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Rotational Force Problem

  1. May 5, 2008 #1
    [SOLVED] Rotational Force Problem

    1. The problem statement, all variables and given/known data
    Consider a conical pendulum with a 89.0 kg bob on a 10.0 m wire making an angle of = 5.00° with the vertical (Fig. P6.9).

    http://img89.imageshack.us/my.php?image=p613qy9.gif


    3. The attempt at a solution
    I am having alot of trouble figuring this out. I have tried to set up force diagrams but I am getting confused. I know that the force is the y direction is just mg but to make it this way it would have to be
    [tex]Tsin(\theta) - mgsin(\theta) = 0[/tex] and I don't understand why it would be mg multiplied by the sine of the angle.
    Also, for the second part I don't understand why the answer is not just mg times the sin of the angle.
    Can someone help me figure out this problem? Thanks.
     
  2. jcsd
  3. May 5, 2008 #2

    alphysicist

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    Homework Helper

    Hi Sheneron,

    I'm not certain which direction you're trying to write that equation for. However, you're right that mg is completely vertical; also, the tension has both vertical and horizontal components.

    For this problem, you'll also need the acceleration (magnitude and direction). What do you get?
     
  4. May 5, 2008 #3
    For the horizontal tension the equation I get is
    [tex]Tcos(\theta) - mgsin(\theta) = 0[/tex]
    For vertical I get
    [tex]Tsin(\theta) - mgsin(\theta) = 0[/tex]

    I am pretty sure those have to be the answers, but I can't see where they came from. I don't understand how to set the force diagram up.
     
  5. May 5, 2008 #4
    Now I am very confused... ma would not equal 0 would it? To find the acceleration you would first have to know the tension, which I am having alot of trouble on
     
  6. May 5, 2008 #5
    ohhhhhhhh I think I got it

    [tex]Tcos(\theta) - mg = 0[/tex]
    [tex]T = \frac{mg}{cos(\theta)}[/tex]

    [tex]T_y = Tcos(\theta) = \frac{mg}{cos(\theta)}cos(\theta) = mg[/tex]

    [tex]T_x = Tsin(\theta) = \frac{mg}{cos(\theta)}sin(\theta)[/tex]

    [tex] F_x = ma_r[/tex]

    [tex]\frac{mg}{cos(\theta)}sin(\theta) = ma_r[/tex]

    [tex]a_r = \frac{mgtan(\theta)}{m} [/tex]
     
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