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Newton's Laws Question I don't understand this

  1. Oct 16, 2006 #1
    In the apparatus shown, m1 = 2.0 kg, theta = 37°, and the coefficients of static and kinetic friction between m1 and the inclined plane are 0.30 and 0.20, respectively.


    Figure 4.38,
    m1 is on an incline, 37 degrees above the horizontal, connected to a pulley which is connected to m2 hanging.

    (a) What is m2 if both masses are at rest?
    between kg and kg

    (b) What is m2 if both masses are moving at constant velocity?
    between kg and kg

    Include friction in all parts.

    Ok, I'm looking at this question and i think first I have to do

    m1gsintheta = m2gsintheta

    But I'm not sure
    Could someone direct me in the right path?
     
    Last edited: Oct 16, 2006
  2. jcsd
  3. Oct 16, 2006 #2
    perhaps you could explain what the apparatus in figure 4.38 depics
     
  4. Oct 16, 2006 #3
    I edited it

    I edited it.
     
  5. Oct 16, 2006 #4
    For part a, you need to apply ur static friction on the object on the incline. Compare the tensions caused by both objects. Note: tension by mass on incline should be less due to the help of static friction.
    Therefore, Tension caused by mass hanging= Tension caused by mass on incline-static friction.
     
  6. Oct 16, 2006 #5
    Thank you!

    Thanks so much!

    Now for part b,
    What's the difference?
     
  7. Oct 16, 2006 #6
    Ok...but there's one more thing

    How do I get a range of values, the question asks

    What is m2 if both masses are at rest?

    Between blank kg and blank kg?

    How would I get a range?
     
  8. Oct 16, 2006 #7
    PArt 2 requires you to use kinetic friction, simply because the masses have started moving. However, you do need to know in which direction are they moving in. Else, its very hard to determine whether m1 or m2 is heavier. Also note: In this part, you will not be able to find an exact value for m2, You can only use the sign < or > than a maximum or a minimum value coz you were not given the speed at which they moved.
     
  9. Oct 16, 2006 #8
    ohh

    thanks so much!!

    Gunblaze you're my hero.
     
    Last edited: Oct 16, 2006
  10. Oct 16, 2006 #9
    I think for a, to get a range, maybe you can try out by calculating one that does not include the static friction and one that does.
     
  11. Oct 17, 2006 #10
    and for part b?

    I would find one that includes kinetic friction vs. one that does not?
     
    Last edited: Oct 17, 2006
  12. Oct 17, 2006 #11
    I'm so sorry but i'm gonna take back what i just said about both part a) and b).

    For part a) basically, static friction do not just apply for one direction. Therefore, one variation will be Tension by hanging mass = Tension by mass on incline - static friction, while the other will be Tension by hanging mass - static friction = Tension by mass on incline. This will actually give you your range of max and min values for m2.

    For b) You need to recall what are the properties of a mass in constant velocity? Basically, they must have no resultant forces on them right? Again, since the masses can move in both directions, kinetic friction do also apply for both directions. Therefore, Tension by hanging mass = Tension by mass on incline - Kinetic friction, if masses are moving in the direction of m2. And Tension by hanging mass - kinetic friction = Tension by mass on incline, if masses are moving in the direction of m1(down the incline).

    Always remember that friction always do apply in 2 directions depending on the direction of motion or the tendency to move in the direction. That's what also got out of my mind as i was helping you solve this question.
     
    Last edited: Oct 17, 2006
  13. Oct 17, 2006 #12
    Thank you again

    Thank you so much
     
    Last edited: Oct 17, 2006
  14. Oct 17, 2006 #13
    no problem. Hope that helps.
     
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