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Electrodynamics problem - force equilibrium

  1. Sep 12, 2016 #1
    1. The problem statement, all variables and given/known data
    charges.jpg

    Hello! I have a question about the following problem:

    Two point masses m1 and m2 are attached to isolating wires to point P. They are both positively charges (charge Q1 and Q2) and in the picture you can see the situation at equilibrium.

    What is the proportion of the masses (m1 / m2)?


    2. Relevant equations


    3. The attempt at a solution

    So this is how I would do it:

    We have the weight force for m1 which is W1 = m1 * g and for m2 we have W2 = m2 * g

    The forces W are is in y-direction

    For other forces of the y-direction we have to break down the tension force into its x and y components.

    For m1 we can say that the tension force in y direction is T1y = T1 * cos (60°) and for m2 the tension force in y direction is T2y = T2 * cos(30°)

    this means that T1 * cos (60°) - m1 * g = 0
    and T2 * cos (30°) - m2 g )= 0

    or T1 * cos (60°) = m1 * g
    and T2 * cos (30°) = m2 * g

    If we devide the both equations we get
    T1 / T2 * cos(60°) / cos(30°) = m1 / m2

    for cos(60) = 1/2
    for cos (30) = sqrt (3) /2

    cos(60)/cos(30) = 1/sqrt(3)

    which leads us to T1 / T2 * 1/sqrt(3) = m1/m2

    how can I get rid of T1/T2 and what is m1/m2?

    Thanks for your help!
     

    Attached Files:

  2. jcsd
  3. Sep 12, 2016 #2

    BvU

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    Gold Member

    Hi Y, :welcome:
    So a this point you need another equation involving these two ratios. Any idea ?
     
  4. Sep 12, 2016 #3
    well the x components are T2 * sin(60°) - T1 * sin(30°) = 0

    which is T2 * sqrt (3)/2 = T1 * 1/2

    so T1 / T2 = sqrt (3)

    if we but that in our first equation then we get m1 / m2 = sqrt (3) * 1/sqrt(3) which should be 1

    and the possible given solutions are:

    A) 3 B) 1/3 C) 1/sqrt(3)

    so I don't know where I did something wrong

    Y
     
  5. Sep 12, 2016 #4

    gneill

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    Staff: Mentor

    Are you sure you've chosen the right trig function?
     
  6. Sep 12, 2016 #5
    ok I see the mistake and if I turn it around it should be 1/3
     
  7. Sep 12, 2016 #6

    gneill

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    Staff: Mentor

    Right. You might have also taken a slightly different approach by considering the torque about point P. Since the assembly is in equilibrium that torque should be zero. You should know the relative lengths of the sides of the common 30-60-90 triangle.
     
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