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Statics of Particles

  1. Aug 25, 2015 #1
    1. The problem statement, all variables and given/known data

    Two tugboats are pulling a barge. if the resultant of the forces exerted by the tugboats is a 5000-lb force directed along the axis of the barge, determine
    (a) the tension in each of the ropes, give then alpha = 45 (b) the value for alpha for which the tension in rope 2 is mininum

    2rh4opu.jpg

    2. Relevant equations

    3. The attempt at a solution
    I've already found an answer to the problem by using the law of sines and the parallelogram rule
    but I would like to know HOW to solve it by breaking it into vector components, I am not quite sure where to start in order to solve it in this manner.
     
  2. jcsd
  3. Aug 25, 2015 #2
    Each of the ropes pulling the barge can be thought of as an vector because they have both a magnitude (the tension) and a direction.

    A vector V of magnitude M and direction [itex]\theta[/itex] above the x-axis can be broken into components:
    Vx = Mcos([itex]\theta[/itex])
    Vy = Msin([itex]\theta[/itex])

    For part a, the idea is that the horizontal components of the two tensions will add up to 5000 lb, while the vertical ones will cancel because they have opposite directions. Given this, you should be able to write down a system of equations and solve for the tensions.
     
  4. Aug 25, 2015 #3
    so to find the x components I would do

    5000 lb * Cos(30)
    and 5000 lb * cos(45)

    ?
     
  5. Aug 25, 2015 #4
    I'm still unsure how to find the magnitude of the original vectors : /
     
  6. Aug 25, 2015 #5

    SteamKing

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    Staff Emeritus
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    Not quite.

    If the tensions in the ropes are called T1 and T2, then the sum of the horizontal components of each tension must equal 5000 lbs.

    In other words, T1 ≠ T2 ≠ 5000 lbs.
     
  7. Aug 25, 2015 #6
    First start by drawing your free body diagram. You have three forces. Figuring out what your angles are is easy when you have a diagram to look at.
     
  8. Aug 25, 2015 #7
    The sum of the forces in the X direction add up to 5000 lbs so I have

    Sum of Forces in X: Cos(30)* (Magnitude of 1) + cos(45) * (Magnitude of 2) = 5000lb

    Sum of Forces in Y: Sin(30) * (Magnitude of 1) - Sin(45)*(Magnitude of 2) = 0

    I'm not sure where to go from here...

    EDIT: I'm really sorry Its been a while since I've done math/physics.
     
  9. Aug 25, 2015 #8
    I DID IT I solved M2 and got 2600 lbs
     
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