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Quick question

  1. Nov 10, 2004 #1
    If you take the wheel as your system, would the tension in the red rope be 10 lbs also? pic attached

    thanks
     

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    Last edited: Nov 10, 2004
  2. jcsd
  3. Nov 10, 2004 #2
    the picture is pending approval, however, I might be able to give some help.

    On a pully, the tension is equal to:

    Tension=mg-ma
    The m is the mass of the block only.
    The a is the acceleration, which is NET FORCE/TOTAL MASS. If you have two masses pulling in two different directions, get the NET force of the system by F=(m2-m1)a. Then, get total mass by adding the masses. For the acceleration, a=net force/total mass.

    Now, for the tension, isolate the system to just the rope and block. T=the weight of the block minus acceleration of the system times the mass of the block. T=mg-ma
     
  4. Nov 12, 2004 #3
    the picture is up
     
  5. Nov 12, 2004 #4

    Diane_

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    Homework Helper

    If the red rope is attached at the bottom, as it appears, then the tension would be zero. Think about it - which way will the force from the brown rope cause the wheel to turn, and which way will the point at which the red rope is attached be moving?
     
  6. Nov 12, 2004 #5
    thanks for the replies

    The wheel is turning to the left...so if i understand, the external force of the red rope would be zero, but if we were to cut the rope, there would be an internal force of 10 lbs? Or am I just crazy for thinking this? :confused:
     
  7. Nov 12, 2004 #6
    If the wheel is not in motion, the red rope would have to exert the same amount of force on the wheel as the brown one, that means 10 lb. If the rope was cut, then the wheel would be rolling towards the left, accelerating at about 100m/s^2.
     
  8. Nov 12, 2004 #7
    So when its in motion the tension in the red rope is zero

    Also, when I said
    I was talking about when we analyze the whole system. When I said "cut" I didn't mean it literally, I was just talking about the analysis of the system. I should have been more clear about that. My apologies.
     
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