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Looking for help with equilibrium physics

  1. Oct 10, 2007 #1
    The system shown below are in equilibrium with m = 4.50 kg. If the spring scale is calibrated in newtons, what does it read? (Ignore the masses of the pulleys and strings, and assume the incline is frictionless.)

    There is a picture on this question but the best thing I can do is to describe it for you.
    There are two equal masses on each side of the table supporting by its' pulleys. Each mass is attached to each end of the spring scale.

    I'm kind of lost on this question. Does the two objects pull the spring scale in both directions, left and right, creating a force of 88.29N (F=2*ma=2*(4.50kg*9.81m/s^2))? Can anyone help me out on this question?
     
  2. jcsd
  3. Oct 10, 2007 #2

    Doc Al

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    I don't have a clear picture of the entire setup, nonetheless...

    The spring scale will read the tension in the string where it's attached. A force equal to that tension will be exerted on each end of the scale, but they don't add.
     
  4. Oct 10, 2007 #3
    Does this mean the force on one mass is really the tension exerted by two masses pulling each other?
     
  5. Oct 10, 2007 #4

    Doc Al

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    I don't quite understand your question. If a mass is attached to a rope, one of the forces on the mass will be the tension force exerted by the rope. If that rope attaches to both masses, then it exerts the tension force on each.

    If possible, can you scan in a diagram?

    Is the table horizontal? Is there an incline involved?
     
  6. Oct 10, 2007 #5
    yeah...here's the diagram
     

    Attached Files:

  7. Oct 10, 2007 #6

    Doc Al

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    Cool. A diagram is very helpful.

    As I had said, the scale will read the tension in the cable. And since everything is in equilibrium, what must that tension equal?
     
  8. Oct 10, 2007 #7
    thanks Doc Al, now I know the answer for this question
     
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