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Static equillibrum problem (FBD)

  1. Apr 14, 2013 #1
    1. The problem statement, all variables and given/known data

    The forearm weighs 20.0 N and has a center of gravity in the position shown in the diagram. The person is pushing DOWN on the weight scale shown in the picture. If the reading on the scale is 70 N, calculate the tension, M, in the triceps muscle.

    I understand how this problem works. My question is with the FBD, and which way the force vectors are supposed to go. My intuition (like in my attachment pic) is that the hand pressing on the scale would be going down (counterclockwise negative direction). But there should also be the reaction force going up? I've talked to people and it seems the correct answer should be with the arrow going up in order to get a positive answer for M.

    2. Relevant equations

    M=0

    3. The attempt at a solution

    What makes sense to me:

    -70(0.4)-20(.15)-M(.025)=0 So, M=-1240

    What I think the answer is:

    70(0.4)-20(.15)-M(.025)=0 So, M=+1000

    So why wont there be a down force from the hand pushing down?
     

    Attached Files:

  2. jcsd
  3. Apr 14, 2013 #2

    tiny-tim

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    Homework Helper

    hi jklops686!! :smile:
    a free body diagram must show all the forces on a body

    in this case, the body is the forearm

    the reaction force from the scale is a force on the forearm, so it goes on the free body diagram :wink:
     
  4. Apr 14, 2013 #3
    brilliant! That's definitely good to remember. Thank you. I was thinking the hand pushing would act as a load force.
     
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