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Homework Help: Tricky question involving masses, a pulley on incline with friction

  1. Mar 4, 2013 #1
    1. The problem statement, all variables and given/known data

    What does M2 need to be in order to move?
    The interesting thing with this question is if it does move, will it move up or down?

    2. Relevant equations

    3. The attempt at a solution
    I thought about this and if it's going to go up, that's one mass. If it goes down, that's another mass so will there be two masses?

    I also though, if it gave the static friction for both, maybe I thought of finding force of static friction for M1:

    After that, I think there will be a force of tension and a parallel force of gravity on M1 and if you add all these up with Ff, it must add up to 0...or not since it is a connected to a pulley that is being pulled (or not), it won't add up to 0?
  2. jcsd
  3. Mar 4, 2013 #2
    Here's a smaller picture. That was too big :P
  4. Mar 4, 2013 #3
    Draw out two FBD for the boxes. One where the friction forces both point in +x and another where they point in -x. I don't get why the question asks if it will move up or down. You should get two masses, one the moves up and that moves down.
  5. Mar 4, 2013 #4
    I found the force of static friction for them and just drew the FBD. So if it starts to move, wouldn't it be kinetic friction now?
  6. Mar 4, 2013 #5
    Kinetic friction is always smaller than static friction so you don't need to worry about it.
  7. Mar 4, 2013 #6
    So here I drew the FBD for both boxes and since I thought there would be two masses possible, I treated right as positive for case #1

    Then I tried to work it out using Fun=ma but since it's in equilibrium, ƩF=0.
    So I plugged in all my given information to develop two equations for this:

    And so I was left with one unknown, FT, so I isolated them and made them equal each other assuming they have the same magnitude. So here I made them equal each other and found the mass!

    But since this is Case #1, so in conclusion the mass it needs to move upwards is 15.9kg?
  8. Mar 4, 2013 #7
    I think my answer looks pretty huge for it to be right :P
  9. Mar 4, 2013 #8
    You put your coefficients of friction in the wrong places. It looks like that was your only mistake though.
  10. Mar 4, 2013 #9
    What do you mean? I substituted them in though because Ffs=μsFn
  11. Mar 4, 2013 #10
    You substituted them into to the wrong equations. If you were to fix that you'll have an answer that is quite smaller.
  12. Mar 5, 2013 #11
    Could you explain that in further detail? And is it true that the forces of tension will be the same for both sides of the pulley?
  13. Mar 5, 2013 #12
    m2 has μs=0.2 and m1 has μs=0.3. You mixed them up though so that should be why your answer is wrong. Yes tensions will always be equal as long as the pulley is massless and frictionless.
  14. Mar 5, 2013 #13
    I am getting 10.95kg but would that seem more reasonable?
  15. Mar 5, 2013 #14
    Actually it looks like your us values were in the right place but you put the masses in the wrong places. Also for your FBD of box 2 you had friction going in the wrong direction.
  16. Mar 5, 2013 #15
    So I did make changes to my mistakes with my coefficients of static friction and this is how I got 10.95kg.

    So on my current one, the directions of my Force of Static Friction is in the wrong direction?
  17. Mar 5, 2013 #16
    So for box 2 I had a mistake there for the Ffs, so it should be going towards right (same direction as the Fg||)?
  18. Mar 5, 2013 #17
    Oh I get it now!!! I started again on a fresh sheet of paper and tried again with what I came up and fixed my mistakes. Can't believe I mixed up the two boxes because box 2 was on the left and box 1 on the right so I ended up mixing them both up. Now, with everything fixed, I managed to get 1.24kg which seems to make more sense since box 1 is 5.0kg.


    And I also fixed the direction of force of friction for box 2 like you said and yeah, I got 1.24kg.
  19. Mar 5, 2013 #18
    Also, a little question, why would the force of friction for box 2 go the other way (like you said)?

    I got 1.57kg if I made the friction the same, 1.24kg when I changed the friction for box 2, like you said.

    So which one would make more sense?
  20. Mar 5, 2013 #19
    Friction is always opposite the motion
  21. Mar 5, 2013 #20
    what if it's static friction where it is equilibrium and there's a string attached to it on an incline, where will the force of static friction point towards to now? for this instance, like the problem here...
  22. Mar 5, 2013 #21
    if m1sinθ1>m2sinθ2, friction will pull down the plane on m2 and up the plane on m1, because in order for system to not move, friction will have to add to force down the plane of m2, and it friction will have to resist the force of m2 trying to go down the plane.
  23. Mar 5, 2013 #22
    So would this be the right diagram then?
    I know that Force of tension goes towards the center. Would the force of static friction goes toward center because they are opposing each other so that they won't move?

  24. Mar 5, 2013 #23
    No the boxes are moving so you want the friction to oppose that motion.
  25. Mar 5, 2013 #24
    Oh!!! I understand now. The force of friction (static) shouldn't oppose each other, because I want to find out the mass it would take just so it would move (up or down) the incline. Right, so this is now my diagram, changed the Ffs for M2.


    I worked that out ealier and got 1.24kg in post #17. So would that be correct?
  26. Mar 5, 2013 #25


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    Another way to look at the problem is to find the range of values for m2 so that the system doesn't move.
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