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Inclined Plane Problem

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data
    A skateboarder slides down a frictionless ramp inclined at an
    angle of 30° to the horizontal. He then slides across a frictionless
    horizontal floor and begins to slide up a second incline at an angle
    of 25° to the horizontal. The skateboarder starts at a distance of
    10 m from the bottom of the first incline.How far up the second
    incline will he go if the coefficient of kinetic friction on the second
    incline is 0.10?


    2. Relevant equations
    Fnet=ma
    a=gsin(theta)


    3. The attempt at a solution
    So I got the first part right-- I found V2 to be 9.9 m/s. So for the second ramp, I used the formula Fnet=F(parallel)-u(Fn).
    ma=mgsin25 deg. -0.1(cos 25 deg.)
    I got an acceleration of 0.33 m/s^2, but the answer at the back is different. Also, I don't know whether g should be positive or negative.

    Thanks in advance!
     
  2. jcsd
  3. Oct 5, 2011 #2

    PhanthomJay

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    The net force is always in the direction of the acceleration. On the 2nd incline, which way does the gravity component act, down or up the incline? Which way does friction act, down or up the incline? Which way does acceleration act? be consistent with your choice of plus and minus signs.
     
  4. Oct 5, 2011 #3

    collinsmark

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    I assume that "F(parallel)" is the component of the gravitational force that is parallel to the ramp? That's fine if it is, I just want to make sure that we're both talking about the same thing.

    For this particular problem, is "F(parallel)" in the same direction as the frictional force or in the opposite direction? Draw a free body diagram if that helps.

    [Edit: PhanthomJay beat me to the response. Hi PhanthomJay!]
     
  5. Oct 5, 2011 #4
    @Phanthom Jay:Everything is downwards, right?
     
  6. Oct 5, 2011 #5
    yeah, F(parallel) is the gravitational force parallel to the incline. It's in the same direction of the frictional force, but I'm not sure if it's right to subtract the force of friction from F(parallel), as the example in the book does, because I got the wrong answer.
     
  7. Oct 5, 2011 #6

    PhanthomJay

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    Why would you subtract them if they act in the same direction? And once you find the acceleration, you still need to find the distance it travels up the plane.

    Hi, Collinsmark!
     
  8. Oct 5, 2011 #7

    PhanthomJay

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    I have to sign off.. :zzz:Collinsmark can take it from here, please, thanks.
     
  9. Oct 5, 2011 #8

    collinsmark

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    Well, in the example in the book was the F(parallel) in the same direction as the frictional force or in the opposite direction?

    Base your formulas (many of which you will be deriving yourself) on your own free body diagrams (FBDs), created for the specific problem at hand. Formulas that apply to one problem don't necessarily apply the same way to other problems. That's why FBDs are so important.

    On a different note, earlier you asked a question about g being positive or negative. The standard gravity g is always a positive constant. It doesn't mean that acceleration has to be positive though. It depends on how you define positive or negative direction. You get to define this, by the way. For example you could define up as being the positive direction for a problem involving a free-falling object. In that case, since the object is accelerating down, a = -g. But note that g is still positive even though the acceleration isn't. Sometimes you'll need to stick a negative sign in front of g to make the acceleration negative, but that doesn't make g itself is negative.
     
  10. Oct 6, 2011 #9
    I don't know, that's what the example in the books does.
     
  11. Oct 6, 2011 #10
    Yeah, Fparallel and Ff is in the same direction. So i shouldn't subtract them but add them together?
     
  12. Oct 6, 2011 #11

    collinsmark

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    Let me ask in a different way. In the example in the book, was an object sliding down a ramp or sliding up the ramp? It makes a difference! The [dry, kinetic] frictional force is always in opposite direction of the velocity.
    There ya go. :approve:

    But if you don't understand why that is, it might help to review vectors. Remember, forces are vectors and they must be added together like vectors. Think of vectors like arrows, both a magnitude and direction. When adding vectors together, you take the head of the first arrow and put right on the tail of the second. The result is from the tail of the first to the head of the second.
     
  13. Oct 6, 2011 #12
    The object is going up.:)
     
  14. Oct 6, 2011 #13

    collinsmark

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    Allow me to reword something I said earlier. In a previous post I said, "The [dry, kinetic] frictional force is always in opposite direction of the velocity." That statement is true when an object is sliding on top of a stationary surface (actually, even if the surface isn't stationary, it's still true as long as the object is moving faster than the surface it happens to be sliding on). But the statement might not apply if there are multiple moving objects involved, all sliding around and all moving. Or it might not apply to a given object if that object is stationary, and some other object is sliding along it. So let me make a more general statement in its place.

    The direction of the [dry, kinetic] frictional force on an object is opposite the direction of the object's *relative* motion, with respect to whatever object that it is sliding against. The frictional force is always in a direction such that the force tries to stop/reduce the object from sliding (but not necessarily to stop/reduce the object from moving, such as when both objects are moving). Frictional force tries to reduce the relative sliding motion, is what I'm trying to say.

    I haven't seen the example in the book, so I can't comment on what's going on there.
     
  15. Oct 8, 2011 #14
    Okay, I get it :) Thankss
     
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