Friction and Ice

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Homework Statement


While walking on ice, one should take small steps to avoid slipping. This is because smaller steps ensure
a)larger friction
b)smaller fiction
c)larger normal force
d)smaller normal force.

The Attempt at a Solution



I think that the answer should be a) or c) as both will help in balancing.
Any ideas?
 

Answers and Replies

  • #2
Larger friction. As the body weight is constant, the normal force will be constant. You can vary the friction however by shifting your weight balance.
 
  • #3
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Wrong again :biggrin:
 
  • #4
Lol. :D Nasty habit.

Whats the solution though? If not larger friction, it must be smaller friction. I dont see how the normal force could change in either case.
 
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  • #5
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Sorry for late reply. I was busy with my final exams.

Back to your question, I think we should look at it not in equilibrium. After all, we move due to forces being unbalanced. Let's view it this way: A person, when taking a step, is like a stick moving around the point of contact with the ground (when putting the other leg on the ground, he's temporarily back to equilibrium, then takes another step - the process is repeated). So the stick's movement is a circular motion around the point of contact.

When he changes from equilibrium to motion, he uses some energy from his muscles. But when he "falls" down, i.e. moving down from the vertical / highest position, since the average walking speed is not slow enough for muscles to act in this situation, it's likely that he doesn't use any energy - what being utilized is usually gravity.

Now some math:
[tex]P = N + Qcos\theta[/tex]
[tex]F = Qsin\theta[/tex]
Assuming that the person is a "uniform stick", in the extreme case where he falls down by gravity only, it can be proved that: [tex]Q = 6P(1-cos\theta)[/tex].
Therefore:
[tex]N = P - 6Pcos\theta(1-cos\theta)[/tex]
[tex]F = 6Psin\theta(1-cos\theta)[/tex]
However the situation is walking on the ice. No one would have that courage to take so big a step and fall, flat on his back. So:
[tex]N \approx P - 3P\theta^2 \approx P[/tex]
[tex]F \approx 3P\theta^3[/tex]
So N changes very little, and F matters. Maybe the answer is (b)? Let's justify it by doing further analysis :wink:

Say, P = 700N, [tex]\theta = 5^o-20^o[/tex].
_ At small angle (small step): [tex]\theta = 5^o[/tex] , N = 684N and F = 1.39N - he's safe!
_ At large angle (big step): [tex]\theta = 20^o[/tex] , N = 462N and F = 86.6N - quite a big step to hospital :biggrin:
Comparing 2 results, it can be seen that N changes "slightly", but there is a huge difference in F. Since what matters is the ratio F/N (as to compare with the static friction coefficient), F has the more influential role here.

This result is intuitive. Since ice is slippery, this implies that friction should be the one that matters. However fat or thin a person is, he still has to carefully walk on the ice. The end remains the same for every careless walkers :biggrin:

P.S.: I'm thinking about whether that person would most likely fall backwards (and be flat on his back) or forwards (and be flat on his face). Maybe you want to give it a try? :wink:
 

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  • #6
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Now some math:
[tex]P = N + Qcos\theta[/tex]
[tex]F = Qsin\theta[/tex]
Assuming that the person is a "uniform stick", in the extreme case where he falls down by gravity only, it can be proved that: [tex]Q = 6P(1-cos\theta)[/tex].
Suppose I start walking from the equilibrium position bringing my right foot forward while the left foot is still back (forming a triangle between legs). The friction on the right foot is backwards ( towards right in diagram ) and the friction on the left foot is forwards ( towards left in the diagram ).
Why did you take only one case with direction of friction towards left?

How did you prove this?
[tex]Q = 6P(1-cos\theta)[/tex]
By using torque equation and the equations above?
 
  • #7
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Suppose I start walking from the equilibrium position bringing my right foot forward while the left foot is still back (forming a triangle between legs). The friction on the right foot is backwards ( towards right in diagram ) and the friction on the left foot is forwards ( towards left in the diagram ).
Why did you take only one case with direction of friction towards left?
When you start walking, you are no longer in equilibrium: one leg is up, while the other, on which friction is exerted, is still on the ground. This is why the direction of friction in my diagram is towards left: there is only one frictional force on the leg.

How did you prove this?
[tex]Q = 6P(1-cos\theta)[/tex]
By using torque equation and the equations above?
Applying the law of energy conservation. Notice the assumption of the extreme case.


EDIT: This is just a model to explain the phenomenon. You can create your own model (e.g. the person starts walking from equilibrium) - just don't be satisfied with my answer; there are better ways to address the problem. I think you should try when having free time. Modeling is one important thing for not only scientists and researchers but also engineers - but sadly most engineers don't learn it.
 
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  • #8
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Applying the law of energy conservation. Notice the assumption of the extreme case.
Can you just write the equation? I still can't figure it out :confused:



EDIT: Modeling is one important thing for not only scientists and researchers but also engineers - but sadly most engineers don't learn it.
I am glad you learned it :approve:
 
  • #9
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Can you just write the equation? I still can't figure it out :confused:
Sorry, there is mistake in my calculation. Anyway here is how I deduced the equation:

Now since I assume that the person "falls" down by gravity, i.e. negligible initial speed & no internal energy used, plus that the person is a "uniform stick", from the law of energy conservation, we have:
[tex]mgR(1-cos\theta) = \frac{1}{2} \times \frac{4}{3}mR^2w^2[/tex]
where R is the distance from the center of the stick to the point of contact, w is the angular speed when the stick reaches to the angle [tex]\theta[/tex].
The centripetal force: [tex]Q = mw^2R[/tex]
Notice that [tex]P = mg[/tex], we have: [tex]Q = 3P(1-cos\theta)/2[/tex]
The result doesn't change much (as compared with the wrong one), so the conclusion remains the same.

I am glad you learned it :approve:
Then you should learn it :smile: Pretty simple: Imagine the situation, proceed explaining and whenever you find it too hard, make assumptions :biggrin:
 
  • #10
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Now since I assume that the person "falls" down by gravity, i.e. negligible initial speed & no internal energy used, plus that the person is a "uniform stick", from the law of energy conservation, we have:
[tex]mgR(1-cos\theta) = \frac{1}{2} \times \frac{4}{3}mR^2w^2[/tex]
where R is the distance from the center of the stick to the point of contact, w is the angular speed when the stick reaches to the angle [tex]\theta[/tex].
The centripetal force: [tex]Q = mw^2R[/tex]
Notice that [tex]P = mg[/tex], we have: [tex]Q = 3P(1-cos\theta)/2[/tex]
The result doesn't change much (as compared with the wrong one), so the conclusion remains the same.
Thanks alot! I got it now :smile:


Then you should learn it :smile: Pretty simple: Imagine the situation, proceed explaining and whenever you find it too hard, make assumptions :biggrin:
I will keep that advise :wink:
 

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