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Hard inclined plane problem

  1. Oct 30, 2009 #1


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    1. The problem statement, all variables and given/known data

    so a block of mass m slides on a frictionless surface of an inclined plane of angle theta which itself has mass M and can slide on a horizontal surface. Assuming no friction, find the accel of the block and the inclined plane..

    3. The attempt at a solution

    well i see that there are two components to the accel of the block and one component to the accel of plane, giving 3 equations...but how do i solve to find the accel of the block and the plane? is there another constraint equation i'm missing?
  2. jcsd
  3. Oct 30, 2009 #2


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    Welcome to PF!

    Hi bon! Welcome to PF! :wink:
    The geometrical constraint is that the block stays on the plane …

    how would you write that? :smile:

    (oh, and I'd go for conservation of energy)
  4. Oct 30, 2009 #3


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    Re: Welcome to PF!

    ahh thanks i think ive got it now..

    just working it through..stuck on the last line..

    the bottom of the fraction should be sin^2(theta) + M/m

    but i have M/m + M/m (sin^2(theta)) - sin^2(theta)...

    i know these are equivalent..but how do i simplify my expression to sin^2(theta) + M/m
  5. Oct 30, 2009 #4


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    actually i think i've made a mistake..

    i can't see why though in my working!

    the constraint is: tan theta = ay (block) / ax (block) + ax (plane), yes?
  6. Oct 30, 2009 #5


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    Hi bon! :smile:

    (have a theta: θ and try using the X2 tag just above the Reply box :wink:)
    Nooo …

    try it with x and y, rather than ax and ay,

    and concentrate on the position of the block relative to a fixed point on the plane. :wink:
  7. Oct 30, 2009 #6


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    thanks.. :)

    so so tanθ = y / xblock - xplane

    or tanθ = y / xplane - xblock??
    which one?

    also then could you differentiate twice to get ratio of accel's? what would the ratio be?

    Many thanks again
  8. Oct 30, 2009 #7


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    (hmm … brackets would help :redface:)

    you decide! :smile:
    A linear combination of x and y will be the same linear combination of ax and ay :wink:
  9. Oct 30, 2009 #8


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    i really don't know which one it is?! how can i decide?

  10. Oct 31, 2009 #9


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    Such problems can be solved by identifying the forces acting on each objects, and noting down their relations.
    In this problem, the forces acting on the block are
    mg*sinθ..along the inclined plane
    mg*cosθ ....perpendicular to the inclined plane
    Normal reaction N due to the wedge. Since wedge is moving horizontally, mgcosθ is not equal to N. The block is moving along the wedge.
    So (mg*cosθ - N) = m*ay, where ay is the acceleration of m along the normal,say y.
    Hence N = (mg*cosθ - m*ay)....(1)
    When M moves horizontally through X, block moves through y along normal such that
    y = X*sinθ. Οr ay = Ax*sinθ ..(2) where Ax is the acceleration of the wedge.
    Forces acting on wedge are
    Weight Mg downward, normal reaction N1 from the ground upward and normal force N due to block. Only component which moves the wedge horizontally is N*sinθ
    So N*sinθ = M*Ax ...(3)
    Using equations (1) and (2) solve for Ax.
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