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Two boxes tied together on an inclined plane. Find common acc and tension

  1. Sep 22, 2011 #1
    Greetings! This is my first post on this forum :)


    phys1.jpg
    ("Fasit" means solution in Norwegian)

    1. The problem statement, all variables and given/known data

    Inclined plane, 30 degrees from horisontal
    Two boxes tied together.
    BoxA = 4 kg, friction coefficient = 0,25
    BoxB = 8kg, friction coefficient = 0,35

    Box A is closest to the bottom of the plane. Box B is slighly above. Not touching.


    A: Find each box' acceleration (the solution implies a common acceleration)
    B: Find the tension between the boxes


    2. Relevant equations

    [tex] a = (g * sin(30)) - (\mu * g * cos(30))[/tex]
    [tex] \sum F = m * a[/tex]


    3. The attempt at a solution
    Regarding Part A
    I found both boxes' individual acceleration using the formula above:

    [tex]A_A = 9.81 m/s^2 * sin(30) - 0.25 * 9.81 m/s^2 * cos(30) = 2.78 m/s^2[/tex]
    [tex]A_B = 9.81 m/s^2 * sin(30) - 0.35 * 9.81 m/s^2 * cos(30) = 1,93 m/s^2[/tex]


    However, the solution is supposed to be [tex]2.21m/s^2[/Tex]

    What am I doing wrong?


    Regarding Part b

    I am getting nowhere trying to find a solution for this part.
    I have tried multiplying the difference in acceleration with box A's downward force, but I am not getting anything near the solution which is supposed to be 2.27 N
     
    Last edited: Sep 22, 2011
  2. jcsd
  3. Sep 22, 2011 #2

    PeterO

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    The boxes are tied together, so Box A cannot accelerate at a greater rate than Box B.

    There will be some tension in the string which will reduce the acceleration of A and increase the acceleration of B so they are equal so you have to consider the forces on the masses, not just their acceleration.
     
  4. Sep 22, 2011 #3

    Doc Al

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    If you look at each box separately, you must include all the forces on them. Including the tension in the rope.

    First find the acceleration by treating both boxes together as a single system. What external forces act on the system?
     
  5. Sep 22, 2011 #4
    Thanks for your reply, Peter.
    I'm not really getting the hang of this newton-chapter we're going through, but your reply seems logical (at least).

    So, the difference in acceleration is
    [tex]2.78m/s^2 - 1.93m/s^2 = 0.85 m/s^2[/tex]
    This acceleration multiplied with BoxA's weight (4kg) should result in a force (3,4N, so it's not the tension force). This force slows down A and speeds up B so that they eventually end up at 2.21m/s^2.

    But there's something in between I dont know how to calculate.
     
  6. Sep 22, 2011 #5

    PeterO

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    As Doc-AL says it is easier to treat the two masses together and add up all the forces helping and hindering the acceleration of one large system.
     
  7. Sep 22, 2011 #6

    Ok, thanks. I have been experimenting somewhat on this single system part. What I did was trying to find a mean friction coffesient, but i got lost and starting googling for help.

    Anyways, thanks for helping me asking the right questions and thinking in the right direction(s).


    the external forces on this system would be the following:
    G, being
    [tex]F_A=(9.81m/s^2 \times 4kg \times cos(30) + F_B = (9.81m/s^2 \times 8kg \times cos(30)) = 102N[/tex]

    R, being the friction force for both boxes:
    [tex]FF_A= (sin(30) \times 9.81m/s^2 \times 4kg \times 0.25) + FF_B=(sin(30) \times 8kg \times 9.81m/s^2 \times 0.35) = 18.63N[/tex]

    As these two forces act in opposite direction, their difference equals:
    [tex]102N - 18.63N = 83,4N[/tex]

    Using the formula F = m*a, I obtain that a=F/m. Using my recently aqquired numbers i get:
    [tex] \frac{83.37N}{4+8kg} = 6,94m/s^2 [/tex]....

    What force did I forget.. ?
     
  8. Sep 22, 2011 #7

    PeterO

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    I think you may have got your sines and cosines back to front this time? In your first attempt, the cosine was in the friction calculation, and the sine was in the "accelerating force".
     
  9. Sep 22, 2011 #8
    edit: the friction was calculated by the means of the normal force, which act in the Y-direction. (i have shifted the axis, so that X-axis is parallell to the plane while the y-axis is normal to it)
     
  10. Sep 22, 2011 #9

    PeterO

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    If the angle was zero, there would be no parallel component.

    Is it sine or cosine that equals zero when the angle is zero? That is the way I remember it? - or do I really mean work it out each time?
     
  11. Sep 22, 2011 #10
    see my edit above.

    I like to remember it as moving in the x-direction is a lot more cozy(cosine) than moving upwards in the sine direction.
     
  12. Sep 22, 2011 #11

    Doc Al

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    :confused:

    What's the component of the weight parallel to the surface? Normal to the surface?
     
  13. Sep 22, 2011 #12
    I think that the normal force is directly oposite of the y-component of the weight ?

    example:
    [tex]9.81 * 4kg * \sin (30) = 19.62N[/tex]
    in the Y-direction poiting downwards
    The normal force should be the same force in the oposite direction, right?
    And the friction force is the normal force times friction coeffisient.

    Have i misunderstood this as well?
     
  14. Sep 22, 2011 #13

    Doc Al

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    As PeterO has already pointed out, you are mixing up your sines and cosines. Answer my questions:
    What's the component of the weight parallel to the surface? Normal to the surface?
     
  15. Sep 22, 2011 #14
    OK, the component of weight parallel to the surface would be mass times gravity times cosine(30) i suppose.

    Normal to the surface should be the same as above with sine instead of cosine.
     
  16. Sep 22, 2011 #15

    Doc Al

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    That's backwards. Consider what the normal component would be if the angle of the incline were 0 degrees. Or 90 degrees.

    (Interesting that you had that correct in your first post. What happened to change your mind?)
     
  17. Sep 22, 2011 #16
    Yeah.. i see. Flat surface = cosine 0 deg = 1 = no change.

    I am gonna do some new calculations now, and hopefully return with a sensible solution to this.

    Thanks a lot for helping me out :)
     
  18. Sep 22, 2011 #17
    New dawn:

    First: Finding the friction force for both A and B:
    [tex]FF_A = 0.25 \times 4kg \times 9.81m/s^2 \times \cos(30) = 8,5N [/tex]
    [tex]FF_B = 0.35 \times 8kg \times 9.81m/s^2 \times \cos(30) = 23,8N [/tex]
    [tex]FF_A + FF_B = 8,5N + 23,8N = 32,3N[/tex]

    Second: Finding the x-component of weight for both boxes:
    [tex]F_A = 4kg \times 9.81m/s^2 \times \sin(30) = 19,6N[/tex]
    [tex]F_B = 8kg \times 9.81m/s^2 \times \sin(30) = 39,2N[/tex]
    [tex]F_A + F_B = 19,6N + 39,2N = 58,84N[/tex]

    Third: Subtracting friction force from weight:
    [tex]58,8N - 32,3N = 26,5N [/tex]

    Fourth: Applying Newton's second law:
    [tex]A = \frac{\sum F}{m} = \frac{26,5N}{4+8kg} = 2,21 m/s^2[/tex]


    Thank you so much, Doc Al and Peter for getting my pieces together
     
  19. Sep 22, 2011 #18
    OK, I am suddenly on a roll now and figured out Problem B as well.
    B: What is the tension between the boxes.



    First step: Determine Box A's desired weight downwards:
    [tex]F_A - FF_A = 19,6N - 8,5N = 11,1 N[/tex]

    Second step: Determine Box A's current weight downwards:
    [tex]4kg \times 2.21 m/s^2 = 8,84N[/tex]

    Third step: Tension between box A and box B has to be the difference in actual and desired weight:
    [tex]11,1N - 8,84N = 2,26N[/tex]


    woho
     
  20. Sep 22, 2011 #19

    PeterO

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    Good Work
     
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