1. Limited time only! Sign up for a free 30min personal 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!

Dynamics. Inclined plane with friction coefficient

  1. Oct 29, 2012 #1
    I've got the following exercise, which I've done 3 times already and I think I haven't made any mistake:



    A body is slid upwards through an inclined plane of angle [tex]\theta=15\degree[/tex]. Determine the friction coefficient of the body with the surface of the plane if the time it takes to go upwards [tex]t_a[/tex] is double the time it takes to go downwards [tex]t_b[/tex] ([tex]t_a=2t_b[/tex]).

    Mi procedure is as follows:

    The normal force: [tex]N=mg\cos(\theta)[/tex].

    1. Upwards:

    During the ascent the x componen of the weight and the friction force act in the same direction, therefore: [tex]a_a=-g\sin(\theta)-\mu g\cos(\theta)=-g(\sin(\theta)+\mu\cos(\theta))[/tex].

    The ecuacion of space as a function of time is: [tex]x=2v_0 t_b -2g(\sin(\theta)+\mu\cos(\theta))t_b^2[/tex] (1).
    The ecuation of speed as a function of time is: [tex]0=v_0 -2g(\sin(\theta)+\mu\cos(\theta))t_b[/tex] (2).

    2. Downwards

    During the descent the x component of the weight and the friction force act in opposite directions, therefore: [tex]a_a=-g\sin(\theta)+\mu g\cos(\theta)=-g(\sin(\theta)-\mu\cos(\theta))[/tex].

    The ecuation of space as a function of time is: [tex]0=x-\frac{1}{2}g(\sin(\theta)-\mu\cos(\theta))t_b^2[/tex] (3).


    Combining (1), (2) and (3) we obtain:

    [tex]\frac{1}{2}g(\sin(\theta)-\mu\cos(\theta))t_b^2=4g(\sin(\theta)+\mu\cos(\theta))t_b^2-2g(\sin(\theta)+\mu\cos(\theta))t_b^2[/tex]
    [tex]\sin(\theta)-\mu\cos(\theta)=4\sin(\theta)+4\mu\cos(\theta)[/tex]
    [tex]\mu=-\frac{3}{5}\tan(\theta)[/tex]

    From here I deduce that a value for [tex]\mu[/tex], for with the conditions are met, doesn't exist. Is it correct?

    Thanks! :)
     
  2. jcsd
  3. Oct 30, 2012 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    From the calculation, I deduce that the object is given an initial push up the plane, then left to the mercies of gravity and friction. That was not clear from the description.
    If so, it seems obvious to me that the descent will last longer than the ascent. So maybe the problem description is wrong and it should be tb = 2ta.
     
  4. Oct 30, 2012 #3
    Adding to that, the relation between ascent time and descent time depends on the push velocity given to the object, which is not specified. Is that the complete problem statement?

    Also, there are several algebraic errors in the solution.
     
  5. Oct 30, 2012 #4
    Oh my.. That might be the mistake.. U_U Didn't notice that (the exercise is wirtten wrongly) Thanks. I'll see if that was the problem.
     
  6. Oct 30, 2012 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No, that doesn't affect the ratio.
    Yes, but I suspect these were in transcribing it for the post. They all seem to cancel out later.
     
  7. Oct 30, 2012 #6
    Thanks for your replies. The problem was the question was wrongly written (tb=2ta, not ta=2tb). That was it :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Dynamics. Inclined plane with friction coefficient
Loading...