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Acceleration and Mu (k or s)

  1. Oct 21, 2007 #1
    the sol'n i'm posting is correct but i'm just having a hard time grasping it

    original question:

    block B with mass 5kg, rests on block A, with mass 8kg, which in turn is on a frictionless tabletop. the coefficient of static friction between block A & B is 0.750. what mass must block C have so that blocks A and B slide together when the system is at rest?

    my question:

    since A is accelerating to the right due to the string, IF there was friction, it would go opposite of acceleration (correct?). now if there was friction, it would go opposite, and since there is static friction at the top of block A or bottom of block B ... on my force diagram B, wouldn't it go opposite of block B's acceleration? or shouldn't acceleration be going the other direction?

  2. jcsd
  3. Oct 21, 2007 #2


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    The [tex] m_c [/tex] is pulling the mass a and b to the right with certain acceleration. In this case, it's the string, the tension that pulls mass a and b. however, since the string is not connected to the mass b, and since the mass b is also moving to the right with the acceration. So, there must be a force that pull mass b as well, what is the force, it's the friction force given by the surface of the roughness mass a. (If you think in this way, it's pretty straight forward)

    (When we talk about friction in this case, it's the friction between mass a and mass b. We only consider this in this case. )
  4. Oct 21, 2007 #3


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    Just to add some lights to your solutions.

    [tex] m_cg-T=m_ca [/tex]

    [tex] T = (m_a+m_b)a [/tex]

    [tex] \mu N= ma [/tex]
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