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Find the Normal Force of a box in an elevator cab

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Elevator cabs A and B are connected by a short cable and can be pulled upward or lowered by the cable above cab A. Cab A has mass 1700 kg; cab B has mass 1300 kg. A 12.0 kg box of catnip lies on the floor of cab A. The tension in the cable connecting the cabs is 1.91×10^4 N. What is the magnitude of the normal force on the box from the floor?"

    2. Relevant equations

    F=ma

    3. The attempt at a solution

    T-mBg=mBa
    T = mB(a+g)
    a+g=T/mB

    N=m(a+g) = m (T/mB)= (12.0)(1.91×10^4/1300) = 176 N (rounded value)
     
  2. jcsd
  3. Oct 28, 2009 #2

    Delphi51

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    Why did you drop the "g" in this step? Other than that, it looks good to me.
     
  4. Oct 28, 2009 #3
    I found that a+g=T/mB
    So, I replaced a+g by T/mB
     
  5. Oct 28, 2009 #4

    Delphi51

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    Homework Helper

    You are correct! I bungled the move from Mb to the catnip.
     
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