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Determine maximum weight and angle for equilibrium

  1. Sep 12, 2015 #1
    1. The problem statement, all variables and given/known data

    The cords ABC and BD can each support a maximum load of 100 lb. Determine the weight of the crate, and the angle θ for equilibrium.
    2. Relevant equations
    ∑F= 0
    ∑Fx=0
    ΣFy=0
    Setting derivative of an expression to zero To find critical point
    Second derivative test for determining whether min or max at a point

    3. The attempt at a solution
    -FBDcosθ + FBCsin22.62 = 0
    -mg -FBCcos22.62 + FBDsinθ
     

    Attached Files:

  2. jcsd
  3. Sep 12, 2015 #2
    I don't think your approach will give you a meaningful answer. I would suggest to assume that the cord forces are all equal to 100lbs.
     
  4. Sep 12, 2015 #3
    If we are to take that approach, then would the crate have to be 100 lb?
     
  5. Sep 12, 2015 #4
    Sorry, I meant assume cord BD is 100lbs.
     
  6. Sep 12, 2015 #5
    So from that, we get.FBD=100cosθi + 100sinθj

    The system is in equilibrium so FBD=FBA+ FBC

    F
    BA=-mgj
    F
    BC= FBCsin22.62i - FBCcos22.62j

    Would I be correct in assuming that the magnitude of the force on BA and BC add to 100 lb?

    If so, where would I go from there?
     
  7. Sep 12, 2015 #6
    One of the tricks here is to realize that BA and BC must be equal. Otherwise the pulley wheel would start to rotate.
     
  8. Sep 12, 2015 #7
    This is a rather simple problem. You get the force along AB from the weight of the crate. Then consider the relationship between the forces along AB and CB. Tip: Imagine the wheel's axle as fixed. The angle ABC can be read off directly from the diagram. With these, you can calculate the vector sum of the forces BA and BC. From this, you immediately get BD and θ. No need to fiddle with component vectors. :cool:

    If you want to challenge yourself, do this one in your head. I just did.
     
    Last edited: Sep 12, 2015
  9. Sep 12, 2015 #8

    haruspex

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    It only makes sense to add magnitudes of vectors if they are in the same direction. As vectors they must add to be equal and opposite to the pull from BD. I assume you know how to add vectors.
    Btw, the 100 lb is irrelevant. You don't need to know any actual force values to solve the question.
     
  10. Sep 12, 2015 #9
    But how do we calculate the weight of the crate?
     
  11. Sep 12, 2015 #10
    I disagree since the 100lbs is the given limiting value you must use it to solve the problem.
     
  12. Sep 12, 2015 #11

    haruspex

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    My mistake - I was only looking at finding the angle. Didn't notice that we're asked for the weight of the crate too.
    Note that it makes the wording a little strange. It doesn't ask for the max weight of the crate, but for the actual weight. So, correspondingly, it should state the maximum actual tension experienced by any of the strings in the set-up, but instead it says 100 lbs is the max they can support.
    Maybe it lost something in the retelling.
     
  13. Sep 15, 2015 #12
    It is not entirely explicit in the problem statement, but what is asked for is the maximum weight the crate may have, without breaking the ropes. We have three force vectors that point along pieces of rope: BA, BC, BD. Any of these forces is obviously proportional to the weight of the crate: Double the weight of the crate, and each of the rope forces will double. You should know by now the relationship between the forces BA and BC, and between BA+BC and BD. Set the highest of the three rope forces to the maximum allowed, find the corresponding crate weight, and you're done.
     
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