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Two blocks sliding down a rough inclined plane

  1. Mar 12, 2012 #1
    *****The problem statement, all variables and given/known data****

    Two blocks, A and B, are sliding down an inclined (20 degrees) plane. Block A is sliding in front of block B with both of them touching. The blocks slide from a distance of 6.50m from the bottom (along the inclined plane, not the surface the plane rests on).

    Block A has a mass of 5.00kg with a kinetic friction coefficient of .150.
    Block B has a mass of 10.00kg with a kinetic friction coefficient of .200.


    Find the Acceleration of the blocks going down the plane.

    What is the final velocity of Block A at the bottom?



    ----------------------------------------------

    Attempt at solution consisted of starting with the FBD's of both blocks.

    From this point here are my questions:

    1) It doesn't seem like Block B is really putting any force on Block A going down the incline. Do I even need to consider it when finding my Net Force values in the x-direction? (I set the incline as the x-axis and perpendicular to it is the y-axis.

    2) What would be the x-components of force?
     
  2. jcsd
  3. Mar 12, 2012 #2
    Given that the blocks slide together down the plane, you may consider the sum of the following equations:

    magsinθ-Ffa+Fb→a=Fra

    mbgsinθ-Ffb-Fa→b=Frb

    → mbgsinθ-Ffb-Fa→b+magsinθ-Ffa+Fb→a=Fra+Frb

    By Newton's third law, those forces cancel each other out since they have the same magnitude but opposite directions, and so:

    (ma+mb)gsinθ-Ffb-Ffa=Fr(a+b)

    Now the force of friction is given by the following equation (with μ being the coefficient of kinetic friction and N the force exerted by the plane on the block):

    Ff=-μN

    (Note that: N=mgcosθ)

    Substitute and you get:

    (ma+mb)gsinθ-(maμa+mbμb)cosθ=Fr(a+b)

    By Newton's second law, use F=ma and complete the problem using cinematics:

    x(t)=x0+v0t+(1/2)at2
     
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