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Finding the mass of a beam by static equilibrium

  1. Feb 11, 2010 #1
    1. The problem statement, all variables and given/known data

    This is actually an experiment we did at class. A horizontal beam was loaded with 100 grams weight at the left end. This beam was supported (with my hands) at 30 centimeters from the left end until balance is restored by putting another weight (168 grams) on the other side, which was found out to be 6.4 centimeters from the point of support (POS) at the right side. I need to find the mass of the beam given this information.

    I'll try my best to illustrate this with text

    |---------X----------|-----------|

    wt here POS other wt here
    (100g) (168g)


    2. Relevant equations

    Torque left = Torque right; or that they will sum up to zero

    T=r x F; but since weight is perpendicular then T=rF or T=rmg (since only the weight is acting on the beam); but then I need to account for the weight of the weight added and the mass of the beam itself so: T=r (mass of beam portion + mass of added obj) g

    to find the mass of the beam portion, we assumed the mass was distributed evenly; hence mass beam portion = Mass beam total * r / total length

    3. The attempt at a solution

    I tried to Torque left=Torque right --->

    r left * (mass beam left + mass added left) g = r right * (mass beam right + mass added right) g

    and removed g

    then I substituted the mass beam (left/right) with total mass beam * r / total length

    giving me

    rleft (Mtotal rleft/Ltotal + madded left )=rright (Mtotal rright/Ltotal + madded right)


    Is this the correct way to go about the problem? The answers I get are very far from the measured mass of the beam.
     
  2. jcsd
  3. Feb 11, 2010 #2
    What is the length of the beam?
     
  4. Feb 12, 2010 #3
    I forgot, sorry it's 98.3 cm.
     
  5. Feb 12, 2010 #4

    rl.bhat

    User Avatar
    Homework Helper

    Weight of the bar acts at the center of the beam. Now apply

    Torque left = torque right and find m.
     
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