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Problem in the C.G. and torque equilibrium

  1. Nov 2, 2017 #1
    1. The problem statement, all variables and given/known data
    A 40.0 kg uniform platform is suspended by two cables, 2.00 m apart, where the centre of the platform lies midway between the cables. A 5.00 kg paint bucket is placed on the platform, 24.0cm to the right of the right cable, as shown in Figure A2.15. What is the tension in each cable?
    2. Relevant equations
    net torque = 0
    I attached an image showing my work so far.
    I started by choosing the left cable as the axis, so T1 is canceled due to not having a lever.

    3. The attempt at a solution

  2. jcsd
  3. Nov 2, 2017 #2

    Your attempt is right. But please specify where do you find the problem.
  4. Nov 2, 2017 #3
    Thank you for your response.
    I guess I am stuck because I dont know how to calculate the C.G.
    Do T1 and T2 affect C.G?
    What about the weight of the platform and its effect on the C.G calculation?
  5. Nov 2, 2017 #4
    When you try to find the torque of the platform (due to its weight), I think you must consider the center of the platform as C.G. as long as you have already calculated the torque of the bucket (due to its weight).
  6. Nov 2, 2017 #5
    So even though there is more weight (due to the bucket) on one side, the C.G is at the center of the platform?
  7. Nov 2, 2017 #6
    The C.G. of the platform itself (without the bucket) is still in its center. The center of the mass must have changed if you had defined both platform + bucket as one system and calculated the torque as (40+5)*9,8*C.G. Then that C.G. of the system platform + bucket would be different (shifted to the bucket side).

    P.S. Remember to use the negative and positive signs correctly when you use the "net torque = 0" equation.
  8. Nov 2, 2017 #7

    I think I get what you suggest.
    What do you think of this solution. I also added my solution for T1 and I used Newton's law for that.
  9. Nov 2, 2017 #8
    It seems complete.

    You can also add Στ=0 and ΣFy=0 in order to show the basic equations that you use.
  10. Nov 2, 2017 #9
    Thank you very much for your help.
  11. Sep 13, 2018 #10

    I know it is an old post but I am having difficulty understanding why the Normal force of the bucket was not taken into account for calculating the torque of the bucket?
  12. Sep 14, 2018 #11


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    The OP should have drawn a free body diagram.

    A free body diagram for a body (eg the plank) only shows one body and the forces acting on the body.

    Does the normal force of the bucket act on the plank or on the bucket?
  13. Sep 16, 2018 #12
    Oh okay thank you, I get it! Because the normal force does not act on either of them, there is no way that it can have torque. So normal force never has torque right?
  14. Sep 16, 2018 #13


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    No that's not correct.

    The normal force doesn't act on the beam so has no effect on it.

    If you were to draw a freebody diagram of the bucket the normal force would be shown and have an effect on the bucket.

    A normal force can cause a torque on an object in the right situation .
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