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Statics table problem

  1. Aug 27, 2014 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=72501&d=1409176178.jpg

    How close to the edge of the 26.0-kg table shown in the figure (Figure 1) can a 64.0-kg person sit without tipping it over?

    2. Relevant equations

    ∑F = 0 ∑T = 0

    3. The attempt at a solution

    forces A (left leg) and B (right leg) are the forces from the two legs of the table and x is the distance of the person from the right edge of the table (what i am trying to find)

    ∑F = A +B - 26g - 64g = 0
    = A + B = 90g
    = A + B = 882

    for the torques I chose leg A as the pivot point

    ∑T = -(.6)26g + (1.2)B -[(1.7-x)64g] = 0
    = -15.6g + 1.2B - (108.8g-x64g) = 0
    = -15.6g + 1.2B -108.8g + x64g = 0
    = -152.88 +1.2B -1066.24 +627.2x = 0
    = 1.2B + 627.2x = 1219.12

    I have 2 equations but 3 unknowns. What am I missing here? please help :(
     

    Attached Files:

  2. jcsd
  3. Aug 27, 2014 #2

    BvU

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    Choosing A as pivot point indicates to me that you are after a value of x in the interval [1.7, 2.2] . At the point of tipping, what can you say about FB ?
     
  4. Aug 27, 2014 #3
    The point of tipping? When it lifts off the ground? It would become 0 right? because the normal force is no longer pushing back up from the ground? But I thought I was not allowed to set it to 0 because I DONT want the table to tip?
     
  5. Aug 28, 2014 #4

    BvU

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    If you don't like the 0, you set it to 10-20 or something, and then take the limit to zero.
    As long as it's > 0 no tipping.
    How realistic do you want this to be ? The legs don't rest on spike tips, etc, etc.
     
  6. Aug 30, 2014 #5
    I wasnt trying to be difficult. I thought about it more and I understand what you are saying now. Thanks.
     
  7. Aug 31, 2014 #6

    BvU

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    I know. Glad to be able to help.
     
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