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Friction MCQ

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data

    While walking on ice,one should take small steps to avoid slipping.Because the smaller step ensures
    (i)larger friction
    (ii)smaller friction
    (iii)larger normal force
    (iv)smaller normal force

    2. Relevant equations
    3. The attempt at a solution

    I have the following idea.While walking we put step such that frictional force by the ground helps us to proceed.Now,friction co-eff of ice being small,ice does not provide us with the friction---in other words,ice does not prevent our leg from slipping beyond our control.And we might fall.So,we make smaller,careful steps,avoiding hurry,so, that we get larger normal force,and we can safely proceed...

    so,in my opinion, the correct result should be (i) and (iii).

    But the answer given is (ii)
  2. jcsd
  3. Sep 16, 2007 #2


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    Homework Helper

    Yeah, I agree with you. I don't see how smaller friction causes less slipping.
  4. Sep 17, 2007 #3
    First I will quote my problem:

    While walking on ice,one should take small steps to avoid slipping.Because the smaller step ensures
    (i)larger friction
    (ii)smaller friction
    (iii)larger normal force
    (iv)smaller normal force

    Now,let us see how do we walk.I will cansider a man walking.His back leg pushes the ground back while his front leg (heel,to be specific) thrusts the ground forward.

    So,the relative motion at the point of contact is backwards for back leg and forward for front leg.

    Ground provides contact force at front and rear legs-these contact forces can be resolved to normal components and horizontal components.Horizontal components are known as friction.

    So,friction acts forward for the back leg (f₁) and backward for the front leg (f₂).

    The force equations would be: N₁+N₂ =mg
    & f₁-f₂ =ma
    As it turns out,it is not only the force equations that matter really.We are also to consider the torque equations.

    Let the difference between the legs in equilibrium be 2L
    Then,we have to have (N₁-N₂)L=(f₁-f₂)h where h is the CM height.

    Clearly less L means less (f₁-f₂) [which is significant for walking].Note that it is not individually f₁ or, f₂ that counts.It is the (f₁-f₂) term that is significant.

    So,less stride ensures less "effective" friction...

    In ice, the term (f₁-f₂) is very small,so,for safe walking L should be less.I mean,less stride ensures the proper condition of walking...

    And,lastly,I will say something why do people fall while walking on ice.I may neglect friction which--->0.
    Then,(N₁-N₂)L=0 and N₁=N₂
    Let the man lift his back leg from ice floor while walking.Then,only the torque -N₂L is working on him that makes him fall facing the heaven...and what if he stands and then starts to walk?

    in that case,there is nothing to drive him forward.I already assumed friction absent...

    So,if he tries to proceed,he will fall according to first law...
  5. Sep 17, 2007 #4
  6. Sep 17, 2007 #5
    That link really does not help too much.And I think my explanation is OK.
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