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Simple Dynamics Problem.

  1. Mar 29, 2013 #1
    1. The problem statement, all variables and given/known data

    I've solved it already, I think. I'm just not sure about the result.

    There is a block (B), which is touching a cart (C) on one side.
    Let an external force, parallel to the surface, ##\vec{F_a}## be applied on B

    mass of B = m; mass of C = M; static friction coefficient between B and C = μ.

    Taking no notice of the ground's friction, what is the minimum value of ##\vec{F_a}## such that the block doesn't fall?


    3. The attempt at a solution

    After drawing the free-body diagram for B, i see:
    ##\vec{F_s}## (static friction force) ##\leq m \cdot \vec{g}##
    and being ##\vec{F_s}=μ \cdot \vec{F_N}## i get ##\vec{F_N}= \frac{m \cdot \vec{g}}{μ}##
    ##\vec{F_a}=\vec{F_N} + \vec{F_f}## the latter being the force applied to C, which makes it move.
    ##\vec{F_f}=\frac{\vec{F_N}}{M} * m## . So,
    ##\vec{F_a}=\frac{m \cdot g}{μ}+ \frac{m^2 \cdot g}{μ \cdot M}##

    is it okay?
     
  2. jcsd
  3. Mar 29, 2013 #2

    ehild

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    You made some little errors.

    ##\vec{F_s}## (static friction force) ##=-m \cdot \vec{g}##

    ##{F_s}\leq \mu \cdot {F_N}##

    The minimum value of Fa is the question. So ##F_a\geq\frac{m \cdot g}{μ}+ \frac{m^2 \cdot g}{μ \cdot M}##

    ehild
     
  4. Mar 29, 2013 #3

    vela

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    Can you explain these two steps? I don't follow what you did here.
     
  5. Mar 29, 2013 #4

    haruspex

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    As vela notes, this is wrong. Try introducing an unknown for the acceleration of the system and developing the F=ma equation for each body separately.
     
  6. Mar 30, 2013 #5

    ehild

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    You meant by Ff the resultant force acting on B instead of C, didn't you?

    haruspex: The OP solved the problem, but made some little errors when typing in. The result for the minimum applied force is correct, except the vector sign.

    ehild
     
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