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Don't get this dynamics problem

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data

    the ends of a rope are fastened to two hooks on a horizontal ceiling. A 100 N crate is attached to the rope so that the two segments of the rope make angles of 35 and 55 degrees with the horizontal. Determine the tension in each segment



    2. Relevant equations

    F=ma

    3. The attempt at a solution

    I don't get what to do.. I know a force on the crate is mg. I don't know where to put T1 and T2 for the segments of the rope. Do I also put mg on the segments of the rope?
     
  2. jcsd
  3. Jan 25, 2010 #2

    rl.bhat

    User Avatar
    Homework Helper

    Draw the diagram. Resolve T1 and T2 into vertical and horizontal components. In equilibrium condition, Horizontal components cancel each other and vertical components add up to mg. Solve these two equations to find T1 and T2.
     
  4. Jan 26, 2010 #3
    I get them as 82N and 58N? Anything close to what you get?

    The crate is already given to you in Newtons, So no need to add gravity as it is already a force... If it was in kg, then you need to mulitply it by gravity.

    My Solution:

    Doesnt even need calculations ;) Although, it only gives a rough estimate...
    Take the diagram and draw it out,

    VectorDiagram1.jpg

    You have one known vector there and 2 angles, 100N with 35' and 55'. So take the 100N as it is vertical and draw it vertically.

    You also know that both the ties are holding it UP, so they will both be in an upwards direction in that picture. So take the 55' and draw it from the bottom of the 100N, and the 35' from the top of the 100N.

    If you've drawn it accuratly and to scale, just measure off the rough figures. Like:

    VectorDiagram2.jpg

    I drew the one I have on a piece of paper at 1mm = 2N so I had a 5cm line as my 100N.

    You can even confirm your results using the Sin Rule:

    [tex]

    \frac{A}{Sin a}

    [/tex]= [tex]

    \frac{B}{Sin b}

    [/tex] = [tex]

    \frac{C}{Sin c}

    [/tex]

    Therefore A = [tex]

    \frac{100 X Sin55}{Sin90}

    [/tex] = 81.92N

    and B = [tex]

    \frac{100 X Sin35}{Sin90}

    [/tex] = 57.36N

    X = Multiply....

    Pretty close for a shoddy protracter and ruler!

    (I apologise for my rubbish drawings as I do not have my tablet with me. And if my equations look weird, I am not sure how formatting fractions works on this forum!)

    Hope it helps and you can understand my methods. Oh, And if anyone spots any mistakes or im generally wrong feel free to bring me up on it :)
     
    Last edited: Jan 26, 2010
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