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Finding mass of an object being pulled on a slope

  1. Sep 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A girl of mass mg = 52 kg is pulling a sled up a slippery slope. The coefficient of friction between the girl's boots and the slope is µs = 0.157; the friction between the sled and the slope is negligible. The girl can pull the sled up the slope with a ≤ a max = 0.015 m/s2 before she begins to slip. Assume the rope connecting the girl to the sled is kept parallel to the slope at all times. The angle of the slope is θ = 8°.

    ONYxHIM.png

    2. Relevant equations

    (All below)

    3. The attempt at a solution

    Net Force on girl:

    T=mg(a+gsin(8)+Us(gcos(8))
    T=52(((0.015+(9.8sin8)+(0.157)(9.8)(cos8))
    T=150.9N

    Net Force on sled:

    ma=T-mgsin8
    m(a+gsin8)=T
    m= (150.9)/(0.015+9.8sin8)
    m= 109.5 kg

    It's wrong. Anything I did wrong? Appreciate any help!
     
    Last edited: Sep 19, 2015
  2. jcsd
  3. Sep 19, 2015 #2

    Orodruin

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    You might want to check the signs in your equations.
     
  4. Sep 19, 2015 #3
    Should I be subtracting the first part so

    T=mg(a+gsin(8)-Us(gcos(8))

    I really am lost now.
     
  5. Sep 19, 2015 #4

    Orodruin

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    So go back to Newton's second law. What does it tell you?

    Edit: You also have a g too much multiplying that parenthesis.
     
  6. Sep 19, 2015 #5
    a=Fnet/m I don't know where to go from here. I already broke into FBD and tried to solve it already. When I redid it I ended up with the same thing.

    Oh, that g is a typo so it's not affecting my current calculation.

    T=mg(a+gsin(8)-Us(gcos(8))
     
  7. Sep 19, 2015 #6

    Orodruin

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    There it is again ...

    Yes, this is a good way to go. So which forces act on the girl (along the slope direction) and in which direction do they act? Which direction do you define as positive? Uphill or downhill?
     
  8. Sep 19, 2015 #7
    T=m(a+gsin(8)-Us(gcos(8)) my bad.

    In the y-direction there is a component of weight and also the normal force. positive would be uphill. so the component of weight is negative and the normal force would be positive. Now I get: Fny=m(a+gcos8) so here I would subtract the x component? adding sounds more correct.
     
  9. Sep 19, 2015 #8

    Orodruin

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    You are trying to go too fast and somehow mess up. You have not said what your "y-direction" is and we are not mind readers. You have also not defined Fny.

    Write down, in turn and for each force acting on the girl: 1) The kind of force. 2) Its magnitude. 3) Its direction along the slope.

    It will help you organize yourself to write these things down.
     
  10. Sep 19, 2015 #9
    I have written it down, the y direction would be the direction of Fn in this case.

    But thanks for the help, I'll do it later. No need to reply after this.
     
  11. Sep 19, 2015 #10

    Orodruin

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    No you have not. How are we supposed to help you figure out where you went wrong when you refuse to show how you have been thinking? To tell me that you have it written down on a paper tells me absolutely nothing about how you have been thinking with respect to the actual problem.
    You also cannot say "the direction of Fn" without defining in thread what Fn is.
     
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