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Forces help

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data
    M1 and M2 are two masses connected as shown (M2 hangs over the table). The pulley is light (massless) and frictionless. Find the mass M1, given that M2 (3.0 kg) accelerates downwards at 2.31 m/s^2, q is 20°, and μk is 0.41.


    2. Relevant equations
    F=ma
    μn=friction


    3. The attempt at a solution
    Well I thought I could find tension using F=3*2.31 but that didn't get me anywhere. I drew a force diagram for m1, but it looks like there's too much missing to be able to continue. Is the tension part at least right?
     
  2. jcsd
  3. Sep 21, 2013 #2

    tiny-tim

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    hi thatguy101! welcome to pf! :smile:

    (we usually use "T" for tension :wink:)
    is that Ftotal = ma?

    what about gravity? :wink:
     
  4. Sep 22, 2013 #3
    so then would I subtract gravity?
    Ft=3*2.31-3*9.81?
     
  5. Sep 22, 2013 #4
    Why not give us the diagram so that we know what q = 20[itex]^{o}[/itex] refers to?
     
  6. Sep 22, 2013 #5
    Sorry. It's suppose to say θ.
    m1 is on an incline of 20°, and m2 is hanging off the incline.
     
  7. Sep 22, 2013 #6
    Ok so my diagram looks like this.
    For m1 I have forces in the y direction as y: Fg*cosθ-Fn=0
    And the forces in the x direction as x:Ft-Fg*sinθ-Ff=a.
    And then for m2 there are none in the x direction but for y I have y: Fg-Ft=ma.
    Am I right so far?
     
  8. Sep 22, 2013 #7

    tiny-tim

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    yes, except in your second equation, you need to check the signs (and you forgot to write the m before the a) :wink:
     
  9. Sep 22, 2013 #8
    Thank you. I figured it out.
    So I had Ft-mg*sinθ+μ*mg*cosθ=ma
    so Ft=mg*sinθ+μ*mg*cosθ+ma
    factor out the m, Ft=m(g*sinθ+μ*g*cosθ+a)
    then solved for m. and since Ft= m2g-m2a, I just put in the numbers and came up with 22.47 N amd just plugged in everything else into the equation above and got 2.38 kg
     
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