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Force of Friction w/o Coefficient of Friction Given

  1. Feb 4, 2012 #1
    I am having trouble determining what the force of friction of the box on a table described as below:

    Two blocks are arranged at the ends of a massless string. M1 is on the table. M2 is attached to another side of the rope hanging off of the side of a table with a massless pulley (90 degrees from the table).

    M1= 4.78KG
    M2 = 3.26 KG

    When M2 has fallen through .396m, its downward speed is 1.29 m/s. Acceleration of gravity is 9.8. What is the frictional force between M1 and the table?

    My first step was to find the acceleration with I figured to be 4.202. (By taking the acceleration formula A=V2-V1/T2-T1. I figured the time in this formula by using the velocity formula and solved for T.

    My next step was to figure out the tension force which I think is where I am getting confused on where to go. I want to find the Tension force and subtract that from the Net force to find the force of friction, but I am having difficulty determining the net force an now I am not so sure I am figuring out the tension force correctly.

    I am calculating the Tension force by taking T=M2(-A)+M2(9.8). Can you please help me finish?
     
  2. jcsd
  3. Feb 4, 2012 #2

    tiny-tim

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    welcome to pf!

    hi stehat! welcome to pf! :smile:

    (btw, always use a small "t" for time, so as not to confuse with tension :wink:)
    i don't follow this: what formula did you use? :confused:
    yes, M2a = M2g - T …

    now do a similar F = Ma equation for the other block :smile:
     
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