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Newton's laws / Friction 2

  1. Aug 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A 4.0-kg toboggan rests on a frictionless icy surface, and a 2.0 kg block rests on top of the
    toboggan. The coefficient of static friction µs between the block and the surface of the toboggan is 0.60, whereas the kinetic friction coefficient is 0.51. The block is pulled by a
    30 N horizontal force as shown. What are the magnitudes and directions of the
    resulting accelerations of the block and the toboggan?


    2. Relevant equations

    Ffr = [tex]\mu[/tex]s Fn = [tex]\mu[/tex]s mg
    Ffr = [tex]\mu[/tex]k Fn = [tex]\mu[/tex]k mg

    Fn = mg

    [tex]\Sigma[/tex]F= ma


    3. The attempt at a solution

    Fn= (m1 + m2) g
    Fn= (2.0kg + 4.0kg) (9.80m/s2)
    Fn= 58.8 N

    Now I calculate the friction of both separately: (im not sure about this part)

    Box:

    Ffr = 0.60 (58.8N)
    Ffr = 35.25N

    Tob:

    Ffr = 0.51 (58.8N)
    Ffr= 29.988N

    Then I attempt the acceleration:

    [tex]\Sigma[/tex]=m1a
    Fr-F=ma
    35.25N-30N = (2kg) a
    5.25N/2kg = a
    2.625 m/s2 = a

    [tex]\Sigma[/tex]= m2a
    F-Ffr =ma
    30N-29.988N = (4kg) a
    0.003 m/s2 = a (or 3 x 10-3)

    This is where I get confuse. When the box is pulled doesn't it stay static while the sled slides on the ice? And what do they mean by direction?

    Any help would be great :)
     

    Attached Files:

  2. jcsd
  3. Aug 14, 2009 #2

    Delphi51

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    Remember the ice is frictionless, so the only friction you need to work out is for the box on the toboggan, Ff = 0.6*2*g = 11.77 N.
    This means the "grip" of the toboggan on the box is 11.77N - the maximum force without slipping, so the maximum acceleration is a = F/m = 11.77/2 = 5.89 N.
    If we are trying to accelerate more than that, we will have a complex problem where the box slides along the toboggan. However, we are only trying to accelerate at
    a = F/m = 30/6 = 5 m/s2, so there will be no sliding.
     
  4. Aug 14, 2009 #3
    But dont we know from the very beginning that the toboggan will not slide just by looking at the static and kinetic friction values.

    Doesn't [tex]\mu[/tex]s mean that it will only move if it is higher than 0.60? And since the [tex]\mu[/tex]k is smaller we know it will not accelerate?
     
  5. Aug 14, 2009 #4

    Delphi51

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    The toboggan must slide on the ice - a force is applied and there is no opposing force.
    I don't understand this. You use the static coefficient if the box is not sliding, the kinetic one if it is.
     
  6. Aug 14, 2009 #5
    Sorry. I have trouble explaining physics.

    Ok, so basically it was a trick question because we actually do not need the [tex]\mu[/tex]k which is 0.51 for the toboggan?
     
  7. Aug 14, 2009 #6

    tms

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    [tex]\mu_s[/tex] is usually larger than [tex]\mu_k[/tex], for any surface. Think of your normal experiences when pulling something. Once you get it started sliding, is it harder or easier to pull than before it started moving?
     
  8. Aug 14, 2009 #7
    Definitely easier to pull after it started sliding. Makes sense! Thanks.
     
  9. Aug 14, 2009 #8

    Delphi51

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    Right, you don't need the kinetic friction this time.
     
  10. Jul 13, 2011 #9

    Delphi51

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    Oops; I made a mistake here! The mass of the toboggan is 4, not 2. So the box does slide along the toboggan and you have to use kinetic friction. Thanks to a sharp student for catching this! I commented on the same question correctly 6 months before making this mistake. Sorry, bumblebeeliz.
     
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