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Static friction

  1. Oct 2, 2003 #1
    Q: There is a mass of a mass A (2kg ) sitting on top of a mass of B ( 4kg), which is sitting on top of a mass of C(3 kg). The static friction between A and B is .5; and between B and C is .5 and between C and the ground is .8. Mass B is tied to the wall. What is the minium force required to pull block C out from under block B?

    A: Well I am really confused about static friction, the Normal force acting on Mass C is 9kg*9.8N=88.2 N, and is u static is .8+.3=.13, so therefore the static force is
    us.* Normal force = Static force =13*88.2 = 88.2N (is that force it would take to move Mass C?)
  2. jcsd
  3. Oct 2, 2003 #2
    sorry I meant that it was .13*88.2N = 11.46N of force, that does not seem right. When you figure the static force, is that the force that it takes to move the object?
  4. Oct 2, 2003 #3


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    Staff Emeritus

    please read what i had to say in another thread: https://www.physicsforums.com/showthread.php?s=&threadid=6603

    Especially the part about drawing a picture.

    Physics is not about plugging in values into formulas and hoping the get the right answer. You must first understand what the problem entails and attack it in an ordially fashion.

    Make the problem simple and break is into two parts - the friction on the bottom of C and the friction on top of C.

    For the bottom of C you have the weight of the whole system times the coefficient of static friction between C and the floor. On the top of C you have the weight of A and B. times the coefficient of static friction between B and C

    Now its just a matter of sum of forces in the x direction.
  5. Oct 2, 2003 #4
    thank you so much, I was lost. You are right about drawing the pictures, it is just hard to do that on the computer.

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