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Homework Help: This one will tackle your physics skills it tackled mine !

  1. Nov 12, 2007 #1
    1. The problem statement, all variables and given/known data

    A mass of 5 kg is tied to two strings of equal length, which are attached to a vertical pole at points 2.o m apart. As the pole rotates about its axis, the strings make a right angle with each other. Also, the mass makes 2 revolutions per second.
    a) the tension in each string
    b) the speed of revolution that makes the lower string go slack

    2. Relevant equations

    F net = F c ( Centripetal force)

    3. The attempt at a solution

    I calculated the total centripal force acting on the mass.
    FNET = Fc
    Fc = mv^2 / r = m ( 2pi r (revolutions) ) ^2 / R
    Fc = 790

    I know that the FNETy = 0 because the object is always forming that angle and is constant at it.
    Therefore, Force in the y direction on the mass = the total of the forces in the y direction for both the strings and of the mass.
    I came up with the equation,
    0 = T1 ( top string) SIN 45 - T2 ( bottom string) SIN 45 - MG
    0 = T1 ( top string) SIN 45 - T2 ( bottom string) SIN 45 - 10 (5)
    0 = T1 ( top string) SIN 45 - T2 ( bottom string) SIN 45 - 50

    I do not know how to get the forces in the x direction for the strings.

    Please help.

    Attached Files:

    Last edited: Nov 12, 2007
  2. jcsd
  3. Nov 12, 2007 #2
    if the net force in the y direction is zero, which component of the strings' forces provides the centripetal acceleration?
  4. Nov 12, 2007 #3
    the only present force is in the x direction
    which is 790 N
    there is no force in the y direction
  5. Nov 12, 2007 #4
    i would suggest drawing free-body diagrams of all the objects involved.
  6. Nov 12, 2007 #5
    ok do you want me to draw them and attach them to the image
    so you want to see a FBD of the mass
  7. Nov 12, 2007 #6
    actually i cant see your image...but it should help you to understand the forces at work.
  8. Nov 12, 2007 #7
    it is a JPG image
    i have put a new one
    which might help you understand better
    i forgot to mention in the picture, that the distance betweeen the two strings on the pole, sorry not wall as i wrote on the picture is 2 m
  9. Nov 12, 2007 #8
    actually they need to be approved before we can see them. i understand the problem fine, and can draw the picture as well. I'm just asking if you have labeled all of the forces at work on the strings and the mass.
  10. Nov 12, 2007 #9
    the forces on the strings
    - gravity
    - the force of the object
    - maybe the reaction force from the pole

    the forces on the mass
    - gravity
    - the two strings pulling it
  11. Nov 12, 2007 #10
    btw how long does it take to approve pictures
  12. Nov 12, 2007 #11
    do the strings have mass?

    also, i have no clue, but you could upload them to photobucket
  13. Nov 12, 2007 #12
    um no consider the strings have no mass
    i haven't done physics with the mass of the strings involved too
    photo bucket?
    what is that
  14. Nov 12, 2007 #13
    photobucket is a site you can upload pics to in order to link them, but yours are approved so don't bother.

    you seem to have the right idea in your drawings. if you look at the forces on the strings, you see that they are the tension which causes the Acent for the object, and the tension which keeps the object from falling due to gravity. therefore the total force on both strings needs to nullify these two other forces.
  15. Nov 12, 2007 #14
    i kind of got what you meant
    but can you help me understand
    how to get the force in the x direction for each string
    so i cant get the total force ( tension) in each string as my question asks
  16. Nov 12, 2007 #15
    here's a hint: what should the tension in the X and Y directions on each string be, since the angle is 45 degrees?
  17. Nov 12, 2007 #16
    is the tension in the upper string
    790 Cos 45
  18. Nov 12, 2007 #17
    if that were the answer, then the tension in each string would be equal. I'm fairly certain this is not the case.
  19. Nov 12, 2007 #18
    yah you are right
    i am blank
    i understand that the tension in each string is the hypotnuse of a 90-45-45 triange
    but i still cant get my head around how do i got about finding it
  20. Nov 12, 2007 #19
    well the y components of the tension are in opposite directions and yet add up to Fg

    what about the x components?
  21. Nov 12, 2007 #20
    no no
    look at the picture
    the y component of tension in the upper string ( US Y ) equals
    the y component of tension in the lower sting ( LS Y ) and the force pulling the weight down which is mg which is 50 N
    US Y = LS Y + 50 N
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