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Find force of friction

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data

    Three blocks are released from rest and accelerated at 1.5 m/s^2. What is the magnitude of the force of friction on the block sliding horizontally?

    2. Relevant equations

    F net = Mass * Acceleration
    (adding up 3 EQ's and canceling out variables)

    *Tension # = Tension sub #
    * Force of gravity = F sub g
    * Force of friction = F sub s

    3. The attempt at a solution

    I made three equations: F sub g - T sub 1 = 4a <---- Block on farthest right
    T sub 1 - F sub s - T sub 2 = 4a <---- block on table
    T sub 3 - F sub g = 2a <--- farthest left block

    *after canceling out variables from combing the EQ's*

    4a+4a+2a = F sub s - T sub 2 + T sub 3

    I do not know how to find the force of friction because i cannot find the tensions.

    I used F sub 1 two times because i thought since the 4kg boxes are equal in mass and are accelerating at the same value, the tension must be equal.

    The answer is 4.6 N. I think I labeled some tension wrong I think.

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Oct 15, 2011 #2

    SammyS

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    Isn't it true that T3 = T2 ?
     
  4. Oct 15, 2011 #3
    if T2 = T3, doesn't that mean they will cancel out after combining the 3 EQ's. giving 10a = -(force of friction)? and then after solving, you get 15 N. but the book says 4.6 N
     
  5. Oct 15, 2011 #4

    SammyS

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    The middle mass (the one on top) exerts a force on the left hand mass, that's equal & opposite to the force the mass on the left exerts on the middle mass.

    ∴, T2 = T3 .
     
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