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Tricky center of mass question

  1. Nov 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Explain how you could measure both your mass and the position of your body’s centre of mass if all you had as experimental equipment was two spring balances (calibrated, of course), and a light, rigid plank, set up as shown in the figure.

    Figure: #1 http://s3.amazonaws.com/docuum/attachments/7166/2011 Final Exam.pdf?1355070977


    2. Relevant equations

    Xcm=ƩMX/ƩM
    Torque=r*F
    F=ma

    3. The attempt at a solution

    I approached it as a torque problem. Since the system is in equilibrium the sum of Torques is 0. The radius of one spring scale is Xcm while the other is L-Xcm. I'm just not sure what the reading of the spring scale represents. I know scales give the normal force, but is this equal to the torque or to Fgrav or what?

    I really want to completely understand the problem so please point me in the right direction, thanks
     
  2. jcsd
  3. Nov 27, 2013 #2
    After reading the problem, I don't think the radius of the scale is part of the problem. I think the scales merely determine the weight at each point. Since you have a light plank, I would assume you would ignore the mass of the plank.
     
  4. Nov 27, 2013 #3
    Oh sorry by radius I was referring to the distance from center of mass to each spring scale
     
  5. Nov 27, 2013 #4

    Doc Al

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    Staff: Mentor

    Good.

    I assume you mean that the center of mass is a distance Xcm from one spring and L-Xcm from the other. Good.

    Those spring scale readings represent the normal forces acting upward on the plank. Identify all the forces acting on the plank (there are three). Then apply the conditions for equilibrium to solve for Xcm and Fgrav, given those normal forces from the springs.
     
  6. Nov 27, 2013 #5
    Are the three forces Fn from each scale and Fgrav from the person onto the plank?
     
  7. Nov 27, 2013 #6

    Doc Al

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    Staff: Mentor

    Yes!
     
  8. Nov 27, 2013 #7
    Does the normal force of one spring scale equal Fgrav-normal force of the other spring?
     
  9. Nov 27, 2013 #8

    Doc Al

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    Staff: Mentor

    Yes.
     
  10. Nov 27, 2013 #9
    Ok so can we equate the torque on one scale as the product of its radius and normal force acting on it?
     
  11. Nov 27, 2013 #10

    Doc Al

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    Staff: Mentor

    Not sure what you mean. You can can compute the torque about some point (the ends of the plank are good points) by multiplying the force in question by its distance to that point.
     
  12. Nov 27, 2013 #11
    i'm still not sure how to approach this. Should I be using torque to solve for the person's mass?

    Isnt the force in question gravity? so torque=d*Fgrav?
     
  13. Nov 27, 2013 #12

    Doc Al

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    You'll use torque equations to solve for Fgrav and Xcm.

    There are three forces. Pick a point--say the left end of the plank. Figure out the torque about that point from all the forces. The clockwise torques must equal the counterclockwise torques in order for there to be equilibrium.
     
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