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Acceleration of Block on an Apparatus

  1. Jan 2, 2008 #1
    [SOLVED] Acceleration of Block on an Apparatus

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

    See the attachment.

    2. Relevant equations



    3. The attempt at a solution

    I am confused about whether the tensions of the two (left and right) parts of the line are the same or not. Since the three blocks are accelerating so it's not a equilibrium situation. Are the tensions the same on either line?

    If block m3 were not there, and instead, m1 and m2 were connected I can use the equations below to solve for acceleration. But I am a bit thrown off by the addition of m3.

    (1) m1*g-T = m1*a
    (2) T-m2*g = m2*a

    Where T is the tension of the either line. But how do with m3, the block on the table?

    PS: HallsofIvy, ShootingStar, I appreciate your help with the previous problem.
     

    Attached Files:

  2. jcsd
  3. Jan 2, 2008 #2

    Doc Al

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    No, the two tensions are not the same. If they were, things wouldn't make much sense: The net force on m3 would be zero and thus no acceleration. Call the tensions T1 and T2 and rewrite your two equations. And add a third equation for m3.
     
  4. Jan 2, 2008 #3
    Okay. So let the left and right lines have tensions T1 and T2 respectively.

    (1) m1*g-T1 = m1*a
    (2) T2-m2*g = m2*a
    (3) m3-(T1+T2) = m3*a

    Would the above be correct?
     
  5. Jan 2, 2008 #4
    [tex] m_3[/tex] is not a force! :smile:
     
  6. Jan 2, 2008 #5

    Doc Al

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    Equation 3 is not quite right. What's the net force on m3? (Be sure to use a consistent sign convention for the acceleration in all three equations.)
     
  7. Jan 2, 2008 #6
    (1) m1*g-T1 = m1*a
    (2) m2*g-T2 = m2*a
    (3) m3*g-(T1+T2) = m3*a

    About the sign convention for acceleration, I thought I needed to change the sign because as block m1 is accelerating say, upward, then m2 must be accelerating downward?
     
  8. Jan 2, 2008 #7
    If [itex]m_1[/itex] is accelerating upward then [tex] T_1-m_1\,g=m_1\,a[/tex] and reverse signs for [itex] m_2[/itex].

    [itex]m_3[/itex] is moving on the table. What that tells you about it's weight [itex] m_3\,g[/tex]?
     
  9. Jan 2, 2008 #8

    Doc Al

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    Equation 3 is still wrong. The weight--which acts vertically, not horizontally--is balanced by the normal force from the table. Only consider the horizontal forces on m3.

    Looks like you changed equations 1 and 2, so they need fixing now. (They were fine before!)

    Previously, in your first two equations you chose to have m1 accelerate up and m2 down when a is positive. Nothing wrong with that. Use that same convention when analyzing m3.
     
  10. Jan 2, 2008 #9
    Other than that the net vertical force on m3 is zero, I'm not sure what else I can extrapolate.
     
  11. Jan 2, 2008 #10
    What Doc Al said.

    "consider the horizontal forces on m3"
     
  12. Jan 2, 2008 #11
    Oh, m3*g is the vertical force. The horizontal force is just the tension.

    (1) m1*g-T1 = m1*a
    (2) T2-m2*g = m2*a
    (3) (T1+T2) = -m3*a
     
  13. Jan 2, 2008 #12
    The first two are ok!
    For the 3rd: are the tensions in the same direction?

    and why -a?
     
  14. Jan 2, 2008 #13

    Doc Al

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    Do both tensions pull in the same direction?
     
  15. Jan 2, 2008 #14
    (1) m1*g-T1 = m1*a
    (2) T2-m2*g = m2*a
    (3) T1-T2 = m3*a

    The tensions are not in the same direction. I suppose because, I assumed +a applied to block m2 then +a must apply to m3 also. Although I have to admit I'm still not quite clear about this concept because the direction of acceleration of block m3 isn't in the direction of m2, the -y direction, rather it's in the +x direction.
     
  16. Jan 2, 2008 #15
    Every block moves at it's "own" axe. If you assume that [itex]m_1[/itex] is going down then [itex]m_2[/itex] must go up and [itex]m_3[/itex] must move to the left. Thus the "+x direction" is to the left.

    As small recipe is that when you are dealing with a system of mases like this one is that you can treat it as "one boby" system. All the internal forces must equal to zero (Newton's 3rd law) and you are left with the externals i.e. the two weight's.
    Thus if you assume that [itex]m_1[/itex] is going down

    [tex]m_1\,g-m_2\,g=(m_1+m_2+m_3)\,a[/tex]

    To put it in other words when you add all the equations the tensions must disapper! :smile:
     
  17. Jan 2, 2008 #16

    Doc Al

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    Good.

    The way to understand this is to realize that since the blocks are all connected by strings they are constained to move with the same magnitude of acceleration. And if m1 moves down, then m2 must move up, and m3 must move to the left. This constraint is applied in your equations by using the same letter (a) to represent the magnitude of the acceleration, and by using a consistent sign convention.
     
  18. Jan 2, 2008 #17
    I see that your equation was just the result of summing the equations 1,2, and 3. But how were you able to come up with it directly?

    As I understand, the left side (m1+m2+m3)*a is the acceleration of the entire system of the 3 blocks. For the left side of the equation, what about m3*g? Did you not include it because m3 is accelerating along the table? I'd appreciate if you could clarify. Thanks!
     
  19. Jan 2, 2008 #18
    Consider a system of N bloks. The net force on the whole system is

    [tex] \sum\vec{F}_{external}+\sum\vec{F}_{internal}=m_{total}\,\vec{a}[/tex]

    The sum [tex]\sum\vec{F}_{internal}[/tex] equals zero by the Newton's 3rd law, thus all the tensions are gone away! :smile:

    The sum [tex] \sum\vec{F}_{external}[/tex] has the three weights and the normal force on [itex]m_3[/itex] which cancels [itex]m_3\,g[/itex].

    Thus you are left with the equation I posted!

    Is this clear enough? My English are pretty poor! :smile:
     
  20. Jan 2, 2008 #19
    Doc Al, that explanation cleared it up for me I think.

    So one direction is negative and the other is positive. I can ascribe a negative sign to the entire movement involving m1 moving down, m2 moving up, and m3 moving to the left, and vice versa. In other words, essentially there are only two directions: positive and negative. Is this reasoning correct?
     
    Last edited: Jan 2, 2008
  21. Jan 3, 2008 #20

    Doc Al

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    Right: There are only two possibilities for the acceleration. I think you've got the idea.

    Here's how I would put it: The key is that you used a consistent sign convention for the acceleration, one that reflects how the blocks are actually constrained to move. You chose a convention such that if the acceleration turns out (when you solve for it) to be positive, that means that m1 is accelerating down and m2 is accelerating up; but if the acceleration turns out to be negative, just the opposite.
     
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