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Find the weight of a sphere connected to a pulley and to wall

  1. Mar 17, 2015 #1
    1. The problem statement, all variables and given/known data

    The system of 2 spheres is in equilibrium. Figure: uPHmi.png . If the weight of the second sphere P2=20N, ABC=60 degrees and BAC=30 degrees, find the weight of the first P1 and the force that is applied on the drum of the pulley.
    2. Relevant equations
    P1=m*a

    3. The attempt at a solution
    Everything is in equilibrium (no acceleration). I think I should solve Newtons 2nd law for P2 to tell me the tension, but I have no acceleration for P2. I don't know how to solve it.
     
  2. jcsd
  3. Mar 17, 2015 #2

    Orodruin

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    You have basically answered your own question. What is the numerical value of the acceleration if you have no acceleration?
     
  4. Mar 18, 2015 #3
    I think it's 0, but it has no sense.
     
  5. Mar 18, 2015 #4

    Orodruin

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    Why not?
     
  6. Mar 18, 2015 #5
    And the force would be 0? This is the answer? Can you give me a clue?
     
  7. Mar 18, 2015 #6

    Orodruin

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    Which force would be zero?
     
  8. Mar 18, 2015 #7
    The tension
     
  9. Mar 18, 2015 #8

    Orodruin

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    Is the tension the only force acting on the object?
     
  10. Mar 18, 2015 #9
    No, it's also the gravity
     
  11. Mar 18, 2015 #10

    Orodruin

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    So what is the acceleration of the object if it is acted upon by a gravitational force mg and a tension force 0?
     
  12. Mar 18, 2015 #11
    g
     
  13. Mar 18, 2015 #12

    Orodruin

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    Exactly, so tension cannot be zero for zero acceleration. What tension do you need for zero acceleration?
     
  14. Mar 18, 2015 #13
    The tension should be 20 N?
     
  15. Mar 18, 2015 #14

    Orodruin

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    Yes, equilibrium means that the resultant force (obtained by adding all forces acting on the object) is zero.
     
  16. Mar 18, 2015 #15
    The numerical answer is in my book. It is 17,3 N (it should be about the force that is applied on the drum). So in order to find it I used the formula P1 according to the drum direction is sin30*20=10 and the P force directed towards the wall is sin60*20=17.3. But in oreder to get the right answer shouldn't it be the vice versa?
     
  17. Mar 18, 2015 #16
     
  18. Mar 18, 2015 #17
    I have this other idea: I found the force applied on the drum with the formula cos 30*20=17.3 N. Then I calucated the P1 knowing that the vector in teh direction of the drum is 17.3 N, I found the P1 which is 17,3*2=34.4 (CB is in front of 30 grades angle.)
     
  19. Apr 13, 2015 #18
    Can you see my last answer and let me know if I got it right?
     
  20. Apr 14, 2015 #19

    haruspex

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    The "force on the drum" is a result of both the tension in the vertical section and the (same) tension in the BC sectIon. You have only calculated the vertical component of the tension in BC.
    The string AC is separate. It may have a different tension from that in BC. You have to figure that out by considering horizontal equilibrium.
     
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