Aug 17, 2012 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The torque of the biceps on the forearm about the elbow does not depend on the position of the forearm. The torque is produced by the component of force which is perpendicular to the arm, and that is Ftsin(55°), the torque is τ=4Ftsin(55).
    The first way is correct.

    ehild
     
  5. Aug 17, 2012 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The bicep makes an angle of (55° + 20°) to fore arm, so the component of the force perpendicular to the fore arm is Ttsin(75°) .

    SammyS
     
  6. Aug 17, 2012 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Is that 55°angle measured from the forearm as in the figure or from the horizontal as written in the text?

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Torque in bicep, please help (Urgent)
  1. Urgent please Help (Replies: 5)

Loading...