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Equilibrium non-uniform Box

  1. Apr 25, 2017 #1
    1. The problem statement, all variables and given/known data
    A non-uniform box sits on two scales as seen below. Scale A reads 10.2N, and scale B reads 19.4N.

    Question: How far from the left end of the box is the center of mass located?

    2. Relevant equations
    τccw=τcw
    τ=f*d(from pivot point)

    3. The attempt at a solution
    If I choose the pivot point to be the left end side of the box I get
    10.2N*d1=19.4N*d2
    I know that at the pivot point the box should be in equilibrium but how am I suppose to solve the equation with 2 unknowns? Do I choose a different pivot point, or am I using the wrong formula?
     

    Attached Files:

  2. jcsd
  3. Apr 25, 2017 #2

    kuruman

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    You need to define x as the unknown distance of the CM from the left end of the box. There are three forces generating torques in this problem, gravity, scale A and scale B.
     
  4. Apr 25, 2017 #3

    gneill

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    Can you say what the total weight of the box is?
     
  5. Apr 25, 2017 #4
    The total weight would be (10.2N+19.4N)/g=3.02kg, right?
     
  6. Apr 25, 2017 #5
    Hmm, would the mg of the box be CW or CCW?
     
  7. Apr 25, 2017 #6

    gneill

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    That would be the total mass. Just sum the forces for the total weight in Newtons. So w = 10.2N+19.4N.
    It's neither until you pick a center of rotation :smile:

    Draw your a FBD with mg located at some (unknown) distance x from your chosen center of rotation.
     
  8. Apr 25, 2017 #7
    Ok, so lets says the center of rotation is the left end point.

    10.2N(0.02m)+19.4N(0.3m-0.074m)=(10.2N+19.4N)x
    10.2N(0.02m)+19.4N(0.226m)=29.6N*x
    x=0.155m

    is that right?
     
  9. Apr 25, 2017 #8

    kuruman

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    That is right.
     
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