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Help: tension of a line with a hanging weight?

  1. Sep 21, 2007 #1
    1. The problem statement, all variables and given/known data
    A horizontal rope is tightly tied to hooks between two walls separated by a distance of 5m. A 100 Newton weight is suspended from the middle fo the rope and it sags so that the middle of the rope is displaced a distance of .25 meters.

    a) what's the tension in the rope?
    b) what are the horizontal and vertical forces are exerted on the hooks?

    2. Relevant equations
    I have no idea how to solve this.

    3. The attempt at a solution

    We've been resolving vectors into their components using
    A*sin(A's θ)= Ayi
    A*cos(B's θ)= Axj

    Resultant = sqrt(sumX^2 + sumY^2)
    Resultant θ = arctan(Y/X)
    Equilibrant θ= Rθ + 180

    I just don't know what to do with the info I have. 100N seems to be the only force I have to work with. It looks like a vector component y.

    I assume I need to fine the force of the hypotenuse vector I need to find it in the direction toward θ1? But I keep going around in circles trying to figure it out.
  2. jcsd
  3. Sep 21, 2007 #2
    I also thought of going about it in a similar fashion to how we did the force table excercise by making a circle and having all the vector components sum to zero? (is that correct on how it should work? But i'm lose at getting the rest of the info i need.

  4. Sep 23, 2007 #3
    When I asked this, I had already turned in the lab with this question unanswered. The ta announced on his website that, since it's easier for him to grade correct answers, he wants to help us. He was absolutely no help. He told me that since I had the angle, I could answer the question.

  5. Sep 23, 2007 #4
    Because the weight is in the middle, it means that the tension forces are equal in magnitude. The system is also at equilibrium. This means two things:
    -The horizontal components are equal and cancel each other out.
    -The vertical components are equal, and the two added together must equal the weight.

    From this, we can draw a triangle to represent this scenario. The only problem is that we don't have the angles. To get the angles, we draw another triangle. This triangles sides will be 0.25m and the 5m, you have to draw it. Grab the angles, transpose them to the force triangle.

    Once you have your force triangle you can answer (a, to get (b, you will need to do some geometry to get one right-angled triangle, from which you can grab the height and length.

    Also remember that force is a vector, so it needs a direction.
  6. Sep 23, 2007 #5
    The angle which you are getting is correct.Let the tensions be [tex]T_1[/tex] and [tex]T_2[/tex]. We have already got the angle.

    So now as the block has no horizontal motion, the x components of the tension in both the strings are equal to each other.

    Now there is no vertical motion also,so the summation of the y components of both the strings is equal to the weight of the block.

    Solve both the equations simultaneosly and you get the answer.
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