1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Friction on an accelerating train

  1. Nov 24, 2006 #1
    I have a question for homework, and this is a high school physics class. The question reads:

    Suppose that you are standing on a train accelerating at 2.0m/s^2. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide?

    I've drawn a free body diagram, and motion in the Y dimension doesnt really matter.
    In the X dimension, I THINK there are 2 forces: Force of the train, and Force of friction. What I've come up with is this:

    Fnetx=0 (Since I'm not accelerating in relation to the train itself. I'm hoping this assumption is right, or everything else is wrong.)
    Fnetx=Ft-Ff
    Ft=Ff
    Ff=9.8mu, and Ft=2m (This is where I'm getting sketchy. Is the force of the train on the person the same as acceleration x the persons mass?)
    9.8m/2m=u
    u=.2

    I'm basically posting to verify my answer, since I'm not entirely sure if the route I took to get to it was right. Thanks for reading!
     
  2. jcsd
  3. Nov 24, 2006 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Looks good to me except this bit;
    It should read; 2m/9.8m , however, your final answer is correct so I assume its a typo.
     
  4. Nov 24, 2006 #3
    Oh yeah, thanks. So the coefficient is .2, and the force on the person is his mass times the trains acceleration, then?
     
  5. Nov 24, 2006 #4

    Doc Al

    User Avatar

    Staff: Mentor

    I don't really understand your reasoning. What's Ft?

    Rather than try to describe things from the accelerated frame of the train, simply describe it from the frame of the tracks. Answer these questions:
    (1) For you not to slide, what must your acceleration be?
    (2) What force provides that acceleration? How does it relate to your weight and the coefficient of static friction?
    (3) Use that to deduce the minimum coefficient of static friction needed.
     
  6. Nov 24, 2006 #5
    Ft is, um, Force of the train. I didn't know what else to label it as. The only thing I was confused about is whether or not the Force of the train on me is it's acceleration times my mass.
     
  7. Nov 24, 2006 #6

    Doc Al

    User Avatar

    Staff: Mentor

    There's no such thing as "force of the train". The only force acting on you (horizontally, at least) is the friction from the floor of the train.

    (You can view things from an accelerating frame of reference, such as the train. But that introduces so-called "fictitious" forces that are artifacts of using such a non-inertial viewpoint. Things end up being close to what you did, but for very different reasons. I suggest forgetting about such advanced methods and sticking to real forces for now.)
     
  8. Nov 24, 2006 #7

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What Doc Al is saying, is that you can also consider this problem from an inertial reference frame. This could be someone stood at the side of the tracks watching you accelerate past at 2m/s2. Now, the force required to accelerate you at this rate is twice your mass, yes? Now, this force is provided by friction between the floor of the train and your shoes; so we can say that [itex]2m = mg\mu[/itex], which is equivalent to your expression and you will obtain the same answer.

    There is nothing wrong with solving problems using non-inertial (accelerating) reference frame provided you understand the implications. In general, it is always advisable to use inertial (non-accelerating) frame of reference to avoid confusion.

    Edit: Looks like Doc Al's got it :wink:
     
  9. Nov 24, 2006 #8
    Yes, actually, I did the problem two different ways, both of which ended up with the same answer. The first way, which i thought was wrong, was this.

    Ff=2m (so now I'm accelerating, and there is no force-of-train)
    9.8mu=2m
    u=2m/9.8m
    u still equals .2
     
  10. Nov 24, 2006 #9
    Alright, coolio, I think I've got both of them now. lol, thank you both.
     
  11. Nov 24, 2006 #10

    Doc Al

    User Avatar

    Staff: Mentor

    Nothing wrong with that. And no need for any mysterious "force of train"!:wink:

    One thing to realize is that [itex]\mu mg[/itex] is the maximum value of static friction. Using that, like you correctly did, gives you the minimum value of the coefficient of friction. (It can always be greater.)
     
  12. Nov 24, 2006 #11
    Are you Alex, by any chance? Because my physics teacher's name is Alex...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Friction on an accelerating train
  1. Friction of a train (Replies: 14)

Loading...