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Slope problem

  1. Mar 9, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    Box is sliding down the slope with angle 30 , and v_0=0. Because coefficient of friction is $$\mu=0.1x$$ where x is distance covered, the box will stop before reaching the end of a slope. Find the time needed for a box to stop.

    I get $$F_1=G \sin \alpha , F_{fr} = \mu \cdot G \cos \alpha$$ and the box will stop when $$F_1 = F_{fr}$$ , that is x=5.77. Ok, so I think I got the distance x, but I have no idea how to find time t. Acceleration is obviously changing, and because of that I'm not sure what formulas can I use, also I have acceleration a as linear function of x (not t) , I tried with some integrating but didn't lead me no where (I was probably going in circles there).
    Any hint on how to find time t?
     
  2. jcsd
  3. Mar 9, 2016 #2

    haruspex

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    That should be the way to go. Please post your working.
     
  4. Mar 9, 2016 #3

    haruspex

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    No, think about that again.
     
  5. Mar 9, 2016 #4
    If the part with F_1=F_{fr} is wrong, I don't think there's a point in posting my working before I realize when will the box stop. Hmm, can you show mistake in my reasoning?
    Since v_0=0 , the box will stop when v=0 , and v depends on acceleration, so acceleration has to be 0 , so F{res}/m=0 , and F_1=F_{fr} .
     
  6. Mar 9, 2016 #5

    haruspex

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    Yes, acceleration will be zero when the two forces are equal, but what does that tell you about the velocity?
     
  7. Mar 9, 2016 #6
    velocity won't change, and acceleration has to be negative in order to v be 0 at some moment?
    Okay, but how do I calculate velocity when acceleration is changing and I don't have velocity in terms of t ? I tried something like this:

    $$ a=g \sin \alpha - \mu g \cos \alpha = g \sin \alpha-0.1 g \cos \alpha x = -cx+d $$
    $$ \frac{dv}{dt} = -cx+d $$
    $$ \frac{dv}{-cx+d}= dt \Rightarrow t= \frac{v}{-cx+d} $$ but doesn't work
     
  8. Mar 9, 2016 #7

    haruspex

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    Right. You don't recognise that equation? What if you write that in terms of ##\ddot x##? You can get rid of the d by a simple change of variable.
     
  9. Mar 9, 2016 #8
    I'm currently learning this on my own, so I might miss some obvious and basic stuff. Do you mean like this

    $$ \frac{d^2x}{dt^2}= -cx+d $$ (Did you meant some kind of substitution y=x-d/c or? )

    EDIT: I did get t=-1/c ln|-cx+d| but I don't have x so that probably not good...
     
    Last edited: Mar 9, 2016
  10. Mar 9, 2016 #9

    haruspex

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    Yes, exactly that.
    What do you know about SHM?
     
  11. Mar 10, 2016 #10
    Simple harmonic motion? Yes, I see that this is the same type of equation. We didn't yet talked about harmonic motion in my physics course (We did in highschool but on that level) so don't be suprised by my ignorance :D , but one more question : when I get x in terms of t, how should I find t when I don't know x?
    Or how do I find x?
     
  12. Mar 10, 2016 #11

    haruspex

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    You can find x without invoking SHM. You only need SHM for the time. Consider work.
    Or, using the SHM solution, you can use the initial velocity as the boundary condition.
     
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