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Just trying not to look like an idiot to my wife

  1. Aug 24, 2015 #1
    0b0dc5bf-4853-43d4-8c76-8ee23604a37b.jpg
    I have never seen anything like this.
    My assumption is that the weight value times the weight position would make up the applied downward force on the scale. And the I plug in the non attached to the scale to balance left and right. But is there more to it?

    1. The problem statement, all variables and given/known data



    2. Relevant equations
    I do not know


    3. The attempt at a solution

    I am guessing that you want both sides to balance out, and that by having the 4lb weight hanging off the the 5th point that becomes a value of 20, then on the other side there is a 9lb weight hanging off the 5th point for 45
     
    Last edited by a moderator: Aug 24, 2015
  2. jcsd
  3. Aug 24, 2015 #2

    Bystander

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    Given those assumptions, and that there's a uniform distance between points, that gives you the torques (leverages) on each arm. What do you want to do next?
     
  4. Aug 24, 2015 #3
    hey, just noticed the title -- hahahah----good luck with that....:(
     
  5. Aug 24, 2015 #4

    HallsofIvy

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    The weight times the distance from the pivot point, your "weight value times the weight position" is the "torque" or angular force. Balancing so that the "weight value times weight position" is the same on both sides means that it does not rotate about the pivot. However, the weight on that pivot point is simply the total weight without regard to distance.
     
  6. Aug 24, 2015 #5

    mfb

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    There is exactly one solution that does not require hanging a weight below other weights while balancing both scales. It also happens to use all the weights which is probably not a coincidence.
     
  7. Aug 24, 2015 #6
    That is a very clever problem ! Got it.
    - Assumed that the points without a dangling line were not available for use.
    - And the six weights between the two pictures are what we are supposed to hang...
    --- on the three dangling lines on the first picture
    --- and the three dangling lines on the second picture.

    Looks like the poster needs to make a bit more of an attempt at solving this problem before we can help though...
     
  8. Aug 24, 2015 #7

    mfb

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    With that assumption I don't see a solution and I think I considered all options.

    Edit: Oops, calculation error. Okay, I can confirm your answer.
     
  9. Aug 24, 2015 #8
    Sent you the answer by PM. Never mind, looks like you got it.

    Edit: Don't want to give the answer away to the student.
     
  10. Aug 27, 2015 #9
    Thank you, for all your input, and direction, makes much more sense now. I was trying to hang two more on the double dot, instead of ignoring it. I have solution now so my wife can go find her geocache.
     
  11. Aug 27, 2015 #10
    A posted answer would be nice...
    Solution:
    Weights available
    1:2:4:4:6:9

    First scale:
    4*5=20... 9*5=45
    5*4=20... 5*4=20
    4*3=12... 3*3=9
    9*2=18... Xx2=0
    6*1=6... 2*1=2
    76...... 76..... 0

    Second scale
    .................. 2*4=8
    4*3=12... 1*3=3
    8*2=16... 4*2=8
    x*1= 0... 9*1=9
    28 .......28........ 0
     
  12. Aug 27, 2015 #11

    mfb

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    There is a second solution if we don't have to use all weights.

    First scale:
    4*5=20... 9*5=45
    5*4=20... 5*4=20
    9*3=18... 3*3=9
    6*2=18... Xx2=0
    0*1=0... 2*1=2
    76...... 76..... 0

    Weight 4 stays unused.
     
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