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Inclines plane problems

  1. Oct 22, 2005 #1
    3. A 15 kg box slides down a frictionless ramp of 25 degrees. What is its acceleration?
    m/s/s

    What is the normal force of a 35 kg box sitting on a slope of 14 degrees? (assume the box is not moving).
    N

    A 45 kg box is sliding at a constant speed down a ramp of 28 degrees.

    What is the frictional force on the box?

    What is the coefficient of sliding friction?

    After an initial push, a box goes up a frictionless incline of 20 degrees.

    What is its rate of deceleration?
    m/s/s

    A 75 kg skier starts at the top of a 15 m incline at an angle of 25 degrees.
    Assuming the coefficient of friction is .15, what is his acceleration?).

    m/s/s

    What is his speed at the bottom of the hill?
    m/s

    can someone do those for me just so i can figure out the logic like with the equations id ont get it at all thanks
     
  2. jcsd
  3. Oct 22, 2005 #2

    arildno

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    The principle used is called Newton's 2.law of motion, which gives rise to an energy balance sheet often more convenient to work with.

    As for how to do this:
    Start by decomposing your forces in the tangential and normal directions.
     
  4. Oct 22, 2005 #3
    i dont know how to do it that why im asking my teacher is the worst teacher ever shes a nerd who is smart as s*** but can teach for her life and im really frustrated right now ive been told to use newtons 2nd law like 20 times now and i dont know how to use it i hae the equations i just dont know how t apply it so i can get the answer
     
  5. Oct 22, 2005 #4

    arildno

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    Okay, suppose your incline has a positive angle [itex]\theta[/tex] to the horizontal.
    That means that the unit tangent vector [itex]\vec{t}[/itex] up along the incline must be [itex]\vec{t}=\cos\theta\vec{i}+\sin\theta\vec{j}[/itex] where [itex]\vec{i},\vec{j}[/itex] are unit vectors in the horizontal and vertical directions, respectively. Agreed so far?

    Furthermore, the unit normal [itex]\vec{n}[/itex] with positive vertical component must therefore be [itex]\vec{n}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/itex]
    Agreed?

    But now, we can, if we wish, write [itex]\vec{i}=\cos\theta\vec{t}-\sin\theta\vec{n}, \vec{j}=\sin\theta\vec{t}+\cos\theta\vec{n}[/itex]
    Verify this!

    This is basically what you need to proceed..
     
  6. Oct 22, 2005 #5

    arildno

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    For example, the force of gravity, given by [itex]-mg\vec{j}[/itex]
    may now be rewritten as [tex]-mg\vec{j}=-mg(\sin\theta\vec{t}+\cos\theta\vec{n})[/tex]
     
  7. Oct 22, 2005 #6
    i got 1 and 2 but the rest i keep getting wrong


    i dont know hwo to get Fk abd Uk
     
  8. Oct 22, 2005 #7

    arildno

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    Post your ideas for one of the problems, say number 3.
     
  9. Oct 22, 2005 #8
    i dont know i have

    A 45 kg ,a ramp of 28 degrees.

    What is the frictional force on the box? ....

    coeficient of friction?...

    i tried to use ma =mgsin0 - ukFn = mgsin0-ukmgcos0

    and got 389.37 for Fk

    and i need fk to get coeeficient so i stoped
     
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