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What is the required coefficient of static friction

  1. Jul 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Two blocks of equal mass are connected by a rope of negligible mass that is passed over a frictionless pulley as shown in the figure. The angle made by the horizontal plance and the ramp is 30 degrees.

    a) If the system is in this configuration and neither block moves what is the required coefficient of static friction to keep the upper block from moving?

    b) If the ramp is replaced with an identical but frictionless ramp, how fast will the blocks be traveling when the lower mass descends by 0.50 meters?


    2. Relevant equations

    F=mg


    3. The attempt at a solution

    a) For the block on the ramp, I said that:

    Normal Force - mgcos30 = Fy1

    I then found the normal force to be 8.48M

    I then set F(fr)=uK (8.48)M

    Please help I am really lost and I feel this is totally wrong.

    Thanks in Advance
     
  2. jcsd
  3. Jul 23, 2012 #2

    tiny-tim

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    welcome to pf!

    hi hockeybro12! welcome to pf! :smile:

    what is the set-up? :confused:

    is the string parallel to the ramp until it meets the pulley, with the second mass hanging straight down?

    if so, what is Fy1 in this equation? …
     
  4. Jul 23, 2012 #3

    CWatters

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    Google the definition of coefficient of friction. It's defined as the ratio of two forces that are at right angles to each other. Work out those two numbers from the problem.
     
  5. Jul 23, 2012 #4
    Re: welcome to pf!

    Im not really sure if what I am doing is correct because I think it is wrong. Basically there is a downward angled ramp angled at 30 degrees. There is a box of Mass M at the top of the ramp and a string attached to it. The string goes parallel to the ramp until the bottom. At the bottom, it goes straight down and is attached to another box of mass M. I hope you can see it better and please help me.
     
  6. Jul 24, 2012 #5

    tiny-tim

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    hi hockeybro12! :smile:

    (just got up :zzz:)

    ah! so i got it back-to-front … the second mass is hanging off the bottom of the ramp! :biggrin:

    ok, but my question is still, what is Fy1 in that equation?​
     
  7. Jul 24, 2012 #6

    CWatters

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    Work out the forces..

    a) acting down the slope (eg parallel to it)
    b) normal to the slope..

    Down the slope you have:

    mg from the mass hanging on the rope and
    mgSin(angle) from the mass on the slope

    mg + mgSin(angle)

    Force acting normal to the slope is just

    mgCos(angle)

    Calculate the ratio.
     
  8. Jul 24, 2012 #7
    Thanks I have figured it out.
     
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