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Homework Help: [Force] Two Prisms

  1. May 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A prism has the mass m with the angle x = 45°(look at the picture) placed on the ground without friction. The other prism, which has the same mass m, is placed above the first prism(the first prism is bigger). A horizontal force F is given to the second prism so the displacement with respect to the first prism is zero( or not moving). What is the friction between the two prism ( in F, m, and g). g = earth gravitational acceleration

    The picture -> http://i28.tinypic.com/nycih4.jpg

    2. Relevant equations
    [tex]F = m.a[/tex]

    3. The attempt at a solution
    [tex]F.cosx - f - m.g.sinx = m.a[/tex]

    Since the prism is not moving, so a = 0

    [tex]F.cosx - \mu.m.g.cosx - m.g.sinx = 0[/tex]
    [tex]F.cosx = \mu.m.g.cosx - m.g.sinx[/tex]
    [tex]\mu = \fric{F}{m.g} + tanx[/tex]

    Then i have no idea what to do next. I haven't even used the mass of the first prism. I also don't quite understand how the forces work so can somebody help me please?
    Last edited by a moderator: May 17, 2008
  2. jcsd
  3. May 17, 2008 #2


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    "Parable of the Surveyors"

    Hi Vermillion! :smile:

    The big prism is "placed on the ground without friction".

    So both prisms are accelerating, with the same acceleration.

    Start again! :smile:
  4. May 18, 2008 #3
    Thanks Tim,

    I corrected the plus-minus mistake. It should be
    [tex]F = mg(\mu + tanx)[/tex]

    Then the acceleration of the prisms is

    [tex]F = (m + m)a[/tex]
    [tex]F = 2ma[/tex]
    [tex]2ma = mg(\mu + tanx)[/tex]
    [tex]a = \frac{g} {2}(\mu + tanx)[/tex]

    Finally, i don't know how to relate the acceleration to the friction. Can you explain it to me?
    Last edited: May 18, 2008
  5. May 18, 2008 #4


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    Hi Vermillion! :smile:

    You're making the very common mistake of assuming that the friction force is µ times the normal force.

    For static friction, it usually isn't!

    µ times the normal force is the maximum for static friction.

    It only applies if the question specifically tells you that the object is "on the point of moving" (or "just about to move", or something like that).

    Here, the question not only doesn't say that … but it doesn't even bother to tell you what µ is (which should have been a clue, shouldn't it? :rolleyes:).

    You have to use Newton's first and third laws, to find the reaction force, R.

    Forget everything you know about friction.

    Draw a new diagram, put in R, and take horizontal and vertical components. :smile:
  6. May 18, 2008 #5
    Do you mean, i should throw in F.Sinx to my friction equation? Which i forgot to do.
  7. May 18, 2008 #6


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    No :smile: … I meant exactly what I said, which was …
  8. May 18, 2008 #7
    Just for quick clarification, you mean like this?

    Horizontal Components
    [tex]\SigmaFx = 0[/tex]
    [tex]F - f.cosx - mg.sinx.cosx = 0[/tex]

    Sorry, i'm not used to talk physics in english.
    Last edited: May 18, 2008
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