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Tension and acceleration

  1. Jun 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Three objects are connected on a table. The rough table has a coefficient of kinetic friction of .350. The objects have masses of 4.00 kg. 1.00 kg and 2.00 kg. The pulleys are frictionless. Draw a free body diagram for each object. A.) Determine the acceleration of each object and their directions. B.) determine the tensions in the two cords.


    2. Relevant equations
    Sum of Force=MA


    3. The attempt at a solution
    I'm not exactly sure how to draw the free body diagrams and relate them to one another. I know the Acceleration and tensions are going to be the same, but I have to make sure my free body diagrams are correct and I'm not sure how to do that. View attachment phy1.bmp
     
  2. jcsd
  3. Jun 22, 2008 #2

    Hootenanny

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    As I said in your other post, I cannot as yet see your attachments, but as before the tensions in the two different ropes are necessarily the same.
     
  4. Jun 22, 2008 #3
    Ok I don't know what's wrong with the attachments. I'll just keep trying.
     
  5. Jun 22, 2008 #4

    Hootenanny

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    The attachments must be approved by a mentor before they are viewable. It shouldn't be too long before they are approved.
     
  6. Jun 22, 2008 #5
    Anybody??? I know the two blocks off the table have forces of T-Mg. and the one in the middle has T1 acting on it and T2 acting on it with the coefficient of friction stating it has .35 more force than normal acting on it. I just need to know how to determine all the forces acting on the middle one.
     
  7. Jun 23, 2008 #6

    Hootenanny

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    Okay, so the first thing to decide is in which direction is the middle block going to move, then you assign a sign convention.
     
  8. Jun 23, 2008 #7
    Ok so the force in x direction would be the tension of the string connected to the 4kg - Tension on the 2 kg. Would the friction be subtracted from this as well?
     
  9. Jun 23, 2008 #8
    Ok I have my Force for the 4 block as T4-M4g, and the force for the 2 block T2-M2g. When I sat them equal to each other because of their acceleration equals one another. Everything subtracts itself and becomes zero. This is not what I need to happen is it? I'm left with 2a=0
     
  10. Jun 23, 2008 #9

    Hootenanny

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    Yes you also need to include the frictional force.
    This is almost correct. For the 4kg block you sign convention has the tension acting in the same direction as the motion and the weight acting in the opposite direction to motion. Does that same logical to you? (The force on the 2kg block is correct).

    Now, once you have re-written the sum of the forces on the 1kg block to include friction you can apply Newton's second law to each block individually to produce a system of three equations.

    Do you follow?
     
  11. Jun 23, 2008 #10
    Yea, I believe so, since the block is moving down the Forces on the 4kg block would be T4+M4G , or M4g-T, this part is confusing me!! I would say M4g-T but I'm so confused on this I have no clue.
     
  12. Jun 23, 2008 #11

    Hootenanny

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    Are the two forces acting in the same direction that the 4kg mass moves or in the opposite direction?
     
  13. Jun 23, 2008 #12
    I was thinking in opposite directions. Because the one block has friction on it holding the 4 block up.
     
  14. Jun 24, 2008 #13

    Hootenanny

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    You are correct! I was just making sure you understood why, guessing is a bad habit to get into :wink:. Okay so let's summarise what you have so far:

    [tex]\sum F_4 = 4g - T_4[/tex]

    [tex]\sum F_2 = T_2 - 2g[/tex]

    Do you follow? Can you now do the same for the 1kg block, taking into account friction?
     
  15. Jun 24, 2008 #14
    yea thats what I have however when I solve for T in them it seems everything subtracts out to be zero? and I end up with Just A. The One block would be T4-T2-F
     
  16. Jun 24, 2008 #15

    Hootenanny

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    Correct, so now we have three forces:

    [tex]\sum F_1 = T_4 - T_2 - \mu g[/tex]

    [tex]\sum F_2 = T_2 - 2g[/tex]

    [tex]\sum F_4 = 4g - T_4[/tex]

    Instead of subtracting the equations, can you apply Newton's second law to each one individually?
     
  17. Jun 24, 2008 #16
    Thats what I believe I'm doing. I have T4=-4a+39.2 i set 39.2-(4A+39.2)=4a everything cancels out and I'm left with nothing.
     
  18. Jun 24, 2008 #17

    Hootenanny

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    I'm not quite sure what you doing. Start by explicitly writing down the three equations and we'll take it from there.
     
  19. Jun 24, 2008 #18
    Ok well i tried it another way pluggin my T4 and T2 in The force of one equation and I came up with the acceleration as 3.23 and plugged that in and found the T4=12.92 and T2=26.06 Would that seem reasonalble?
     
  20. Jun 24, 2008 #19

    Hootenanny

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    I'm afraid I wouldn't be able to say what would be reasonable. However, I would suggest plugging those values into all the original equations to see of they are consistent.
     
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