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Forces in equilibrium

  1. Mar 23, 2014 #1
    We had a lab 2 weeks ago that dealt with predicting the "known unknown" masses (we know it's 250g and we have to experiment to try and get to 250g as close as possible) within a pulley system.

    All three trials we had for part 1, where the known unknown was placed in the middle, we only needed to calculate for its y-component. I was just wondering why didn't we need to calculate the x-component?
     
  2. jcsd
  3. Mar 23, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Hi santoki. Seems you forget to attach the diagram showing the pulley & rope system you were using.
     
  4. Mar 23, 2014 #3
    Oh sorry. Here is the system for trial 1:

    qyVspeb.jpg
     
  5. Mar 24, 2014 #4

    NascentOxygen

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    Let's label as point A that point on the M-shaped rope where the unknown mass is hanging from.

    First step, resolve tension T1 into its horizontal and vertical components.
    Repeat this for T2

    Since the system is in equilibrium, what can you say about the sum at A of the horizontal components of all the forces acting there? See whether you can form this into a mathematical expression.

    And about the sum of the vertical components at A? Do the same for that.

    See any clues yet towards answering your original question?
     
  6. Mar 24, 2014 #5
    I was thinking that the masses hung at the sides served only to change the direction of the force applied and the unknown force is the equilibrant force which is equal in magnitude but of opposite direction to the resultant force. So if I add T1y and T2y together, the magnitude of this resultant vector will be the weight of the unknown mass.

    Am I thinking of it correctly?
     
  7. Mar 25, 2014 #6

    NascentOxygen

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    That's it in words, so can you express this mathematically?
     
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