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Looking for help with torque

  1. Jul 23, 2008 #1
    A 55g mouse runs out to the end of the 17cm -long minute hand of a grandfather clock when the clock reads 10 minutes past the hour.

    What torque does the mouse's weight exert about the rotation axis of the clock hand?
    in Nm

    Im using the eqn. (torque)=rFsin(theta)
    I put

    (torque)=(.17m)(.539the mg)sin(150) and im wrong...I think its the angle or the force....

    The mouses weight should be Mg right? since its the only acting force?
    And if a clock is at 2 then its 30 degree above the horizontal and thaking the angle perpendicualr to that is 150 degrees....
    I d K.....
     
  2. jcsd
  3. Jul 23, 2008 #2

    Doc Al

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    Re: Torque

    Looks right to me. Why do you think it's wrong?
     
  4. Jul 23, 2008 #3
    Re: Torque

    well my answer is .09 Nm and it says im wrong....
     
  5. Jul 23, 2008 #4

    Doc Al

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    Re: Torque

    Redo your calculation. The formula you had before looked OK to me, but that answer does not.
     
  6. Jul 23, 2008 #5
    Re: Torque

    I got 4.58e-2...still wrong..I dont see what im doing wrong?
     
  7. Jul 23, 2008 #6

    Doc Al

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    Re: Torque

    That looks OK to me. Do they want you to include direction? (Like clockwise or counterclockwise?) Those online systems can be picky (and sometimes wrong).
     
  8. Jul 23, 2008 #7
    Re: Torque

    NOpe they just want it in N m.....I hate online homework...thanks for your help.
     
  9. Jul 23, 2008 #8

    alphysicist

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    Re: Torque

    Hi Lance WIlliam,


    The minute hand is 30 degrees above the horizontal, so that's right. However, I don't believe the sin(150 degrees) in your equation is correct. Do you see what it needs to be?
     
  10. Jul 23, 2008 #9

    Doc Al

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    Re: Torque

    At 10 minutes past the hour, the minute hand will be 30 degrees to the right of the vertical or 60 degrees above the horizontal (making the angle between "r" and "F" 150 degrees). Or did I misread the problem?
     
  11. Jul 23, 2008 #10
    Re: Torque

    if its 10past past the hour its at 2 on the clock so since 3 is a horizontal on the clock the minute hand its 30degrees above it.In toque the (theta) used is the angle between the r vector and the F vector thus they are perpendicular to each other...I get 150 degrees....maybe im doing it wrong...
     
  12. Jul 23, 2008 #11

    alphysicist

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    Re: Torque

    Hi Doc Al,

    At 10 minutes past the hour, the minute hand has made 1/6 of a revolution, so I believe it's 60 degrees from the vertical.
     
  13. Jul 23, 2008 #12

    Doc Al

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    Re: Torque

    D'oh! You're right, of course. I was thinking of the minute hand pointing to 1 o'clock, which is 5 minutes past the hour. :redface:

    (Good thing I use a digital watch!)
     
  14. Jul 23, 2008 #13

    Doc Al

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    Re: Torque

    Yes, the r vector is 30 degrees above the horizontal, but the F vector (the weight of the mouse) points down and is thus 90 degrees below the horizontal. They are not perpendicular. (And if they were, that would be 90 degrees, not 150!)

    Given that, what's the angle between r and F?
     
  15. Jul 23, 2008 #14
    Re: Torque

    I get 60 degree as my angle between the two vectors
     
  16. Jul 23, 2008 #15

    Doc Al

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    Re: Torque

    That will work fine, for your purpose. (Since [itex]\sin\theta = \sin(180 - \theta)[/itex].)

    But I'd say to draw yourself a picture. The r vector is 30 degrees above the horizontal, while F is 90 degrees below. What would the angle be between those two vectors?
     
  17. Jul 23, 2008 #16
    Re: Torque

    THANKYOU for your Help! :)
     
  18. Jul 23, 2008 #17

    Doc Al

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    Re: Torque

    You are most welcome!

    Thanks to alphysicist for catching the error.
     
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