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Tension/Force problem

  1. Sep 9, 2012 #1
    First time taking physics in many years, so please forgive the simplistic nature of this problem.

    A 2.0 kg ball tied to a string fixed to the ceiling is pulled to one side by a force F. Just before the ball is released, (a) how large is the force F that is holding the ball in position and (b) what is the tension in the string?

    Also given: the angle between the string and the ceiling is 30 degrees.

    I understand everything up to the point that I've found the components of T and calculated:

    sum of x forces = Tcos30 (since in eqb sum of x forces=0)
    sum of y forces = Tsin30 (since in eqb sum of y forces =0)

    In the solutions manual, they are then taking F/w (force vector in positive-x direction/ weight vector in negative-y direction) = Tcos30/Tsin30

    I don't understand why they are dividing F by w. I see that this will eliminate T so that you can solve for F, however, the decision to divide F by w seems arbitrary. Either that or I'm fundamentally missing something. Any insight?
     
  2. jcsd
  3. Sep 9, 2012 #2

    Doc Al

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    Staff: Mentor

    What you have are not the sum of forces, but just the x and y components of the tension.

    What other forces act on the ball?
    Write equations for ƩFx = 0 & ƩFy = 0.

    (I presume that the given force F is horizontal?)
     
  4. Sep 9, 2012 #3
    Yes, the force F is horizontal.

    ƩFx = F - Tx = 0
    so F = Tx
    since Tx = Tsin30, F=Tsin30

    ƩFy= Ty - W = 0
    so Ty - w = 0
    Ty = w
    Tsin30 = w

    So, in this instance, it turns out that the x and y components of the tension force are equal to F and W, respectively. So that's all I should be using to solve, right? Confused on the next step from here.
     
  5. Sep 9, 2012 #4
    What we have here is this

    c3be4894.jpg

    now all forces must cancel each other

    so u get 2 equations and then solve for force


    Tension can be found similarly

    Tsin30 = mg....equation (1)
    F= Tcos30....equation (2)

    m is given so find tension using equation 1
    then put that value in equation 2 to get force
     
    Last edited: Sep 9, 2012
  6. Sep 9, 2012 #5
    Welcome to the forum kxk010!
    You need to solve for F. The problem is of course that you don't know T so you have two unknowns.
    You have to figure out a way to get rid of one of the two variables (T) .One way to go is to divide one by the other, as your solutions manual did, a substitution would also be fine.
     
  7. Sep 9, 2012 #6
    Edit.
     
  8. Sep 9, 2012 #7
    Ah, ok. So really you can take these 2 equations that you've "generated" and use them in any way necessary to isolate and solve for variables.
     
  9. Sep 9, 2012 #8

    Doc Al

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    Staff: Mentor

    Is the 30 degrees the angle between the string and the ceiling, as you said before? If so, rethink those components--you have them reversed.
    Once you get those equations fixed, you just have to solve them simultaneously for the unknowns T & F. There are several ways to do that--take your pick. One way, like they show in your solutions manual, is to just divide the two equations by each other. That will eliminate T and allow you to get F. Try it and see.
     
  10. Sep 9, 2012 #9
    You're right, I mistyped and F=Tcos30. Sorry about that. I seem to be stuck in this rigid mindset from other science classes that there is a sequential set of steps to solve every problem. Definitely finding thats not the case and there are multiple options (within limits) for the same problem. Thanks for all the help!
     
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