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Dashing up the Slope

  1. Sep 23, 2007 #1
    A girl of mass m_G is walking up a slippery slope while pulling a sled of unknown mass; the slope makes an angle theta with the horizontal. The coefficient of static friction between the girl's boots and the slope is mu_s; the friction between the sled and the slope is negligible. It turns out that the girl can pull the sled up the slope with acceleration up to a without slipping down the slope. Find the mass of the sled m_S. Assume that the rope connecting the girl and the sled is kept parallel to the slope at all times.

    [​IMG]
    [​IMG]

    Find the mass of the sled m_S.
    Express the sled's mass in terms of the given quantities and g, the magnitude of the acceleration due to gravity.


    I would like to think it involves using something like:
    Code (Text):

    cos(theta)=x/(Mg-Ms)
     
    What other variables and I missing?
     
  2. jcsd
  3. Sep 23, 2007 #2

    mgb_phys

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    Your force diagrams are correct - so assuming have done the trig right you should be there.
     
  4. Jul 9, 2008 #3
    I don't understand what you are saying that "x" represents in your equation, this isn't a given variable.
     
  5. Jul 9, 2008 #4
    sin(theta)*(m_G+m_S)=mu_S*m_G*cos(theta)
    m_S=m_G(cot(theta)-1)

    is this correct?
     
  6. Jul 9, 2008 #5

    tiny-tim

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    use Newton's second law

    Hi Nitrag! :smile:

    It's no good picking a formula you're familiar with, and hoping it fits.

    Hint: use Newton's second law along the slope:

    mGa + … = … ? :smile:

    EDIT: vipulsilwal, you're doing it again! :frown:

    Don't try to answer the question … just offer help.

    (And when you're not sure of the answer, wait until the end, and then ask!
    Remember, it's not your thread. :smile:)​
     
    Last edited: Jul 9, 2008
  7. Jul 9, 2008 #6
    ok sorry.,,
    usual way of starting such problems is
    frictional force is acting only on girl .
    how to determine its direction?,,jst see where girl would slide if there was no friction force.
    friction force is preventing that to happen...so it is in upward direction...

    now, balance upward n downward forces...m_S would be the maximum one.
     
  8. Jul 9, 2008 #7
    I have already done the force diagram and free body diagrams and such, as well as identified the action reaction pairs. I think their is an error in your calculation of m_s though, wouldn't it have to include "a" somewhere in there?
     
  9. Jul 9, 2008 #8
    i admit it was wrong...
    m_G*a have to be include in upward direction..

    thanks chipper for correction
     
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