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Force and static friction

  1. Mar 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Block b with mass 5kg rests on block a with mass 8kg which in turn is on a horizontal tabletop. there is no friction between block a and the table top but the coefficient of static friction between block a and b is 0.75.A light string attached to block a passes over a frictionless pulley and c is suspended from the other end. what is the largest mass the block c can have so that a and b still slide together when the system is released from rest


    2. Relevant equations

    w=mg
    fs=usmg
    f=ma

    3. The attempt at a solution
    I figured i needed to work out what force is required to move a and b together

    w=mg
    w=(5+8)x9.8
    =127.4N

    I then worked out the static force
    fs=usmg
    = 0.75x5x9.8
    =36.75

    I then decided that this was the maximum force that the two blocks could be moved with

    There for for c we know acceleration = 9.8
    F=ma
    36.75=mx9.8

    m=3.75kg

    im not sure if this is right as i dont think my logic makes much sense
    any help would be much appreciated thank you very much
     
  2. jcsd
  3. Mar 19, 2009 #2
    That is where you've gone wrong. If there is a net acceleration, surely the block B would slide over A, right?
     
  4. Mar 19, 2009 #3
    sorry im not sure then how to correct my error
    any hints would be greatly appreciated thanks
    i am so totally confused with this whole topic
     
  5. Mar 19, 2009 #4
    What you've been asked to find out, is the max. value of m so that the system is still in equilibrium.
    What is the max. static frictional force acting on B?
     
  6. Mar 19, 2009 #5
    i thought the max static force was 36.75N as calculated in the above post
    but then how do i calculate the mass of block c if you do not no the acceleration.
    sorry i must seem so dumb
     
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