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Homework Help: Two pulley system

  1. Apr 11, 2008 #1
    http://users.on.net/~rohanlal/Q40.jpg [Broken]
    Does the tension in the rope from B to A and from A to C remain the same?
    To determine the value of theta when the system is in equilibrium do I need to calculate the sum of all acting x and y components? If so would B and C be negative and A be positive?
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Apr 11, 2008 #2

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    Not in general.

    At a point which is in equilibrium, you always have to equate the sum of all forces along any direction to zero. The same goes for a system in static equilibrium, but considering all the forces at one time may not always turn out to be so fruitful

    In this case, the weight of A is balanced by the resultant of the two tensions, which in turn are equal to the weights of B and C respectively. (The pulleys are assumed to be frictionless, and I've just given you the method of solving the problem.)

    No. (I presume you mean their weights.)
     
  4. Apr 11, 2008 #3
    if B and C are not negative then what is? something has to be negative otherwise the sum of the system won't be equal to zero.
    do i just use a right triangle to solve for the resultant?
     
    Last edited: Apr 11, 2008
  5. Apr 11, 2008 #4
    are B and C positive and A negative?
     
  6. Apr 11, 2008 #5

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    There are reaction forces on the pulleys, acting upward. That gives your negative.

    I didn't quite understand your 2nd Q. Do you mean how to solve for the forces at where A is hanging from? Break up into horizontal and vertical components, and then equate the sums to zero.
     
  7. Apr 11, 2008 #6
    does this look right
    Bsin(theta)+Csin(phi)-A=0
    Bcos(theta)-Ccos(phi)=0
    so i dont even have to find the resultant tension i just solve for theta?
     
  8. Apr 12, 2008 #7

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    That is correct, but you should show the steps how you got there.
     
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