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Urgent Question - Tensions and acceleration

  1. Oct 10, 2011 #1
    Hello,

    I have an urgent question due soon, and I have no idea how to solve it.

    bio6dv.jpg
     
  2. jcsd
  3. Oct 10, 2011 #2
    I have all the answers and they turned out to be right. I just need help with the last one.

    Here is what I did:

    Fk = MuK N
    Fk = 0.41 (167)(9.8)
    Fk = 671.006

    F = ma
    671.006 = 167 a
    a = 4.018 m/s^2

    However, it's not the right answer. Please help!
     
  4. Oct 10, 2011 #3

    berkeman

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    Staff: Mentor

    I don't see a cos(θ) term in your equations. That is needed when calculating the effective normal force on the ramp.
     
  5. Oct 10, 2011 #4
    Ok so let me redo this.

    Fk = MuK N
    Fk = 0.41 (167)(9.8)cos52
    Fk = 413.11 N

    F = ma
    413.11 = 167 a
    a = 2.47 m/s^2

    The answer is still wrong. I am pretty sure I messed up because I didn't use the M2 in part c that they give me.
     
  6. Oct 10, 2011 #5

    berkeman

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    You are working the first problem of the 3, correct?

    The first equations where you calculate Fk are correct.

    But I don't understand what you are doing in the 2nd set of equations. The force of tension up by the rope on the M2 bucket is balancing what force down on the bucket?
     
  7. Oct 10, 2011 #6
    I'm solving on the last problem. Sorry for the confusion.

    I am using the second set of equations to solve for acceleration of M1.
     
  8. Oct 10, 2011 #7

    berkeman

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    Ah, no wonder I was off in the weeds :tongue2:

    The mass that gets accelerated (by the difference in the forces) is the sum of the two masses. Does that help?
     
  9. Oct 10, 2011 #8
    I don't understand. So I add the 2 masses in the first set and second set of equations?
     
  10. Oct 10, 2011 #9

    berkeman

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    I'm kind of de-synchronized from what equations you have right now. Maybe re-summarize them?

    In the 3rd question, you have more force than needed to break M1 loose and cause the M1 & M2 combined masses to accelerate to the right & down. The acceleration is defined by the sum of the forces (they are in opposite directions), i.e., the tension in the rope, and by the sum of the two masses, since that is what is being accelerated.
     
  11. Oct 10, 2011 #10
    It's giving me the coefficient of kinetic friction for M1. So I do:

    Fk = Muk*N
    Fk = 0.41(167)(9.8)cos52 = 413.11 N

    Then, to find the acceleration, I use the F = ma equation. so you said add the two masses for m.

    413.11 = (167+231.3) a
    a = 1.04 m/s^2 but it's wrong
     
  12. Oct 10, 2011 #11

    berkeman

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    The 413.11N is the force due to friction on M1. It is not the sum of the forces acting on the two bodies together.

    The better way to write the equations would be like...

    [itex]\sum F = M_{total} a[/itex]

    The sum of the forces is the friction force on M1, and what force on M2?

    The total mass is what? And that gives you what resulting acceleration for the system of M1 and M2 tied together?
     
  13. Oct 10, 2011 #12
    so ∑F = 413.11? if not, how do I find the force on M2?
     
  14. Oct 10, 2011 #13

    berkeman

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    If you draw a FBD for M2, the tension up is due to the frictional force from M1. What is the force down on M2 due to?
     
  15. Oct 10, 2011 #14
    Gravity.
     
  16. Oct 10, 2011 #15

    berkeman

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    Ding ding ding. :biggrin:

    So draw the FBD for both masses, and show the forces that act on the system. The imbalance in those forces is what causes the acceleration of the two masses together.

    Show us your work...
     
  17. Oct 10, 2011 #16
    so Fgrav = mgsin52 = 231.3(9.8)sin52 = 1786.22 N

    so ∑F=Mtotal*a

    (1786.22-413.11) = (167 + 231.3)*a

    a = 3.45 m/s^2 ? It's not recognizing my answer still
     
  18. Oct 10, 2011 #17
    Oh, I should subtract the two forces to get a net force?
     
  19. Oct 10, 2011 #18

    berkeman

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    There is no sin() involved in the force of gravity on M2. It points straight down, directly opposing the tension in the rope due to M2's frictional force...
     
  20. Oct 10, 2011 #19
    This problem is so frustrating. After so many attempts, and my answer is still wrong. please tell me where I am messing up.

    Fgrav = mg = 231.3(9.8) = 2266.74 N

    so ∑F=Mtotal*a

    (2266.74-413.11) = (167 +231.3)*a

    a= 4.65 m/s^2

    I am only trying to find the acceleration of M1...therefore I am only dividing by 167 kg/ Please help, I can't see where I am messing up!
     
    Last edited: Oct 10, 2011
  21. Oct 10, 2011 #20

    berkeman

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    231.3kg * 9.8m/s^2 does not quite equal 2263.53, according to my calculator.

    And you still only have on of the two masses instead of the total mass on the RHS of your last equation.
     
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