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Henry Sincosky

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data

    In about 1915, Henry Sincosky of Philadelphia suspended himself from a rafter by gripping the rafter with the thumb of each hand on one side and the fingers on the other side (Fig. 6-22). Sincosky's mass was 82.0 kg. If the coefficient of static friction between hand and rafter was 0.700, what was the least magnitude of the normal force on the rafter from each thumb or opposite fingers?

    http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c06/pict_6_22.gif

    2. Relevant equations

    Fnet = Ma
    Fs,max = UsFn

    Us = Coefficient of static friction

    3. The attempt at a solution

    With Normal force and Gravitational force pointing down and Static Friction pointing up on the free-body diagram, I for Fs - Fn - Fg = 0 [Zero acceleration]. Then I put Fg = Fs - Fn and substituted Fs with UsFn. Got the answer 2680 [rounded to 3 sig figs] but was the incorrect answer.

    I have a feeling that it's my free body diagram that's wrong.
     
    Last edited: Mar 15, 2009
  2. jcsd
  3. Mar 15, 2009 #2
    i think that for the fingers of the hand, your normal force would be pointing up from the board and your frictional force to the side.. because friction always opposes motion, and if the fingers were to slide off the plank they would move sideways.. thats if i understand the situation correctly...
     
  4. Mar 15, 2009 #3
    Maybe this picture will help:

    http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c06/pict_6_22.gif
     
  5. Mar 15, 2009 #4
    oh ok.. ya i was thinking about it wrong... so for this pic i think that the frictional force points up, the normal force to the side and gravity down...
     
  6. Mar 15, 2009 #5
    I was wondering, why normal force to the side?
     
  7. Mar 15, 2009 #6
    well normal force is always perpendicular to the surface of contact, so from the board to his fingers...
     
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