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Friction and relative motion 2

  1. Nov 13, 2008 #1
    1. The problem statement, all variables and given/known data

    A sled weighing 60 N is pulled horizontally across snow so that the coefficient of kinetic friction between sled and snow is uk = 0.100. A penguin of 70 N rides on the sled. If the co. static friction between the penguin and the sled is us = 0.700 find the maximum horizontal force that can be exerted on the sled before the penguin begins to slide off.


    2. Relevant equations

    fs = us R, fk = uk R.

    3. The attempt at a solution

    Please see the attached file for the force diagrams.

    Forward force on the sled = F - fs - fk

    In the absence of the static friction, penguin will slide backward with this force. But the penguin expereinces a force due to static friction fs.

    F - fs - fk = fs

    F = 2 fs + fk = 2 us R1 + uk R2 = 2 us m1g + uk (m1+m2)g = 111 N

    Book answer is 104 N.

    Can you check my force diagrams please?

    Thanks.
     

    Attached Files:

  2. jcsd
  3. Nov 13, 2008 #2

    tiny-tim

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    Homework Helper

    Hi Nuha99! :smile:

    I can't see your diagrams yet, but nor can I see anywhere in your equations where you've worked out the acceleration that the penguin can take.

    Hint: start by applying Newton's second law to the penguin, and find that acceleration. :smile:
     
  4. Nov 13, 2008 #3
    Attached file is not validated yet.

    I did not find the acceleration of the sled. I found the net forward force on the sled. According to my understanding this net forward force is what will make the penguin to slide off.

    Forward force on the sled = F - fs - fk

    Note that fs is the static frictional force on penguin between penguin and the sled surface. By newton's 3rd low, same but opposite force should act on the sled. This is the origin of fs in the above expression.

    Now for the penguin to be stable, static frictional force = F - fs - fk

    fs = F - fs - fk

    Apparently some thing is not right. I solved this for F.
    See my first post.

    Thanks.
     
  5. Nov 13, 2008 #4

    tiny-tim

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    Hi Nuha99! :smile:

    If there is acceleration (which there is),

    then the force diagram must include accleration …

    and so must the equations.
     
  6. Nov 13, 2008 #5
    That does it. Thank you.
     
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