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2 blocks, frictions, and no numbers

  1. Sep 28, 2008 #1
    Block A, with mass mA, is initially at rest on a horizontal floor. Block B, with mass mB, is initially at rest on the horizontal top surface of A. The coefficient of static friction between the two blocks is μs. Block A is pulled with a horizontal force. It begins to slide out from under B if the force is greater than:

    A. mAg
    B. mBg
    C. μsmAg
    D. μsmBg
    E. μs(mA+mB)g

    I came up with a couple of equations (that may not be right).

    f = μmBg
    f = F - mAaA


    So I tried substitution and solving for F, but I can't get the right answer (E).
    So what's the right way to do this problem?
     
  2. jcsd
  3. Sep 28, 2008 #2
    The equation for frictional force is the coefficient of friction times the normal force. What is the normal force on block A?
     
  4. Sep 28, 2008 #3
    It would be mBg, I guess.
     
  5. Sep 28, 2008 #4
    Really? Draw a free body diagram.

    Don't forget that block B is sitting on top of block A.
     
  6. Sep 28, 2008 #5
    I did.

    I have force F acting on mA.
    mAg is directed down
    mBg is directed down, and then up as a normal force
    friction is in opposite direction of the 2 blocks

    What am I doing wrong here??? =(
     
  7. Sep 28, 2008 #6
    The problem is that block A has block B on top of it. That means that the weight of block A is (Ma + Mb)g. What that means is that since block A is not accelerating down, it's normal force must be the same as the weight. Now do you understand?
     
  8. Sep 28, 2008 #7
    Ok, I see now. So E says that F = μNA
    But what does that mean? I thought friction is supposed to be the same for both blocks in this problem. This is getting more confusing.
     
  9. Sep 28, 2008 #8
    The coefficient of friction is the same. The frictional force depends upon the normal force though. Because A has a greater normal force than B, it would produce a greater frictional force.
     
  10. Sep 28, 2008 #9
    I see. Thanks for the explanations. Now I really need to study hard for that test.
     
  11. Sep 28, 2008 #10
    No problem :) Glad I could help. Good luck!
     
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