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Homework Help: Block on a moving incline

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data

    A small block of mass m rests on the sloping side of a triangular block of mass M which itself rests on a horizontal table. Assuming all surfaces are frictionless, determine the magnitude of the force F that must be applid to M so that m remains in a fixed position relative to M (that is, m doesn't move on the incline).

    2. Relevant equations

    F = ma

    3. The attempt at a solution

    I find the forces on block m due to gravity and the force F. So the net force on that block is:
    Fcos(theta)+mgsin(theta)=ma

    Since the block m isn't moving, I set the net force on it equal to zero (?):
    Fcos(theta)+mgsin(theta)=0

    And I solve for F:
    F= -mgtan(theta)

    The actual answer is:
    F=(m+M)gtan(theta)

    What am I missing? I've been scratching my head for awhile now...
     
  2. jcsd
  3. Feb 20, 2010 #2
    Draw a FBD for the small block, and prove that its horizontal acceleration must be equal to
    [itex]\rm a_x = g tan\theta[/itex], if its vertical acceleration is to equal zero, i.e. [itex]\rm a_y =0[/itex]. Please note, there is no F in the FBD diagram for the small block, only the externally applied N and mg. You could now do a FBD for the large block, making sure you include the action-reaction pair, to determine the necessary F. But a quicker way would be to isolate both blocks together and apply [itex]\rm F = (m+M) a_x[/itex], since the two move as one.

    So, step one, draw a FBD for the small block and figure out how the contact force N breaks up into vertical and horizontal components. Let me know how it's going, and we'll see if you need some more hints.
     
    Last edited: Feb 20, 2010
  4. Feb 20, 2010 #3
    Isn't the normal force perpendicular to the surface of the incline? Why would there be x and y components of it.

    Thanks
     
  5. Feb 20, 2010 #4
    This will help you to get started:

    http://img32.imageshack.us/img32/8211/blocks1d.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  6. Feb 20, 2010 #5
    I gotcha.

    x-dir: Nsin(theta)=ma
    y-dir: Ncos(theta)-mg=0

    a=gtan(theta)

    How come the net forces will not be the same if I decide to split mg into x and y components instead of splitting the normal force? How do I know which to split?
     
  7. Feb 20, 2010 #6
    Good job!

    If you look at the picture, the system is going to accelerate in the horizontal direction (+x for my picture). Therefore the net force on the system is a vector pointing in, for my picture, the +x direction as well. You can make a coordinate system anywhere you want, but then you would have to break the applied F up into components as well. The coordinate system I chose makes solving the problem as simple as possible. For example, consider solving circular motion problems with x-y cartesian coordinates, you can do it, but it just makes the problem more difficult, as opposed to using r-[itex]\rm \theta[/itex] unit vectors (cylindrical coordinates) which vastly simplifies things.

    Now that you've got a handle on this problem, try to see if you can get the required result by isolating the large mass M, and using Newton's 2nd postulate, [itex]\rm {\bf F}_{net} = M {\bf a}[/itex]. Here's the picture:

    http://img69.imageshack.us/img69/894/blocks2.jpg [Broken]

    Note: (1) Newton's 3rd postulate: the force N must be applied to the mass M, equal magnitude and opposite direction, since M applied N on little m. This is the action-reaction pair I mentioned earlier. (2) I moved the contact force [itex]\rm \eta [/itex], the contact force from the floor, a little to the left so you could see where Mg and [itex]\rm \eta [/itex] are being applied.
     
    Last edited by a moderator: May 4, 2017
  8. Feb 20, 2010 #7
    Okay I think I understand now...pick a coordinate system and stick with it...

    For the larger block:
    x-dir: F-Nsin(theta)=M(gtan(theta))
    y-dir: Ncos(theta)-mg=0

    Plugging N=mg/cos(theta) into the x-direction net force:
    F-mgtan(theta)=Mg(tan(theta))
    F=Mg(tan(theta))+mgtan(theta)

    F=(M+m)gtan(theta)

    Great! Thanks a lot!
     
  9. Feb 20, 2010 #8
    [itex]\rm \eta-Mg-NCos\theta = 0[/itex] (this is from the FBD for M, but you don't need it)

    Yes, good job! You needed information from doing two FBDs, for (1) m and M, or (2) m and (m+M) as a system. When you do (m+M) as a system, the action-reaction pair doesn't show up, it's "internal" to the system -- they cancel by Newton's 3rd postulate.

    Now, since you're becoming a physicist, you should look at your result for [itex]\rm a_x = g tan \theta[/itex], and ask if it makes physical sense when [itex]\rm \theta = 0^o[/itex], and [itex]\rm \theta = 90^o[/itex]. :cool:
     
  10. Feb 20, 2010 #9
    Wait so how do you find the normal force of the larger block? Haha.

    Thanks!
     
  11. Feb 20, 2010 #10
    That was a serious question, by the way. :smile:
     
  12. Feb 20, 2010 #11
    Oh? It was? lol.

    I think I wrote it down earlier. Up there somewhere...hang on, there it is:

    https://www.physicsforums.com/latex_images/25/2590137-0.png [Broken]

    from the summation [itex]\rm F_{y_{net}} = M a_y[/itex]
     
    Last edited by a moderator: May 4, 2017
  13. Feb 21, 2010 #12
    I meant: What is the magnitude of lower-case n, the normal force on the larger block?

    Because now if I plug that equation into the net force equation for the x direction, I still have an unknown variable n...

    EDIT: Nevermind. I don't need to use that equation (N-Mg-Ncos(theta)=0) because that's not what I want when finding F anyways...so that's what you mean by "you don't need it" haha...my bad
     
    Last edited: Feb 21, 2010
  14. Feb 21, 2010 #13

    ideasrule

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    Homework Helper

    Look at the equation n-Mg-Ncos(theta)=0. You don't know N, so look at the other equations you wrote down to find it.
     
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