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Homework Help: Friction Forces (Static Friction)

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A boy has stacked two blocks on the floor so that a 5.0 kg block is on top of a 2.0 kg block.

    1. If the coefficient of static friction between the two blocks is 0.40, and the coefficient of static friction between the bottom block and the floor is 0.22, with what force horizontal should the boy push on the upper block to make both blocks in the system start to slide along the floor.

    2. If he pushes too hard, the top block starts to slide off the lower block. What is the maximum force with which he can push without that happening?

    2. Relevant equations

    Fs = micro(s) * N

    3. The attempt at a solution

    1.

    Fs (between 2 blocks) = 0.40 * (5+2) * 9.8 = 27.44 N

    Fs (between bottom block and the floor) = 0.22 * 2 * 9.8 = 4.312

    Fnet = 27.44 + 4.312 = 31.752 N => wrong answer

    2.

    Fs (maximum) = 27.44 N => wrong answer

    Please help! Thanks!
     
    Last edited: Feb 23, 2010
  2. jcsd
  3. Feb 23, 2010 #2
    yes u r correct
     
  4. Feb 23, 2010 #3
    How about singnificant fingure here????
     
  5. Feb 23, 2010 #4
    No. I'm just submited the answer, but it's INCORRECT. Please help!
     
  6. Feb 23, 2010 #5
    This question is quite long to write out so ill briefly explain: You have to isolate the the top block and get the acceleration of just that block. Then, you use this acceleration for the bottom and the top block m1+m2 = 7kg total and you shud get the maximum force needed.

    Hope this helps:)
     
  7. Feb 23, 2010 #6
    Fs = 0.40 * 5 * 9.8 = 19.6 N

    a = 19.6 / 5 = 3.92 m/s^2

    so the maximum force = 3.92 * 7 = 27.44 N which is incorrect!

    What am I doing incorrect here?
     
  8. Feb 23, 2010 #7
    and the first question is INCORRECT too.

    "Fnet = 27.44 + 4.312 = 31.752 N => wrong answer"

    Please help!
     
  9. Feb 23, 2010 #8
    You forgot to plug in friction b/w the floor = 0.22(7kg)(9.8) = 15N, therefore

    Fnetx = Fa - Ffriction, where Fa = max force needed, therefore
    Fa = 3.92m/s2(7kg) + 15N = 42.44N :)
     
  10. Feb 23, 2010 #9
    How about the first question "what force horizontal should the boy push on the upper block to make both blocks in the system start to slide along the floor ? "

    What did I do wrong here?

    Thanks!
     
  11. Feb 23, 2010 #10
    Just submited the 42.44 N for the second question and it's INCORRECT. I'm confused...
     
  12. Feb 24, 2010 #11
    Sorry , but that solution is if the bottom block is moved without the top block moving, my mistake. I am no expert at this, just trying to help. Ya, so in 1) If you want to get the system moving by just pushing on the top block then the force needed for that is just the force of friction of the two masses: 0.22(7kg)(9.8) = 15N, since they will move in equilibrium with each other.

    For 2) He does not want the top block to move, so you isolate the top block to get the force of friction = 0.4(5kg)(9.8) = 19.6N, which is the max force since beyond this force you will accelerate the top block.

    Hope this helps:)
     
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