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Force applied to incline plane

  1. Aug 13, 2014 #1
    1. The problem statement, all variables and given/known data

    The problem is described and drawn here:

    https://drive.google.com/file/d/0B-HFS9bOMNAcMjNacXlha0Z6cVk/edit?usp=sharing

    Edit: The picture seems not to be showing up, so the address is: https://drive.google.com/file/d/0B-HFS9bOMNAcMjNacXlha0Z6cVk/edit?usp=sharing

    Possibly this clickable link will work: Problem Picture

    2. Relevant equations

    Newton's Laws of Motion, gravitational force is mg, and the redirection of force on an inclined plane.

    3. The attempt at a solution

    In the picture above--basically, I just am not sure how the force of the mass is directed on the plane. I think I get the rest of the important stuff: The acceleration of the system is a, so you can use F = ma since it will only accelerate in the horizontal direction, and each individual object will also have that same acceleration, etc.
     
  2. jcsd
  3. Aug 13, 2014 #2

    jedishrfu

    Staff: Mentor

    Where's the picture?
     
  4. Aug 13, 2014 #3
    Trying to make it appear, but it doesn't seem to be working. I've added a link, maybe that will work.
     
  5. Aug 13, 2014 #4

    jedishrfu

    Staff: Mentor

    Look at your force diagram and the forces on the block: gravity is pulling it downward but the inclined plane prevents it from falling freely instead it breaks up the gravitational force into two components:

    One force component should be parallel to the inclined plane and one should be vertical to the inclined plane.

    The one force component parallel is the force that moves the block

    The vertical force has no effect unless friction is to be considered in the problem.
     
  6. Aug 13, 2014 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The force of gravity, which is straight down, can be converted to a component parallel to the inclined plane and one that is perpendicular to it. It is the component perpendicular to the plane that acts on it. Further, the horizontal force pushing from the side can be converted to such components and, again, it is the perpendicular component that acts on the plane.

    (I agree with jedishrfu though I think his use of "vertical to the inclined plane" rather than "perpendicular to the inclined plane" can be confusing.)
     
  7. Aug 13, 2014 #6

    jedishrfu

    Staff: Mentor

    Yes, sorry for the poor choice of words it should have been vertical. I was trying not to give too much away.
     
  8. Aug 13, 2014 #7
    I'm a bit confused about this. The applied force is purely horizontal, so as far as it acts on the plane, I would think it acts in the horizontal direction.

    Do you mean that, as the plane acts on the block, the action is purely perpendicular to the incline? So the force the plane applies to the block is perpendicular to the incline, and then to relate that to the fact that the block is stationary relative to the plane, I would have to find the vertical component of that?
     
  9. Aug 13, 2014 #8

    olivermsun

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    Science Advisor

    The applied horizontal force is equal to the sum of one force along the incline and one force into/out of the incline. It's easier to break up the force this way so that you can figure out the force applied to the block in the along-incline direction.

    Similarly, downward gravity is equal to the sum of one force along the incline and one force into/out of the incline.

    To keep the block at the same place on the incline, you need the forces along the incline to cancel each other.
     
  10. Aug 13, 2014 #9
  11. Aug 13, 2014 #10

    jedishrfu

    Staff: Mentor

    Compare your diagram to this one:

    http://www.clickandlearn.org/images/incline.gif
     
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