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Tension of wire

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data

    Rod with mass m depends on the two wires are ideal for the same length. Wires connected at the end of the rod. What is the tension in one of the wires as soon as the second string broke?

    2. Relevant equations

    question options are:
    0.25mg
    03.mg
    0.5mg
    1mg
    2mg

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

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  3. Sep 4, 2011 #2

    PhanthomJay

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    Hi, dannee, welcome to Physics Forums!

    Forum rules require that you list what you think might be the relevant equations, and that you make some attempt at a solution, before we can provide asssistance. Do you have any thoughts on this problem?
     
  4. Sep 4, 2011 #3
    hey jay, first thank you very much for welcoming.

    this problem at first looked quite easy, it's clear that before the broke each tension of wire is equel to 0.5mg. now i guess that after broke the rod is starting a circular motion, but i find it hard to relate it to the tension force. thanks in advance for any help !
     
  5. Sep 4, 2011 #4

    PhanthomJay

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    Yes, after one of the string breaks, the rod accelerates due to the unbalanced torque acting about the pivot. What is the initial net torque about that pivot, and what equations are you familiar with that relates torque and acceleration?
     
  6. Sep 6, 2011 #5
    i know that the torque of rod is ML^2/12, which leads to kinetic energy of 0.5iw^2,
    i've tried to compre kinetic energy at the potenital energy, but i figured out that it's not a good idea as the rod keeps its speed even after it reaches the lowest point.

    any ideas?
     
  7. Sep 6, 2011 #6

    PhanthomJay

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    You are confusing Torque with Moment of Inertia. The torque (or moment) of a force about a point is is equal to the product of the force times the perpendicular distance from the (cg of the ) force to that point. The Moment of Inertia of a thin rod that rotates about its end can be found with a web search.. Then you need to relate torque and I with the angular acceleration. And find the linear acceleration, and use Newton 2.
     
  8. Sep 6, 2011 #7
    should i use r x F = I * a ?

    it leaves me with a = equaltion with t, r and mg

    but the answer should be only with mg
     
  9. Sep 6, 2011 #8

    PhanthomJay

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    r, F, and I are known, solve for the angular acceleration, alpha. First list your expressions for Torque, I, and r.
     
  10. Sep 6, 2011 #9
    a=(r * F)/I

    a=(mg-T)/(0.33*m*r)

    i don't know how to go from here
     
  11. Sep 6, 2011 #10

    PhanthomJay

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    The equation is T= Iα, where α is the angular acceleration.

    Let's go one step at a time. What is the torque about the pivot produced by the weight of the rod, assuming that its weight acts at the midpoint (center of gravity) of the rod? Express your answer in terms of m, g, and L, where L is the length of the rod. Remember that torque is force times perpendicular distance.
     
  12. Sep 6, 2011 #11
    hey, thanks again for your help

    when I used 'T' symbol i meant for the tendency, not torque :)

    i'm still not so familiar with the subject.

    is it legal to use the equal torque=I*alpha,torque=F*r -> F*r=I*alpha ?

    Moment of Inertia is (m*L^2)/3, then again i'm stucking with equation which leads to an answer with L, and it's no good enough for solution.
     
  13. Sep 6, 2011 #12

    PhanthomJay

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    Eeks, :eek: T for Torque, T for tension, t for time, and now T for 'Tendency"?

    You must go by steps. You first need to find the Torque produced by the rod's weight. Again, what is the value of Torque that you find using Torque = Force times perpendicular distance (hint: the force is 'mg' and the perpendicular distance from the weight force from its center of gravity to the pivot is ____?
     
  14. Sep 6, 2011 #13
    Torque = Force * (L/2)
     
  15. Sep 6, 2011 #14

    PhanthomJay

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    Yes, and what is the force in terms of m and g?
     
  16. Sep 6, 2011 #15
    sum of force coordinate y: T(tension of wire)-mg
     
  17. Sep 6, 2011 #16

    PhanthomJay

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    That is the sum of forces in the y direction, yes...you'll need this later....but first, you need to note the force which causes the torque (rotation) about the pivot point....that force, F, is just the weight of the rod....which is equal to ___? Then you get the torque, which is F times L/2 , and once you get the torque, solve for α, which is α = Torque/I, where I is ml^2/3. Thus, α = ___?
     
  18. Sep 6, 2011 #17
    i think i got it, F=mg on the center of rod,
    then: torque=mg*0.5L

    then alpha = (0.5 * mg * L) / (0.5 * m * L^2)

    means alpha = (1/L)

    where do i go from here?
     
  19. Sep 6, 2011 #18

    PhanthomJay

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    Yes, correct!
    you are going too fast...
    alpha = .5mgL/((mL^2)/3)
    alpha = .5g/(L/3)
    alpha = (g/2)/(L/3)
    alpha = 3g/2L

    You must watch your algebra, or all else is lost.
    This problem is of a higher level than basic static problems. Be sure to master statics (Newton 1) before moving on to dynamics (Newton 2).
    Now that you have alpha, the angular acceleration, you want to find the linear acceleration of the center of mass, so you can use it in Newton's 2nd Law using the net force you calculated earlier in the y direction. So do you knw the basic equation that converts angular acceleration , alpha, into linear acceleroation, a?
     
  20. Sep 7, 2011 #19
    that's a formula which i familiar with. alpha = a/r, means:

    3g/2L = a/L, a= 3g/2

    now i go back to coordinate y: tension-mg=ma, tension=m(g+a), tension=m*2.5g

    but i think i've missed something again as the answer is 0.25mg ?

    many many thanks for the help !
     
  21. Sep 7, 2011 #20

    PhanthomJay

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    You have missed 2 things....first, in your calculation for 'r', you used 'L'...but you are looking for the acceleration of the center of mass, which is not L units away at the tip, but rather______??

    And second, the acceleration is initially vertically down, so which is greater, the tension in the rod or the weight of the rod??
     
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