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Mass on a friction less incline. What force is needed to stop the block from sliding.

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data

    A block with mass 10kg rests on a smooth, friction less ramp that is inclined at an angle of 45 degrees with the ground. How much force must be applied in the direction parallel to the ground to prevent the block from sliding down the ramp???

    IT TURNS OUT THAT THE MAGNITUDE OF THE FORCE THAT MUST BE APPLIED IN THE DIRECTION PARALLEL TO THE GROUND IN ORDER TO PREVENT THE BLOCK FROM SLIDING DOWN THE RAMP IS EQUAL TO THE WEIGHT OF THE BLOCK (IN NEWTONS). Explain why this is the care. Will this always be the case for a block on a friction less ramp?

    2. Relevant equations

    I found the answer (please check) as 69.3N, the teacher gave us a follow up page saying the force should be equal to the weight of the block in newtons..Did I do this wrong?? The follow up page also asks to explain why the parallel force to the ground is equal to the weight of the block in Newtons. I dont get that.

    3. The attempt at a solution

    Weight = 10.9.8 = 98N
    W vector <0,-98>
    Vector u + Vector V = Vector W
    F=-u
    magnitude of F= magnitude of U
    98 sin 45 = 69.3 N
    A force of 69.3 N parallel to the plane will keep the weight from sliding.

    3. The attempt at a solution
     
  2. jcsd
  3. Sep 11, 2011 #2

    lanedance

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    Re: Mass on a friction less incline. What force is needed to stop the block from slid

    note the force is applied in a direction parallel to the ground and you must also account for the reaction force, perpendicular to the ramp surface

    have you drawn and FBD with all forces?
     
  4. Sep 11, 2011 #3
    Re: Mass on a friction less incline. What force is needed to stop the block from slid

    I drew the fbd with the block sitting on the origin. 3 forces acting on the block Gravity/Normal force and F app.

    what am i missing?
     
  5. Sep 11, 2011 #4

    lanedance

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    Re: Mass on a friction less incline. What force is needed to stop the block from slid

    lets use your notation, so there are three forces, lets call their magnitudes as follows
    W=mg the weight of the mass (vertical)
    N the reaction force (unknown, at 45degrees)
    F applied force (unknown, horizontal)

    in vector notation we have:
    W = <mg,0>
    N = N<cos(45),sin(45)>
    F = <0,-F>
     
  6. Sep 11, 2011 #5

    lanedance

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    Re: Mass on a friction less incline. What force is needed to stop the block from slid

    now do the force balance
     
  7. Sep 11, 2011 #6

    lanedance

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    Re: Mass on a friction less incline. What force is needed to stop the block from slid

    Also leave numbers out for as long as possible, only substitute in at the end - it generally makes things easier to understand
     
    Last edited: Sep 11, 2011
  8. Sep 11, 2011 #7
    Re: Mass on a friction less incline. What force is needed to stop the block from slid

    So fnet=w-n-f?

    This is not sure where I'm headed with the equations you gave. I understand those are the forces acting on the mass, but I am looking for the force that keeps the block from sliding. Which the teacher said should be equal to the mass in N of the weight. I so not understand that concept.
     
  9. Sep 12, 2011 #8

    lanedance

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    Re: Mass on a friction less incline. What force is needed to stop the block from slid

    I don't understand your equation, what is fnet?

    If the block is stationary the net force is zero in both the horizontal and vertical directions

    go back to post #4
    - use the sum of vertical forces =0, to find N in terms of W
    - then use the sum of horizontal forces = 0 to find F in terms of N

    this will give the result you require
     
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