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HI! Physics question

  1. Jun 13, 2005 #1
    Hello all,
    I'm new to this site and think it's a great idea to have a Physics site to help with, b/c we all know physics is the devil.... (j/k) :)

    Anyway I have this problem which I think i figured out but want to be sure.



    Suppose a block of mass 25 kg rests on a horizontal surface and the coefficient of static friction is .22

    a). What is the maximum possible Fs that could act on the block.

    This is what I got, Fs=m*g*Us
    x=(25)(10)(.22) = 55N

    b). What is the avtual Fs that acts on the block if an external force of 25N acts horizontally on the block.

    ME:
    55-25= 30M


    And this problem:

    2). A 5kg block rests on a horizontal plane a force of 10N applied horizontally causes the block to more horizontally at a constant velocity. What is the coefficient of friction b/w the block and the plane, assume G is 9.8 m/s^2.

    My answer doesn't make sense because the coefficient is not between 0 and 1.0 :uhh:

    Anyway:

    ME:
    (9.8)(5)/10= about 5

    Any ideas?

    Thanks in advance!!
     
  2. jcsd
  3. Jun 13, 2005 #2

    OlderDan

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    For 1b) the block is not moving. What must be true of the sum of the forces acting on the block?

    #2 is upside down.
     
    Last edited: Jun 13, 2005
  4. Jun 13, 2005 #3
    Okay, so for 1b.)

    since it's at equilibrium, the sum of the forces must be equal. (err, must equal 0)?

    Fs= (uS)(N)?
    Fs= (.22)(25)
    Fs=5.5?

    Is this good or am I still missing something.

    and thanks for correcting my error on 2. :blushing:
     
    Last edited: Jun 13, 2005
  5. Jun 13, 2005 #4

    OlderDan

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    The coefficient of static friction tells you the maximum possible force of friction acting on a stationary object. The actual frictional force can be any value between zero and the maximum. It is often less than the maximum, and that is true in this case.
     
  6. Jun 13, 2005 #5
    thanks, i read the book and didn't get this concept, thanks again for clearing it up.
     
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