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Tension and centripetal forces

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data

    A stone of mass 4 Kg is attached to a string , its length 10 m .it moves in a horizontal circular path. The tension force of the string becomes 160 N then calculate its velocity.


    3. The attempt at a solution

    Its quite easy to solve when supposing the tension force = centripetal force while this is not given in the problem so how to solve it??

    If tension = centripetal force

    them 160 = m V^2/r
    160= 4 * V^2 /10
    then V = 20 m/s

    so my problem that"it is not necessary for the tension force to be equal to the centripetal force" so could you help me??

    thanks in advance
     
  2. jcsd
  3. Apr 6, 2010 #2

    Doc Al

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    Hint: The string is not horizontal.
     
  4. Apr 6, 2010 #3
    I dunno how this could help
    can u explain more please
    Thanks so much
     
  5. Apr 6, 2010 #4

    Doc Al

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    Analyze the forces on the stone. Hint: The only acceleration is horizontal. Use this to figure out the angle the string must make with the horizontal.
     
  6. Apr 8, 2010 #5
    mmm I think it is simpler than this
    but i think gravity hasn't any effect on the body because the body is moving in a horizontal circular path
    and gravity is vertical and centripetal force is horizontal so making an angle of 90 degrees

    then gravity hasn't any effect on the body , so the the centripetal force is equal to tension force

    i dunno if this right or wron
    but If F resultant (centripetal force) = Tension - gravity

    and since gravity has no effect on the body

    therefore centripetal force = Tension force

    right?
     
  7. Apr 8, 2010 #6

    Doc Al

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    Gravity certainly has an effect on the body. If you don't balance the gravitational force, the stone falls.

    You are given the tension in the string. That tension does two things: It balances gravity and provides the centripetal force. What angle must the string have in order to balance the stone's weight.
     
  8. Apr 9, 2010 #7
    dunno
     
  9. Apr 9, 2010 #8

    Doc Al

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    Figure it out! You know the tension and you know the weight of the stone. What string angle gives zero net force on the stone in the vertical direction?
     
  10. Apr 9, 2010 #9
    90 degrees?
     
  11. Apr 9, 2010 #10

    Doc Al

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    Not the angle that gives zero string force in the vertical direction, the angle that gives zero net force on the stone in the vertical direction. Another way of putting it: Gravity acts down on the stone. The string must exert a vertical force on the stone that just equals its weight. Solve for the angle that does that.
     
  12. Apr 12, 2010 #11
    I made this simple trial
    i think nothin wrong with it but still dunno why Fc = Ft bec theta is unknown
    [PLAIN]http://img535.imageshack.us/img535/9569/tensionp.jpg [Broken]
    or in order to make Fc sin theta = Fg
    then theta should be 90 degrees
     
    Last edited by a moderator: May 4, 2017
  13. Apr 12, 2010 #12

    Doc Al

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    Your diagram has the string slanting upwards towards the stone, but it should slant downwards. The string is pulling the stone up, not down.
    This is not correct, but what's stated in your diagram is correct.

    Ft cosθ = Fg

    Ft sinθ = Fc
     
  14. Apr 12, 2010 #13
    Sorry


    [PLAIN]http://img192.imageshack.us/img192/5111/tension.jpg [Broken]
    yes its not correct but how to solve the prob. without knowing that Fc =Ft??
     
    Last edited by a moderator: May 4, 2017
  15. Apr 12, 2010 #14

    Doc Al

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    You are given Ft and the mass. Start by solving for the angle.
     
  16. Apr 12, 2010 #15
    hmmm
    I should have replaced fg by f normal??

    anyways
    cos theta = 10/160
    theta = 86 dregrees

    160* sin 86 =Fc
    therefore Fc = 159.6

    but i found another problem
    the raduis length is not the same as string length :(
     
  17. Apr 12, 2010 #16

    Doc Al

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    There's no normal force here.

    What's Fg? Redo this.


    Figure out the radius from the string length and the angle.
     
  18. Apr 13, 2010 #17
    sorry its 40 N
    then theta = 76 degrees
    sin 76 = r /10
    r = 9.7 m
    Fc= Ft costheta = 160 * cos 76 = 155 N
    therefore
    Fc = m Vo^2/r

    by substituting Vo = 19.3 m/s approx.

    right?

    i know but i'm just confused bout the direction
    the string makes an angle of 76 degrees with the reaction of the ball to gravity
    so i thought of it as -Fg (and this inverts everythi in the problem upside down) so why do we neglect the direction ??

    thanks very much
     
  19. Apr 13, 2010 #18

    Doc Al

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    Looks good.

    The string makes an angle of 76 degrees with the vertical. I don't know what you mean by "reaction to gravity" or why you think some direction is neglected.

    Realize that in the vertical direction there is no acceleration:
    ΣFy = 0
    Ftcosθ - mg = 0

    Thus: Ftcosθ = mg
     
  20. Apr 15, 2010 #19
    Yes got it
    I'm just havin some prob with drawing free body diagrams as they don't teach us how to draw them at school.
    Thanks
     
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