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Dynamic application of torque with center of mass

  1. Nov 6, 2005 #1
    ok here is the problem

    <br>A uniform meter stick of mass M has a half-filled can of fruit juice of mass m attached to one end. The meter stick and the can balance at a point 33.0 cm from the end of the stick where the can is attached. When the balanced stick-can system is suspended from a scale, the reading on the scale is 2.70 N.
    <br>(a) Find the mass of the meter stick.
    <br>(b) Find the mass of the can of juice.

    Well I am pretty sure that the side with the can on it... will be of greater mass than the other side...

    And I also know that the sum of the torques should equal zero, right?

    Well i can find the "mass" at the center of mass to be.... m = F/a m=2.70/9.81 = .259 kg

    what I did at first was divide the mass in half and tried to set the meter stick in portions and add the totals assuming the portions were equal in mass around, but somebody told me that was wrong... I am somewhat lost... please help!
     
  2. jcsd
  3. Nov 6, 2005 #2

    lightgrav

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    the sum of torques AROUND THE AXIS OF ROTATION
    (the point of suspension) is zero.
    How far from the suspension point is the center-of-mass of the meter-stick?
    How far is the can?

    By the way, the Force by gravity on a mass is F = m g ...
    this setup has acceleration = 0 , so you'd better NOT divide by a !
     
  4. Nov 6, 2005 #3
    well the center of mass of the meter-stick is 17.0 cm to the left of the AXIS OF ROTATION... and the can is 33.0 cm to the right of the AXIS OF ROTATION... and torque = r * F... so then the sum of torques = 0 = (r1*F1 +r2*F2) right??? so...... 0 = .17*F1 + .33*F2 ..... but i know the Force at the AXIS OF ROTATION is 2.70 N but where would I use that?
     
  5. Nov 6, 2005 #4

    lightgrav

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    one of those x-coordinates has to be negative, usually the one on the left.

    so m1 is twice as heavy as m2.

    you know that F1 + F2 = - 2.7 N
     
  6. Nov 6, 2005 #5
    so assuming F1=(2)F2 then F1=1.8 N and F2=.9N so...... then would i be able to use gravity (a=9.81 m/s2) to figure out the mass or no? if so.. the answer would be... m1 (meter-stick)=.1835 and m2 (can)=.09174 is this right?
     
  7. Nov 6, 2005 #6

    lightgrav

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    You calculated, NOT assumed, that F1 = 33/17 F2 .
    so, it's not exactly a factor of 2 ... so re-do with 33 (not 34) /17.

    (again, gravity is a field with units [N/kg] ; nothing accelerates here!)
     
  8. Nov 6, 2005 #7
    well... F1 = 1.94 (F2) so.... FTotal= 2.70N = [1.94(F2) + F2] = 2.94(F2)
    F2=.918N and F1= 1.782N
    the answer wants it in kg... so how would I transform this into kg if I cannot divide by gravity?
     
  9. Nov 6, 2005 #8

    lightgrav

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    You DO divide by gravity ... dividing by g[N/kg] leaves [kg]

    what you do NOT do is divide by an acceleration (a) ,
    since the acceleration in this case is = 0.
    gravity is "g" , NOT "a" ! Do NOT confuse cause (g) with effect (acceleration)
     
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