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Calculating the mass of an object pivoting on a point

  1. Jan 13, 2013 #1
    1. The problem statement, all variables and given/known data

    A pivot is 0.50m from the end (called A) of a 2.2m uniform beam. A load of 8.0kg at end 'A' will cause the beam to be balanced on the pivot. What is the mass of the beam?

    A-----------------------
    ......Δ

    2. Relevant equations

    τ = d*f

    3. The attempt at a solution

    im close to understanding this but i just dont see how they deduce the mass of the beam.

    from what i understand, there are a couple critical things to realize:
    1) the clockwise torque = counterclockwise torque because its balanced (which means the torque on both sides are equal)

    8kg*9.8=78.4N (this is the counterclockwise force 0.5m away from pivot, thus producing 39.2N*m torque

    so the clockwise torque will be equivalent. the end of the beam to the right is 1.7m from the pivot and i dont know exactly how relevent that is.

    2)additionally, i later realized that the center of gravity will be at 1.1m (halfway due to uniform beam) and im not sure but i suspect there is 9.8N pushing counterclockwise here.

    i just dont see how i take all this information and deduce the mass of the beam. i would really appreciate some help, preferably a step by step solution using words rather than numbers.
     
  2. jcsd
  3. Jan 13, 2013 #2
    OK
    Well you are on the right track.
    On one end labelled A of the beam there is the 8kg mass and the mass of the beam from pivot P to A, of length 0.50m. Both the 8.0kg and mass of beam PA provide a torque.
    From the pivot P to the other end, label it B, there is the mass of the beam of length 1.70m, providing a torque.
    Torques on both sides of the bean must be equal for the beam to balance.
    Does that help?
     
  4. Jan 13, 2013 #3

    TSny

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    Hello agm1984. Welcome to PF!

    You've got the counterclockwise torque correct.

    The key point (that you are close to getting) is that the entire weight of the beam can be considered as acting through the center of gravity of the beam for the purpose of calculating the torque due to the weight of the beam. So, with that in mind, see if you can determine what the weight of the beam must be to provide the clockwise torque that will balance the counterclockwise torque.
     
  5. Jan 13, 2013 #4
    thanks for your reply. im not sure if i understand now or not....

    from your reply i set up the equation (first: 8kg*9.8N=78.4N)

    removing all the units for clarity sake

    this below is essentially Fd + Md = Md (where F is the 8kg, and M is the mass of the beam, and 'd' is distance)
    (78.4)(0.5) + M(0.5) = M(1.7)
    when i solve for M i get 32.666 repeating Newtons, and dividing by 9.8 i get 3.333333 repeating kilograms

    but i still dont understand where im at. but i do notice that we have 3.3kg on both sides. do i just add the two sides together 3.33 + 3.33 to get 6.66 (rounded up to 6.7kg which is the answer in the book?)

    if so is that how you set up the question for a solution?
     
  6. Jan 13, 2013 #5

    haruspex

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    Two things wrong with that. You're treating the parts of the beam each side of the pivot as equal in mass to the whole beam; and you're treating those masses as concentrated at the ends of the beam.
    There is a simpler way. You can treat the whole beam in one go as acting at its mass centre. Logically splitting it into separate masses divided at the pivot will produce the same answer, if done correctly.
     
  7. Jan 13, 2013 #6
    i dont understand what you are saying

    oh ok wait maybe i do

    so there the force of 8kg 0.5m..... and since the center of gravity is at 1.1m can we say that all of the balancing force is at 1.1m away from A? (which is 0.6m to the right of the pivot point)

    so the equation would be like.....
    -> Fd = Fd
    -> (78.4N)(0.5m) = (FN)(0.6m)
    -> 39.2 = 0.6F
    -> F=65.333repeating Newtons

    -> F/9.8=kilograms

    kilograms=6.66repeating (which is the answer when taken to two sig figs)

    AHA i get it now... that is to say i might... if i interpretted that correctly and my answer isnt a cooncidence
     
    Last edited: Jan 14, 2013
  8. Jan 13, 2013 #7
    i'd appreciate if someone could confirm that my last attempt is the logically correct way. i just bolded the portion of my last post that i am most concerned with.
     
  9. Jan 14, 2013 #8

    TSny

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    Yes, it's correct. Good.
     
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