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Rope around Pole, friction intuition

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data

    a rope wraps around an angle theta around a pole. You grab one end and pull with a tension T. The other end is attached to a large object say a boat. If the coefficient of stating friction between rope and pole is mu, what is the largest force the rope can exert on the boat, if the rope is not to slip around pole.

    2. Relevant equations
    -


    3. The attempt at a solution
    I know the solution to the question. What I am curious about is what is the intuition behind the friction in this problem. Which way does the friction force act, and how would you put it on your free body diagram. I need some hints/ intution.

    Thanks
     
  2. jcsd
  3. Sep 19, 2010 #2

    Delphi51

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    Homework Helper

    The friction force can go either way! But it always opposes motion.
    So, if you are holding the boat and it is trying to get away, Ff helps you pull on the boat. If you are trying to make the boat move toward the pole, Ff hinders your pull making it harder to pull the boat.
     
  4. Sep 19, 2010 #3
    Another important thing to remember is that the force of friction can only act tangent to the surface in question.
    The normal force is always normal to the surface, and the force of friction is always tangent to it.

    Two other things to keep in mind is that for the critical case, the rope is just on the verge of slipping, so for every pair of friction and normal forces, there holds [tex]dF_{friction}=\mu_s dN[/tex]

    As for an initial approach, I suggest you look at a small arc subtending an angle [tex]\Delta \theta[/tex], or if you're comfortable working with the differentials straight away, a differential angle, [tex]d\theta[/tex], and apply Newton's Second Law to it, claiming equilibrium and using the small angle approximations: [tex]\sin{x} \approx x[/tex], [tex]\cos{x} \approx 1[/tex] for small [tex]x[/tex]
     
    Last edited: Sep 19, 2010
  5. Sep 19, 2010 #4
    Thanks- That's very good intuition and reminders. I did look at a small arc dtheta and my y balance looks like this:

    * T(theta +dtheta) + cos (dtheta/2) -T(theta)cos(dtheta/2)- Friction=0

    My issue was whether it would be -Friction or +Friction.
    If I think about the direction motion to be towards the boat moving away, then normal force is positive on the x balance, and so friction is negative on *.

    Did that make sense?

     
  6. Sep 20, 2010 #5
    Yep, that makes perfect sense, but you made two small mistakes, it should be [tex]T(\theta + d\theta)\cdot \cos{\frac{d\theta}{2}} - T(\theta)\cdot \cos{\frac{d\theta}{2}}=dF_{friction}[/tex]
    The force of friction is infinitesimal, since you're looking at the contribution of a tiny arc segment, the same goes for the normal force, it too is infinitesimal.

    Note that since you're dealing with small differences, you can write: [tex]T(\theta+d\theta)-T(\theta)=dT[/tex]
     
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