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2D Statics Problem

  1. Sep 5, 2014 #1
    1. The problem statement, all variables and given/known data

    The truck is to be towed using two ropes. If the resultant force is to be 910N directed along the positive x axis, determine the magnitudes of force F1 and F2 acting on each rope and the angle θ of F2 so that the magnitude of F2 is a minimum. F1 acts at 20° from the x axis

    attached is a picture.

    2. Relevant equations

    ∑Fx = 0
    ∑Fy = 0

    LOS: sinα/A = sinβ/B = sinγ/C

    LOC: A*A = B*B + C*C - 2BCcos(θ) . where θ is the angle oppsite side A

    3. The attempt at a solution

    For the force to only be along the x-axis the y components sum to zero and the x components sum to 910N

    Problem is: I got 2 equations but 3 unknowns:

    (1) F1cos20 + F2cosθ = 910.
    (2) F1sin20 - F2sinθ = 0

    F1 is the magnitude of the vector F1 and F2 is the magnitude of the vector F2

    I've been fiddling with LOS and LOC simplify to 2 unknowns but no luck and I need to make some progress on this homework, does any one have any insight?
     

    Attached Files:

  2. jcsd
  3. Sep 5, 2014 #2

    Simon Bridge

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    Use your knowledge of geometry and physics to find another equation.
    You have listed two "relevant equation"s you have not used yet.
     
  4. Sep 5, 2014 #3
    Thanks I'll keep working and let you know what I come up with
     
  5. Sep 6, 2014 #4

    Orodruin

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    What would be the solution if you were given the angle θ?
     
  6. Sep 6, 2014 #5
    I would just solve (1) and (2) and get the magnitudes of the F1 and F2.
     
  7. Sep 6, 2014 #6

    Orodruin

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    So doing that would give you F2 for a given θ. How could you use this expression to find the minimum value of F2?
     
  8. Sep 6, 2014 #7
    Here's what I have so far...solving for F1 in (2) and plugging that into (1) I found:

    F2 = 910sin20/(sinθ + sin20cosθ) and this will be minimized when (sinθ + sin20cosθ) is as large as possible.

    If theta is 90 then F2 is equal to 910sin20. If theta is 0 then F2 is equal to 910 which I have a hard time accepting. Also I know the denom will be zero when theta is ~-19°....

    turns out theta = 90 gives the correct answer but Im not sure why.

    I think maybe 90 is just the only value that maximizes the denominator. or I just got lucky..
     
    Last edited: Sep 6, 2014
  9. Sep 6, 2014 #8

    Orodruin

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    I think you need to recheck your algebra. Your expression for F2 is not the same as I got. Once you have the correct expression you should be able to use a trigonometric identity.

    It is true that F2 should be equal to 910 N when θ = 0. When this happens F2 cannot counter the y component of F1 so the y component of F1 must be zero. This means that F1 = 0 and F2 must supply all of the force.

    Hint: Your denominator should be zero when θ is 160° (or equivalently -20°, which will give a negative force, i.e. a push rather than pull). Then the forces are in the same direction and cannot have zero y component unless they cancel.
     
  10. Sep 6, 2014 #9
    Orodruin,

    Thanks for your guidance on this problem it taught me a lot and I probably would have been lost with out it! But I am curious about your equation now... because the answers I got using the equation I found for F2 gave the correct answers. Not only that but it is derived directly from (1) and (2)!

    Best Regards,
    Ockham
     
  11. Sep 6, 2014 #10

    Orodruin

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    When inserting (2) into (1), you lost the cos20° from (1). Putting θ to 90° in your equation gives the right answer because it happens that you denominator then has the same value as the maximal value of what should be there. It does not maximize your denominator. The reason it gives the correct answer for θ=0 is that then the sine is zero and that the cosθ term is correct.
     
  12. Sep 7, 2014 #11
    Ok so I did get lucky but I see what I did wrong!

    If I would have gotten the right equation for F2, my next step would have been to take the derivative , set it equal to zero and then I would have found my angle. Which is 70 degrees!

    And the trig ID you where referring was the sin(a + b) = sinacosb + sinbcosa ..which would have made taking the derivative somewhat simpler.
     
    Last edited: Sep 7, 2014
  13. Sep 8, 2014 #12

    Orodruin

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    Exactly. You could also use your knowledge about the sine function having maximal value one when the argument is 90 degrees. Of course it will give you the same result.
     
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