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Find the tension in the rope pulling a box at an angle

  1. May 28, 2016 #1
    1. The problem statement, all variables and given/known data

    9EFoh9O.png

    2. Relevant equations


    3. The attempt at a solution

    Here is what i have done so far:

    Fx = cos 45 T - Ff =ma
    Fy=sin 45 T -mg+ Fn =ma
    I found ff= uk * Fn = 0.5 *sin 45 +mg
    fn = sin 45 +mg

    Isolated for T in the x and y:
    Tx = ma + Ff / cos 45 = 13.404
    Ty = ma +mg - Fn / sin 45 = 83.938

    found the resultant for T: sqrt( (13.404)^2 + (83.938)^2 ) = 85 N????


    What am i doing wrong? i've been stuck on this for 3 +hours now with no help :(

    the answer is meant to be E.
     
    Last edited: May 29, 2016
  2. jcsd
  3. May 29, 2016 #2

    andrewkirk

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    Presumably ##F_y## denotes the vertical component of net force on the box. That '=ma' should not be there. The acceleration is strictly horizontal, so ##a## shoud not appear in this equation about vertical components. Also, note that writing ##F_y## this way, in which the positive direction is up, it will be negative, because there cannot be a net upward force on the box, otherwise it would move upwards off the floor rather than horizontally along it.
    I don't know what 'uk' is. Assuming that 'ff' denotes frictional force, that force will be ##0.5F_y##, not what is written here.
     
  4. May 29, 2016 #3
    In the question, they stated the kinetic friction as 0.5 , which is what uk is.
    and frictional force, ff, is uk * normal force (fn), therefore Ff = uk * Fn ? right?
    I made the second equation as you said:

    Fy=sin 45 T -mg+ Fn =0

    solved for T and got -0.9998

    i must be doing something wrong, im just not sure what
     
  5. May 29, 2016 #4

    haruspex

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    We cannot tell where you went wrong if you do not post your working.
    The two equations had up to that point look correct.
     
  6. May 29, 2016 #5
    okay i now understand that I got the normal force equation wrong it is:
    Fn = mg -T sin 45

    why is the tension force negative? isn't Ty in the upward positive direction?
     
  7. May 29, 2016 #6

    haruspex

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    Yes, I understood that you had got that equation correct, but that still does not show me how you got from there to a wrong answer. Please show all your steps.
     
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