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Where does the friction come from?

  1. Mar 7, 2014 #1

    A uniform ball rolls smoothly from rest down a ramp at some angle.

    So for equation 2, why friction equals to -Icom*acom,x/R2 as shown?
    And does the equation on the right hand side come from the combination of equations 1) and 2)?
    I really can't figure it out...

    Thank you very much!

    Attached Files:

  2. jcsd
  3. Mar 7, 2014 #2


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    If there was no surface friction, for example, if this were a treacherous ice-coated pathway heading downhill from your door, then there would be no force to spin the ball, and the ball would simply skid downhill without turning, like the wheel on a bike when you have the brakes locked on. For the ball to roll around there must be friction so the ball grips the surface and experiences a "turning force". A fine rubberized surface on both ball and slope would work well, giving plenty of grip. A highly polished timber surface would NOT work, there would be insufficient friction for the ball to grip and be made to rotate, and without friction the ball would career downhill at greater speed (much greater, usually).
  4. Mar 7, 2014 #3


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    Dear Enoch,

    You have a tendency to avoid using the template by posting the exercise as a picture (or even somewhere far down a pdf), mostly also including some equations. Then you fire off one or two (well formulated, to the point) questions and top it off with an advance thanks or by throwing up your hands in despair.

    Not good PF practice, but some helpers fall for it anyway. Laziness isn't such a bad quality for a physicist: better to think a little longer and avoid unnecessary work.

    This minus sign business is recurring in your questions, so you should really invest some time in getting a grasp on that. It's a good investment: saves you lots of time later on and earns you higher scores.

    The bottom line is that we mostly calculate numbers as results: speed, acceleration, force, work, etc. But more often than not they are vectors and/or the outcome of expressions with vectors. A vector is a gizmo with a magnitude and a direction. After you agree that up is positive, gravitational acceleration ## \vec g ##, pointing down with magnitude ##|\vec g| = 9.81 m/s^2## appears in a calculation as e.g. ## v(t) = v(0) - g t^2## m/s. Where we replace g by 9.81 if we need a number result.

    In the ball on slope case, ##f_s## is pointing to the right, ## mg\sin\theta## to the left.
    I think the little ##x## on the right means that the positive x-direction is to the right, but that doesn't matter. The point is that ##f_s## and ##a_{\rm com}## have opposite signs.

    If your book were a little more thorough, they would write ##\alpha, \tau, a_{\rm com}, R## as vectors too, not only the various ##\vec F##. But perhaps they considered that too difficult at this point in the curriculum.

    And NO, the combination of 1 and 2 is not enough to extract ##a_{\rm com}##, as you implicitly already grasped. You need the "hint" in the top left box (to get a sum /difference) and you need uncle Isaac to bring in M.

    Now show me you are as smart as I think you are by posting the steps to find whatever is being asked for (which from now on you will carefully formulate under 1. The problem statement, all variables and given/known data).

    And: why this title for the thread? Is the O2isn answer adequate, or do you need more?
  5. Mar 8, 2014 #4
    Dear Bvu,

    Sorry for my bad work and thank you for your reminder.

    That was because my exam was near and I was afraid that the question posted would appear in the exam. That's why I asked in such a lazy way. I'm sorry for this and I admit that I did't start studying for the exam early enough.

    In fact, the question comes from some homework of a university. And I did try to understand what's happening in the question. I've even found out the related lecture powerpoint through search engine for more detailed solutions but it was not enough.

    So the question is:
    A uniform ball, of mass M=6kg and radius R, rolls smoothly from rest down a ramp at angle 30.0o.
    a) The ball descends a vertical height h=1.2m to reach the bottom of the ramp. What is its speed at the bottom?
    b) What are the magnitude and direction of the frictional force on the ball as it rolls down the ramp?

    I understand part a,
    and for part b, which is something I actually want to ask,
    the solution of the homework gives the equation on the right on the picture I posted.
    I don't understand this.
    And the picture I posted actually comes from the related lecture powerpoint.
    It helped a little, but as you see, I'm still very confused.

    So actually I want to ask part b of the question.
    But I was lazy and mistakenly thought it would be simpler for me to ask that way.

    Anyway, what NascentOxygen answered did help me to understand the concept more though it isn't really the answer I'm looking for.

    And what I thought of after reading part b was that Ffriction = Fgsintheta...
    Surely, I was completely wrong.

    I understand equation 1 on the picture,
    and the former part of equation 2,
    and that's all.
    Can you please guide me a little bit?
  6. Mar 8, 2014 #5
    And for the topic, I'm sorry it's not clear enough as I didn't quite have time to think for a better one. And you're right, I need something more...
  7. Mar 8, 2014 #6


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    So did you or did you not get an answer for part a) ? Can you show what you got and how you got it ?

    Because if you assume Ffriction = Fgsintheta, then there is no net force downward along the slope, so no acceleration.

    Then: the left part of eq 1) ##\tau = I \alpha## apparently ("##\rightarrow##") gives rise to ##Rf_s=I_{\rm com} \alpha##. Which you understand, right? Around the center of mass as center of rotation, the torque is ##Rf_s## and the moment of inertia is ## I_{\rm com}##. ##\quad \tau = I \alpha## is the angular equivalent of ##F = m\ a##.

    The left part of eq 2) you also understand. I do not. I would say that if a is down the slope and if ##\alpha## is positve (counter-clockwise), and there is no slipping, then one should write ## a_{\rm com} = - \alpha R##. But perhaps they state the left part as a general condition of no slipping and add a little x and a minus sign on the part after the arrow to put it right, hence the ## f_s= {I_{\rm com} \alpha\over R} ## (from eq 1, the part after the arrow), after substituting ## \alpha = - a_{\rm com}/R ## becomes $$ f_s= - {I_{\rm com} \ a_{\rm com}\over R^2} $$Now in the top left box it says ##\vec F_g\sin\theta ## and ##\vec f_s## determine the linear acceleration down the ramp. The former is in the negative x direction, the latter in the positive. The resultant is ##f_s - mg\sin\theta## and that they work out to the ## a_{\rm com} = ## ... on the lower right.

    Enoughenough from me, over to you.
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