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Angular Kinetics

  1. Sep 28, 2011 #1
    A 35N hand and forearm are held at a 45 degree angle to the vertically oriented humerus. The centre of gravity of the forearm and hand is located at a distance of 15cm from the joint centre at the elbow, and the elbow flexor muscle attach at an average distance of 3cm from the joint centre. How much force must be exerted by the forearm flexors to maintain this position?

    Okay this is my attempt at the question:

    Wt= 35N
    dwt=0.15m
    df=0.03m

    fm(sin45) x (0.03m) = (35N) X (0.15m)
    fm (0.02) = 5.25
    fm= 262.5

    Rv= 262.5 sin 45 - 35N
    Rv=185.62 - 35
    Rv=150.62

    Rh= 262.5 Cos 45
    Rh= 185.62

    R= SqRoot (185.62)sqr + (150.62)sqr
    R=239.04 N

    But my answer is wrong can anyone see where I went wrong? My answer should be 192.5
     
  2. jcsd
  3. Sep 28, 2011 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    I'd have said:
    fm(cos45) x (0.03m) = (35N) X (0.15m) x cos45

    I haven't figured out why you calculate R. The question doesn't seem to ask for it.
     
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