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Boat pulling two barges

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  1. Oct 23, 2014 #1
    1. The problem statement, all variables and given/known data
    A boat pulls two barges down the river. The barge connected to the boat (barge A) has an inertia of 2x10^5 kg, and the other barge (barge B) has an inertia of 3x10^5 kg. the resistive force between barge A and the water is 7x10^3 N, and the resistive fore of barge B to the water is 9x10^3 N. The common acceleration of all three boats is .4 m/s^2. Even thought the ropes are huge, the gravitational exerted on them is much smaller than the pulling forces.

    2. Relevant equations
    (a). What is the tension in the rope that connects the boat to barge A?
    (b). What is the tension in the rope that connects the two barges?
    (c). Repeat steps a and b for the case in which the order of the barges is reversed.

    3. The attempt at a solution
    So for part (a) I took the contact force of the rope to barge A which is 2x10^5kg * .4m/s^2= 8x10^4 N and added the negative of the resistive force; 8x10^4N - 7x10^3N= 7.3x10^4N. I wasn't sure if I should also subtract the contact force of the second rope on barge A which would cancel out the two contact forces and the answer would be the resitive force of barge A and the water.
    (b). Contact force of the second rope on barge B; 3x10^5kg * .4m/s^2 = 1.2x10^5N; subtract the resistive force of barge B to the water; 1.2x10^5N - 9x10^3N = 1.11x10^5N. Again not sure if I should add the contact force of the first rope on barge B.
    (c). They would be the same unless I do subtract/add the parts I'm not sure about.
     
  2. jcsd
  3. Oct 23, 2014 #2
    What is this "contact force"? Are you translating from another language?

    Anyway, do you know about free body diagrams? Newton'e second law? Net force?
     
  4. Oct 23, 2014 #3
    Contact force is what we are being taught. It is the a force that one object exerts on another object only when the two objects are in physical contact. It is a part of Newton's third law which states whenever two objects interact they exert on eachother forces that are equal in magnitude but opposite in direction. For example FC1,2 = -FC2,1
    We use contact and gravitational forces in our free body diagrams; Newton's second law- the vector sum (net force) of all forces exerted on an object (contact and gravitational forces) is equal to the rate of change of momentum of the object.
    We use a textbook that was released this year that introduces different concepts than other physic textbooks.
     
  5. Oct 23, 2014 #4
    OK. But then why do you equate the "contact force" from the rope with mass*acceleration for the barge?
    This "contact force" is not the net force on that barge. There are other forces acting on it.
    The "net" in the net force is essential. Thus is why a free body diagram is of great help.
    You draw the barge under question (for example) and all the forces acting on it. Then you write the net force and equate with m*a.
     
  6. Oct 23, 2014 #5
    Oh, we learned that Net force of barge B for example = ΣFbarge B = all the forces acting on barge B so we would do ΣFbarge B = FCrope 2, barge B + FCwater, barge B + FGEarth, barge B + FCbarge B, Earth + FCrope 1, barge B
    I know it probably sounds weird, it did to me at first as well, but it's the way the book asks for it lol. And now that I wrote all of that out I think I know the answer now, so thank you! :)
     
    Last edited: Oct 23, 2014
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