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Kinematics of Rigid Bodies

  1. Jul 24, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-7-23_22-28-18.png

    2. Relevant equations


    3. The attempt at a solution
    When the barrel starts to tip, the normal at point A should be 0 Newtons, and then all weight would be on point B.
    I'm guessing if the barrel would tip, it wouldn't be sliding across the floor so kinetic friction is not used, instead we should use static friction.

    I take the moment about G:

    [(0.45m)(Ffriction)] + [(0.1m)(Ftension)] - [(0.25m)(90kg ⋅ 9.8 m/s2)] = 0

    Friction Force = Fnormal ⋅ μs = 352.8 N

    Solving for tension of string = 617 N

    Therefore the max weight of mass C is :

    T = mg
    T/g = m
    617N / 9.8 m/s2 = 63 kg

    Is this correct?
     
    Last edited: Jul 24, 2016
  2. jcsd
  3. Jul 24, 2016 #2

    jbriggs444

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    You can ease the calculations by choosing a different axis about which to compute moments. For instance, point B might yield a simpler equation. But no matter. You've solved the more difficult equation for moments about G. Those calculations look correct. But there is a problem...
    You've guessed that the barrel will tip (if it tips) without sliding. You've computed a maximum tension based on that assumption. Knowing that tension, you are now in a position to test whether that guess was correct.

    So... If the barrel is subject to a rightward force of 617 Newtons (per your calculation), will it slip?
     
  4. Jul 24, 2016 #3
    yes, it will slide because I calculated static friction to be only 352 N.
     
  5. Jul 24, 2016 #4

    jbriggs444

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    Good. Now what is the new condition for tipping or not?

    [Note that your choice for the reference axis about which to compute moments has just become helpful -- we need not worry about any angular momentum associated with the sliding motion of the cylinder. Congratulations on the choice]
     
  6. Jul 24, 2016 #5
    there has to be a force that causes rotation of the body about an axis, but i'm not sure how to find the magnitude of force that will stop the sliding and cause a tipping.
     
  7. Jul 24, 2016 #6

    jbriggs444

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    There is nothing that precludes tipping while sliding.
     
  8. Jul 24, 2016 #7
    Then I believe the acceleration caused by weight C has to be much greater than the deceleration caused by kinetic friction to cause a tipping while sliding?
     
  9. Jul 24, 2016 #8

    jbriggs444

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    Yes. That is likely true. But the value of the acceleration does not enter in to the calculations. Putting the reference axis at the center of mass of the sliding object makes that acceleration irrelevant to the question of whether the object tips or does not. Hence my earlier congratulations on having chosen a helpful reference axis.
     
  10. Jul 24, 2016 #9
    since the object is now sliding, kinetic friction is used:

    Fkinetic = 308.7 N

    Moment about G :

    -(308.7N)(0.45m) + (882N)(0.25) - (0.1) T = 0

    T = 815.85 N

    therefore the max weight of C is 83.25 kg

    (I believe the extra 20 kg difference from my first attempt's calculation is enough to cause a tipping while sliding)
     
  11. Jul 24, 2016 #10

    jbriggs444

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    Without carefully checking the arithmetic, I believe that the result is correct.
     
  12. Jul 24, 2016 #11

    haruspex

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    The arithmetic is right. Could have been simplified by leaving g as a symbol which cancels out.
    Quibble: the mass of C is 83.25kg, not the weight.
     
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