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A Plane, A pulley and two boxes

  1. Apr 25, 2004 #1
    Hi everyone here is the picture of the this plane/friction problem.

    http://phasedma.com/uploaded/Physics problem.JPG

    The question asks what are the minimum and maximum values of m1 in the figure to keep the system from accelerating...

    Take µk = µs = 0.50
    ------------------------


    Ok when drawing my free body diagrams I have come up with this method of solving the problem... tell me if you agree.

    The force of friction can either be up or down the slope, if m2 = 0 or sufficiently small, then m1 would tend to slide down the plane. so Ffr would be directed Up the incline.

    We know that newtons second law for the y direction (i chose my xy coordinate axes along the plane-- IE horizontal x being the plane) shows that

    Fnormal - m1*g*cos(30) = m1*ay = 0

    since theres no y motion Fnormal = m1*g*cos(30)

    Now for the x motion... For the first case (smallest m1) f = ma shows that

    m1*g*sin(30) - Ftension - Ffr = m1*ax <---- x direction

    since we want ax to be 0, we can solve Ftension since thats related to m2.

    Since Ffr can be AT MOST µs * Fnormal= µs*m1*g*cos 30 the minumum value m2 can have to prevent motion (ax = 0) is (after dividing by g)

    m2 = m1 * sin(30) - µs*m1*cos(theta).

    And then finding the max value wouldnt be much more difficult from there since we already set up our equations..

    Am I correct here? If you are willing can someone work it --- what range do you get for the mass?

    Thanks for you help. Mechanics gets soooo tough!
     
  2. jcsd
  3. Apr 25, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Some of your thinking seems OK, but you are getting a little mixed up. First of all, your diagram shows m2 (the hanging mass) as being fixed at 5 kg, so I don't know why you are solving for m2!

    The smallest value of m1 would just prevent it from sliding up the plane. Taking up the plane as positive, the forces on m1 (I just call it m) are:
    -mg sin(30) -μmg cos(30) + T = ma = 0
    (note that T must equal 5g)
    You can solve this for the minimum value of m.

    The maximum value of m1 would just prevent it from sliding down the plane. The forces on m in this case are:
    -mg sin(30) +μmg cos(30) + T = ma = 0
    You can solve this for the maximum value of m.

    The difference between the two cases, as I'm sure you realize, is that the friction acts in different directions.
     
  4. Apr 25, 2004 #3
    Interesting, I see why the tension is 5g, and I get a minimum value of 5.35 kg which seems appropriate... but my max value is nearly 75kg!!!

    Is that correct? Or is there something else that we should consider for max value...
     
  5. Apr 25, 2004 #4
    I think that's correct. Know that 75 kg also causes more friction.
     
  6. Apr 25, 2004 #5
    Thanks urban... but wow-- that is a huge difference.
     
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