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What am I missing here? Mechanics

  1. Nov 4, 2008 #1
    1. The problem statement, all variables and given/known data

    Hello guys :)
    I'm a physics student from Israel and I couldnt find a decent physics forum to help me answer some hard to solve exercises (btw, forgive my english if I fail to express myself well).

    The exercise:
    (hopefully this works..)
    [​IMG]
    [​IMG]

    The mass of object A is 2kg
    The mass of object B is also 2kg
    The mass of object C is yet unknown
    The friction coefficient between object A and B is μ=0.808
    (no friction between A and the rail)


    1) I need to find the maximum value for the mass of object C so that all 3 object will accelerate in the same acceleration (answer:Mc=3.5kg)

    2) Now it is known that the mass of object C=4kg. They are asking for the acceleration of object A,B and C (answers: the acceleration of objects A and C is 2.67 m/s^2, acceleration of object B is 2 m/s^2)

    2. The attempt at a solution

    1) What I tried to do is solve the problem by using 3 equations:

    object B-
    T=fk (fk=friction) => fk=μ*N=0.808 * mgCos30=14N
    so: T=14N

    objects A+B-
    T-MgSin30=MgSin30*a when M=4kg and T=14
    so that: a=-0.3 m/s^2

    object C-
    Mc*g - T = Mc*a when Mc=The unknown mass of object C and a=-0.3
    so that: Mc=1.5kg

    But, the right answer by the book is 3.5kg

    2) Tried to solve but with no luck.

    Can someone please help me understand what is it that I'm doing wrong? I'm in a point where I'm completely helples, I have no clue..
     
    Last edited: Nov 4, 2008
  2. jcsd
  3. Nov 4, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hello Abukadu! Welcome to PF! :smile:

    If T is the tension, then your equation is wrong …

    there is no tension on B …

    the only forces on B are the friction the normal force and the weight.

    Try again! :smile:
     
  4. Nov 4, 2008 #3
    Besides that my other 2 equations are right?

    Hmm.. I'm trying to imagine how object B behaves in this situation.
    I know that in this case the normal force is there for me to find the friction (since i have μ, and I know that the normal force is equal to mgCos30=2*10*sin30=17.32). Now, is object B accelerating? how can i create an equation for object B when there is no force pulling him to the right?
     
  5. Nov 4, 2008 #4

    tiny-tim

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    B is accelerating at the same rate as A and C.

    Call it a.

    And there is a force on B pulling it to the right, it's … ?

    Hint: use good ol' Newton's second law for B to find a.

    Then use Newton's second law for A to find T.

    Then … :smile:
     
  6. Nov 4, 2008 #5
    Object B-

    -Fk = Mb*a ?

    Feeling kinda dumb right now with this guessing game.. I'm in a point where I'm trying things I know arent true just for the chance I'll get to the right answer :uhh:
     
    Last edited: Nov 4, 2008
  7. Nov 4, 2008 #6

    tiny-tim

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    Yes :smile: (but why the minus sign?)
     
  8. Nov 4, 2008 #7
    I figured it's a minus sign because it is opposing the direction of the acceleration.. Anyway, I tryed that option before :) and yet no luck..

    Object B-
    Fk=Mb*a

    Fk=μ * N=0.808 * mgCos30=0.808 * 2*10*Cos30=13.99

    therefor:
    13.99=2a
    a=6.99 m/s^2

    Object A-
    T-Mg=Ma now i figured the mass is 4kg because object B is on top so:
    T-40=4*6.99 -> T=67.98

    Object C-
    Mc*g-T=Mc*a -> 10Mc - 67.98 = 6.99Mc -> Mc=22.5kg which is,of course, incorrect!

    I'm not sure what is the scientific term to call it, but did I place a wrong variable or something like that?

    By that way, Thanks a lot for your help! I'm realy going to adopt this forum as a second home for my physics course and share it with my friends :approve:
    You dont have a problem with foreigners, right? :tongue2:
     
  9. Nov 4, 2008 #8
    Since it looks like tiny-tim logged off (maybe he had to a teach class! I just finished!), I'm looking at it now. I like your above equations for A & C. But you're missing one force in B that acts parallel to the slope. There's the friction force, but also part of the weight wants to oppose forward motion up the slope... include that.

    P.S. One of the reasons I like PF is that I can help students in all different parts of the world (at all hours of the day!). Welcome! :smile: It was a big boost to my procrastination when I wrote my dissertation! :rofl:
     
  10. Nov 4, 2008 #9
    Okay so you mean I should do something like this:
    mgSin30 + Fk = ma -> 2*10*sin30 + 13.99 = 2a -> a=12 m/s^2 ?
    because I still dont get the right answer.. :(

    BTW
    "It was a big boost to my procrastination when I wrote my dissertation! "
    Even if I was a native speaker I dont think I could understand that sentence :D
     
  11. Nov 4, 2008 #10
    I think there's a sign error wrong in your diagram for B.

    Let's examine the concept of static friction a bit (I think maybe I misled you a bit originally, probably hopping in too quick...)

    Static friction is present to KEEP the block from sliding relative to the other block... as Tiny-tim is hinting: "And there is a force on B pulling it to the right, it's … ?"

    It's the static friction! The static friction has a maximum value (given by the normal force and the coefficent). This value can be reached by: some of it being used to keep the block from sliding down due to weight. The rest of it can be used to keep the block from sliding relative to the other block as the other block accelerates beneath it... i.e. to give your acceleration.

    I think set it up like this for B: ma + Fg sin theta <= Fk (with <= being less than or equal.. you want the "equal part to give the maximum value for the acceleration)
     
  12. Nov 4, 2008 #11
    Okay I started all over again..

    C:

    ∑F_y = T-m_C*g = m_C*a
    ∑F_x = 0

    B:

    ∑F_y = N_2 - m_B*gcos(a) = 0
    ∑F_x = f_s = m_B*a

    f_s:

    f_s = μN_2

    A:

    ∑F_x = T - m_A*gsin(a) - f_s = m_A*a
    ∑F_y = N1 - N2 - m_Agcos(a) = 0

    And still.. the results turns out to be 2.23kg, and not 3.5kg.
    Can you please try to solve the question and tell me if you get the same answer?
     
  13. Nov 4, 2008 #12
    I have some intermediate answers.... what did you get for your acceleration?
     
  14. Nov 4, 2008 #13
    6.997 m/s^2
     
  15. Nov 4, 2008 #14
    basically, what you need to understand here is that, the frictional force between the blocks A and B is slowing down A, but accelerating B. Also there is no difference in the MAGNITUDE of accelerations, so acceleration of the system = (Driving force/ Inertia)
     
  16. Nov 4, 2008 #15
    you still have the "x" part of b wrong -- you didn't include the weight parallel to the slope.
     
  17. Nov 4, 2008 #16
    Note: since I have to go, I'll let you know: the a I get is 2 m/s^2. I think it then all works out.
     
  18. Nov 4, 2008 #17
    B-
    ∑F_x; f_s - mg*sin30= m_B*a
    that way I get the acceleration of 2 m/s^2
    but still the final result is Mc=2.3 kg and not 3.5kg ..
     
  19. Nov 4, 2008 #18
    C:

    ∑F_y = T-m_C*g = m_C*a
    ∑F_x = 0

    B:

    ∑F_y = N_2 - m_B*gcos(a) = 0
    ∑F_x = f_s - mgSin30 = m_B*a

    f_s:

    f_s = μN_2

    A:

    ∑F_x = T - m_A*gsin(a) - f_s = m_A*a
    ∑F_y = N1 - N2 - m_Agcos(a) = 0

    is it possible that after i find a=2 m/s^2 and I'm moving on to object's A ∑F_x equation, object A has a different f_s than the one I calculated for object B?
    it turns out that in order to get the right answer, object A's f_s has to be twice has big. maybe its some sort of a coincidence, I dont know..
     
  20. Nov 4, 2008 #19

    tiny-tim

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    Hi physics girl! Hi Abukadu! :smile:

    Yup … it's definitely 2 m/s2, and, like physics girl, I get 3.5kg.

    Hint: you should get a tension of 28N. :smile:
     
  21. Nov 4, 2008 #20

    tiny-tim

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    sorry!

    Hi Abukadu! :smile:

    You beat me by a minute!
    ah … I think I misled you :redface: when I said:
    To make the friction an internal force (so that you can ignore it), you need to use Newton's second law for A-and-B combined … that's much simpler than using it on A separately! :wink:
     
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