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Homework Help: Force and Kinetic friction problem

  1. Nov 2, 2008 #1
    A sled weighing 100 N is pulled horizontally across snow so that the coefficient of kinetic friction between sled and snow is .275. A penguin weighing 50 N rides on the sled

    to start with the penguin digs in his claws so he is firmly attached to the sled. What value of F do you need for the sled and penguin to move at constant speed??

    after a while, our penguin gets tires od holding on with his claws. Now the coeffient of static friction between penguin and sled is .750 find the maximun horizontal force F that can be exerted on the sled before the penguin bgins to slide off?
     
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  3. Nov 2, 2008 #2

    tiny-tim

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    Hi bmandrade! :smile:

    Show us what you've tried, and where you're stuck, and then we'll know how to help. :wink:
     
  4. Nov 2, 2008 #3
    Ok well i've been doing some diagrams
    In my diagrams for 1st part I show that there is a force pulling the sled and opposite to that there is kinetic friction. Force due to gravity is pointing down and a normal pointing up.
    for the penguin I show a force which is opposite to the one that pulls the sled. Fg downwards and a normal upward.

    For sled
    I found the normal to be 100 N since its the same as Fg which 100 N. I multiplied this number times the Kinetic friction to find the force due to kinetic friction (100 N * .275 = 27.5) I knwo that the sum of forces on the horizontal direction are going to equal ma
    and mass is 100N/g = 10 Kg (g=10m/s^2)
    so
    sum of all forces in x = F - 27.5 = 10kg (a)

    now im stuck
     
  5. Nov 2, 2008 #4

    tiny-tim

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    Excellent … :biggrin:

    and constant speed means the acceleration, a, is … ? :smile:
     
  6. Nov 2, 2008 #5
    acceleration = 0
     
  7. Nov 2, 2008 #6
    does that mean that the minimum value for the sled to move will be 27.5 N????
     
  8. Nov 2, 2008 #7

    tiny-tim

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    … poor little penguin!

    Yes … except you forgot the penguin!! :smile:
     
  9. Nov 2, 2008 #8
    ok... I dont know how to deal with the penguin
     
  10. Nov 2, 2008 #9

    tiny-tim

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    Feed it fish, of course! :rolleyes:

    I meant, you forgot its weight! :smile:
     
  11. Nov 2, 2008 #10
    lol


    mmmm i not sure but do I add penguins weight to the force???
     
  12. Nov 2, 2008 #11

    tiny-tim

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    You add the penguin's weight to the sled's weight to calculate the normal force, to calculate the friction force. :smile:
     
  13. Nov 2, 2008 #12
    oh so its the weight of the whole system
    in that case the Force = 41.25 N
    becasue Force due to friction is 150(.275) = 41.45 N


    so then I use this force to solve part 2 right?

    if im correct then the Static friciton between penguin and sled is
    (.750*50)= 37.5 N

    so the sum of forces in x direction = force of kinetic friction - force= 0??????
     
  14. Nov 2, 2008 #13

    tiny-tim

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    Yes. :smile:
    No … it's a different F … you start again.
    Yes. :smile:
    Nooo … for the penguin, the sum of forces in x direction = force of kinetic friction = mass x acceleration. :wink:
     
  15. Nov 2, 2008 #14
    so for part 2 the force will be different is that force going to be 27.5 or no??

    and for the last part where you said no i dont know how to do that because there is no acceleration
     
  16. Nov 2, 2008 #15

    tiny-tim

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    No … the force is unknown … it's the (different) F which you have to calculate.
    ah … yes there is … the penguin won't come off without acceleration, will it? :smile:
     
  17. Nov 2, 2008 #16
    ok so can you help me figure this out because i have no clue how to do it
     
  18. Nov 2, 2008 #17

    tiny-tim

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    The penguin will start to slide when the acceleration is enough to balance the maximum possible friction force …

    so that gives you the acceleration of the penguin …

    the acceleration of the sled-plus-penguin is the same (just before it starts to slide) …

    so that gives you the force F on the sled-plus-penguin.

    Have a go! :smile:
     
  19. Nov 2, 2008 #18
    sorry but i still dont get it

    this is what i have
    the Fs= 37.5 N
    the sum of F is x = Fs - F = ma

    but i dont know what F is so i cant solve for a
     
  20. Nov 2, 2008 #19

    tiny-tim

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    No (and what's x? :confused:) … it's only Fs = ma … the friction is an internal force, between the penguin and the sled …

    so the friction will only show up in a Newton's second law equation for either the penguin on its own, or the sled on its own …

    in this case, do the penguin on its own, to find a, then do the penguin-plus-sled, to find F.
     
  21. Nov 2, 2008 #20
    ok so

    the penguins acceleration will be .75 m/s^2
    i used fs=ma so 37.5 = 50a -------> a = .75 m/s^2


    and i use this value for F=ma

    which will be F = (.75m/s^2) * 150N = 112.5?
     
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