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Tension/acceleration between 2 Masses

  1. Mar 28, 2009 #1
    1. The problem statement, all variables and given/known data
    two masses are connected by a thing string running over a massless pulley. m1 slides on a 35deg ramp with a coefficient of kinetic friction of 0.40, while m2 hans from the string. what is the acceleration of the masses?
    m1 = 1.5 kg
    m2 = 3 kg

    2. Relevant equations
    Fnet = ma

    3. The attempt at a solution
    im not quite sure how to relate the object that is on an incline to the object that is hanging and has its forces going up and down?

    would the net force equation just be:
    Fnet = (m1+m2)a
    m2g - Ff - T = (m1 +m2)a
     
  2. jcsd
  3. Mar 28, 2009 #2

    Doc Al

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    Staff: Mentor

    Rather than attempt to treat both masses together in one equation from the outset, analyze the forces acting on each mass separately. Apply Newton's 2nd law to each, then combine the two equations to solve for the acceleration.
     
  4. Mar 28, 2009 #3

    Delphi51

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    Homework Helper

    You wouldn't put in the Tension when including all the forces like this. Tension is put in when you use separate equations for each mass.

    You do have another force to add to Ff acting on m1 - the component of the force of gravity that is acting down the ramp.
     
  5. Mar 28, 2009 #4
    when you say combine do you mean equate or add the two equations??
    Here i equated by isolating T and substituting into the other equation.

    so for mass1:
    Fnet = ma
    ma = T-(downsloping component of gravity) - Ff
    = T- 14.7sin35 - 5.88
    T = (1.5)a +2.55

    for mass2:
    Fnet = ma
    mg - T = ma
    T = mg - ma
    T = 29.4 - 3a

    then:

    29.4 - 3a = (1.5)a + 2.55
    a = 5.97 m/s^2

    does that seem right?
     
  6. Mar 28, 2009 #5

    Doc Al

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    Staff: Mentor

    I just meant solve them simultaneously. There are several ways to do that. (Adding them to eliminate T might be the easiest.)
    Perfectly fine.

    Good.
    How did you solve for the friction?

    Good.

    Right idea, but check your numbers.
     
  7. Mar 28, 2009 #6
    ** sorry i dont know where i got 5.88 from?? but i think it should be:
    in the question it told me that the coefficient of friction is 0.40.
    Ff = μFn

    Fn = 14.7cos35
    = 12.04 N

    Ff = (12.04)(0.4)
    = 4.816 N
     
  8. Mar 28, 2009 #7

    Doc Al

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    Staff: Mentor

    Much better.
     
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