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Hibbeler 12-231

  1. Feb 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Attached


    2. Relevant equations
    Attached



    3. The attempt at a solution
    Attached. Couple of questions/confirmations:

    it should be vbcos(45)i - vbsin(45)j, but this is the same thing as cos(45)=sin(45)
    vb should be directed at B
    How do you solve equations 1 and 2? Trial and error?
    vb/r will always be 5m/s in this case?
     

    Attached Files:

  2. jcsd
  3. Feb 23, 2014 #2

    lewando

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    Try squaring both sides of 1 and 2 then add the two equations together.
     
  4. Feb 23, 2014 #3
    Thanks! Then I get vb^2 = 29 + 20sin(theta).

    Plugging sin(theta) = [ vb^2 - 29 ] / 20 into (2) I get two solutions for vb, one being negative. I hope I am correct in saying that this is the one to discard as the magnitude of a vector cannot be negative.

    However in a previous problem, the relative velocity magnitude comes out negative in the solution, which doesn't make sense
     
  5. Feb 23, 2014 #4

    lewando

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    Not following how you arrived at that. Squaring eq. 1 should give you:

    (0.707vb)2 = 25cos2θ

    No substitution into (eq. 2)2 needed. Just add the 2 equations.

    However you did it, did you at least get vb = 6.21 m/s?
     
  6. Feb 23, 2014 #5
    Yes I got vb = 6.21 or vb = -3.382 from the quadratic. I substituted because there are 2 unknowns there.

    When I add the squares of the equations,

    vb^2 (cos^2(45) + sin^2(45)) = 25 ( cos^2(theta) + sin^2(theta) ) + 4 + 20*sin(theta)

    vb^2 = 29 + 20sin(theta), still 2 unknowns
     
  7. Feb 23, 2014 #6

    lewando

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    I guess I should have been clearer.


    Sorry, big EDIT:
    Before squaring eq. 1, isolate the sinθ on the right-hand side.

    Before squaring eq. 2, isolate the cosθ on the right-hand side.


    Now you can square them.

    Adding the 2 equations gives you sin2θ + cos2θ on the right hand side.

    And everyone knows what that is equal to. Now you have a quadratic in vb only.
     
  8. Feb 23, 2014 #7
    Excellent, that still works out the same, so I take the positive magnitude.

    In other problems such as 12-227, the magnitude of the relative velocity comes out negative. If this is not wrong, this means that the overall sense of direction of the vector needs to be reversed.

    In this problem here, we get a quadratic with a negative velocity magnitude. I assume we discard this not because it's negative, but because it's smaller than vb/r as vb = vr + vb/r. Hope this is correct.
     
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