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Gravitational torque

  1. Mar 11, 2014 #1
    1. The problem statement, all variables and given/known data

    f2k7ep.png

    2. Relevant equations

    torque = r*Fsin(x)

    3. The attempt at a solution
    2yo7nrq.jpg

    I really need help on walking through this problem...didn't really teach it in class...I tried watching a video and you're supposed to sum the forces * some perpendicular distance to the pivot? Yet what if the force is already perpendicular?
     
  2. jcsd
  3. Mar 12, 2014 #2

    CWatters

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    Check your force diagram for missing and incorrect forces acting on the beam.

    T2 is the tension force acting on the mass m (eg not acting on the end of the beam).

    Your equation has force T1 acting at what distance from the pivot? Lets say the beam has length L so d = Lcos(60). Then T1 acts at h where h= Lsin(60) from the pivot.

    The net torque on the beam should sum to zero because it's not rotating.

    The net forces in x and y on the beam should also sum to zero because it's not translating.
     
  4. Mar 12, 2014 #3
    How does this look?
    2mepzk1.jpg

    Also, how should I approach c? Is it just the weight of the beam?
     
  5. Mar 12, 2014 #4
    What are the forces acting on the weight of mass M ?
     
  6. Mar 12, 2014 #5

    CWatters

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    There is a force on the beam at the pivot.
     
  7. Mar 12, 2014 #6
    Could you please explicate?
     
  8. Mar 12, 2014 #7
    Gravity and the tension of cable 2.
     
  9. Mar 12, 2014 #8
    Good...So considering that the weight is in equilibrium ,what is the relation between them ? You may refer tension as T2 and weight Mg .
     
  10. Mar 12, 2014 #9
    There is also a vertical force from the weight of the beam no? Which is M/2*G*sin60.

    Also, how would I find the forces on the beam by the pivot?
     
  11. Mar 12, 2014 #10
    No..that is not correct.The weight of the beam is acting on the beam ,not on the weight .There are only two forces acting on the weight ;T2 and Mg

    So,what is the relation between them ?

    Let us call the force exerted by pivot as P .The horizontal component as Px and vertical component as Py .

    Can you write the torque equation for the beam with respect to the pivot ?

    Note :Please type your work.Do not paste an image.
     
  12. Mar 12, 2014 #11
    They are equal.

    torque = px*sin60 + py*cos60 = 0?
     
  13. Mar 12, 2014 #12
    Yes. So T2=Mg

    Simply wrong .

    The pivot force does not feature in the torque equation of the beam about the pivot . It would have played a role if we were writing torque equation about the center of mass .

    List all the forces acting on the beam .
     
  14. Mar 12, 2014 #13
    Just gravity?
     
  15. Mar 12, 2014 #14
    It is only one of the forces acting on the beam .What about T1 and T2 ? Do you think any of them is acting on the beam ?
     
    Last edited: Mar 12, 2014
  16. Mar 12, 2014 #15
    I apologize for my lack of knowledge and thank you so much for helping me.

    I suppose T2 does pull the beam in the vertical direction and T1 pulls it in horizontal direction?
     
  17. Mar 12, 2014 #16
    Yes...

    So,basically there are four forces acting on the beam ; T1,T2,Mg/2 and P .

    Do you agree ?
     
    Last edited: Mar 12, 2014
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