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Inclined plane motion

  1. Sep 13, 2016 #1
    1. The problem statement, all variables and given/known data
    14.7_01.PNG

    2. Relevant equations
    f = mu_k * N
    F=ma
    3. The attempt at a solution
    14.7_02.PNG
     
  2. jcsd
  3. Sep 13, 2016 #2

    phinds

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    This is the second post you have made with just your working out a problem but no question. The natural assumption is that you are asking for verification of your result, BUT ... you could be after something more specific, so it would be a good idea to ask whatever it is that you actually want to know.
     
  4. Sep 13, 2016 #3
    Yes, I have something to ask.
    First, the y-component of F is part of the force contribute to the normal force, right?
    I wanna know if I need to consider the friction when the box is pushed up to the inclined plane, which gives: F cos 20 - mg sin 20 -f...
    then when the box is sliding downward, there's another friction against the motion... then the sum of force in x-dir will be:
    [F cos 20 - mg sin 20 -f] +f = F cos 20 - mg sin 20

    Am I correct? so then, I use this divide by m to get a, and integrate to get v

    Thanks a lot
     
  5. Sep 14, 2016 #4

    gneill

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    Staff: Mentor

    It looks like your method for finding the net force acting along the slope is okay. But you may want to recalculate the friction force; The result doesn't match what I think it should be. Perhaps a calculator slip? I see that you've made one of the angles 30° in your work, but that's probably just a typo, right?

    Once you've found the acceleration you shouldn't need to integrate if you know your basic kinematic formulas.
     
  6. Sep 14, 2016 #5
    Oh hey, I made an mistake, as you said, wrote 30° instead of 20°... and the answer for f is actually wrong.

    What is the formula for the sum of force in x -direction (my x coordinate is not horizontal) ??
    Cuz, I need to find v, what will be a better way if not integrating acceleration?

    Thanks a lot
     
  7. Sep 14, 2016 #6

    gneill

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    Staff: Mentor

    You've been finding the components of the various forces along the x and y directions using the trig functions. So you already have them in hand. The friction force is directed along the slope, for example.
    If you have the net force acting on mass M along the direction of the slope, what's the acceleration of M in that direction? Knowing acceleration the rest is basic kinematics.
     
  8. Sep 14, 2016 #7
    It is force applied - weight - friction? but the mass's motion is actually opposite of the direction of force applied
     
  9. Sep 14, 2016 #8
    I mean: net force = force applied - weight - friction,
    then divide m to get a
     
  10. Sep 14, 2016 #9

    gneill

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    That's right, assuming you mean the components of F and g force that are directed along the slope. It's the net force that causes the mass to accelerate. Note that it's entirely possible that F will be insufficiently large to prevent the mass from sliding downslope rather that upslope! You should check the magnitudes of the upslope F component against the downslope g component prior to "adding" you friction. Friction always opposes the direction of motion.
     
  11. Sep 14, 2016 #10
    Yes, I meant the net force in x dir.

    As we push the box, up the slope, there is a friction opposing, but the force applied is not strong enough, so the box slide down. Is there a friction (in -x direction) opposing the motion?
     
  12. Sep 14, 2016 #11

    gneill

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    If the box slides downslope then the friction will oppose that motion. There's only ever one friction force operating between surfaces and it always opposes the motion that is taking place (or, in the case of static friction, trying to take place).

    If the motion is downslope, the friction will oppose this and be directed upslope.
     
  13. Sep 14, 2016 #12
    So the true net force equation will then be: Fx = Fcos20 - mgsin20 +f
     
  14. Sep 14, 2016 #13

    gneill

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    That looks good.
     
  15. Sep 14, 2016 #14
    Then I can divide mass to get a..
    and then integrate a = get v?
     
  16. Sep 14, 2016 #15
    which dv/dt = F/m

    ∫ dv/dt *dt= ∫ F/m dt

    v-v0 = a*t
     
  17. Sep 14, 2016 #16
    which v0 = 0 (at rest)

    so v(t=2) = a*2 m/s
     
  18. Sep 14, 2016 #17

    gneill

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    Sure, but it's easier to apply standard kinematics equations. Remember SUVAT?
     
  19. Sep 14, 2016 #18
    Nope, Like a= vr +vθ thingy?
     
  20. Sep 14, 2016 #19
    OHH v=u+at
     
  21. Sep 14, 2016 #20

    gneill

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    Like v = vo + at. You know the acceleration, you know the time. The initial velocity is zero. So...
     
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