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Time for the bead to slide

  1. Aug 30, 2016 #1
    1. The problem statement, all variables and given/known data
    In the figure shown, friction force between the bead and the light string is ##\dfrac{mg}{4}##. Find the time in which the bead looses contact with the string after the system is released from rest.

    2. Relevant equations
    Weight of the bead ##(m_1)=mg##
    Friction acting on the bead=##\dfrac{mg}{4}##
    Relation between the acceleration of the block ##(m_2)## and the string going over the non-movable pulley

    3. The attempt at a solution
    FBD of ##m_2##
    From the above FBD we get the following equation:-

    FBD of the bead
    I was having trouble figuring out whether tension would be acting or not on the bead, so I thought about the need for the string to be taught(as the dtring is ideal) hence there should be tension acting on the bead.
    From the above FBD of the bead we get the following equation:-

    We, see that there are only two equations and there are ##3## variables, so the equations are not solvable. What is wrong with the solution above, if there is anything please point out.
    Last edited: Aug 30, 2016
  2. jcsd
  3. Aug 30, 2016 #2
    Are you sure :smile: ?

    Tension does not act on the bead . Friction does . Also note that there is a nice relationship between friction acting on the bead and tension in the string .

    The acceleration of the bead is independent of the acceleration of the string . The acceleration of the bead is quite simple to calculate as all the forces acting on the bead are given in the problem .
  4. Aug 30, 2016 #3
    From what I have learned till now about constraint motion the acceleration of any point on the string(the string which goes over the non-moveable pulley) would be as stated in my solution.
  5. Aug 30, 2016 #4
    Think carefully , if left block moves up by distance 'x' , then how much length of string (left and middle string combined) loosen up ? How much string on the right side needs to go down in order for the string to remain taut ?
  6. Aug 30, 2016 #5
    If the block goes up by a distance ##x## then a length ##2x## of the string remains slacken so to make the string taut again a length of ##2x## string gets shifted to the rightmost string. Is that correct and btw that's how I had made the constraints.
  7. Aug 30, 2016 #6
    Good :smile:


    Your constraint relation says that if the block goes up by 'x' , then right string moves down by 'x/2' .
  8. Aug 30, 2016 #7
    Oh my God, now that I see it was a blunder on my part, it took me sometime to figure out that despite the right reasoning I ended up with the wrong constraint. And I am very sorry if I sounded even a bit rude in my last reply, I had no such intention
  9. Aug 30, 2016 #8
    This is quite a common mistake .
    Last edited: Aug 30, 2016
  10. Aug 30, 2016 #9
    Why is the tension not acting? According to me the tension is acting because if there is friction present between the bead and the string this means that there is contact between them and what I thought after thinking this much was that as tension is a force which follows Newtons third law then as the bead pulls at the string with its weight hence the ring also pulls via the tension.
  11. Aug 30, 2016 #10
    Which of the two ( Tension or friction ) is a contact force between the string and the bead ? As the bead slips , friction acts on it . And because of friction , tension exists in the string . Would there be any tension in the string if there were no friction between the string and the bead :wink: ?

    I repeat what I said in post#2 - There is a nice straightforward relationship between friction acting on the bead and tension in the string .
  12. Aug 30, 2016 #11
    Okay, so as the bead is not fixed to the string as it can slide on it, so the tension doesn't act on it instead the friction acts on both the string and the bead, so the straightforward relation ship you were talking about is ##T=f=\dfrac{mg}{4}##. Am I right.
  13. Aug 30, 2016 #12


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  14. Aug 30, 2016 #13
    Right .
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