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I'm I on the right track? Tension question

  1. Oct 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Three masses as shown below are resting on a horizontal, frictionless surface. The blocks are connected by massless, frictionless ropes. A force F of 60 N acts on the system as shown. Find the tensions in each of the other two ropes. m1= 10kg, m2 = 20 kg and m3 = 30 kg.

    2. Relevant equations
    ΣF=ma

    3. The attempt at a solution
    ΣF=ma
    60N=(10kg+20kg+30kg)a
    1=a
    Tension of the rope between m1 and m2
    Ft=(10kg+20kg)1m/s
    Ft=30N
    The second part confuses me, wouldn't it be the weight of (m1+m2) since m3 is pulling both? Or is the first part completely irrelevant?
     

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  2. jcsd
  3. Oct 11, 2015 #2

    JBA

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    Your first equation is fine, but your second is in error. Ask yourself: How much tension is removed from the 60N load as you proceed away from that load point.
     
  4. Oct 11, 2015 #3

    haruspex

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    Can you explain in detail why the tension would be that? Did you draw a free body diagram for m1?
     
  5. Oct 11, 2015 #4

    JBA

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    The total tension on the rope to m1 is 60N = (m1 + m2 + m3)a, so what would the tension on the rope be if m1 were removed.
     
  6. Oct 11, 2015 #5
    Would it be Ft=m(a×g)?
     

    Attached Files:

  7. Oct 11, 2015 #6
    Oh okay thank you, that's what I originally thought it was :)
     
  8. Oct 11, 2015 #7

    haruspex

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    Eh? Why would you multiply two accelerations together? If you meant +, it still makes no sense.
    Your FBD is ok, but what do you deduce from it regarding horizontal forces and accelerations?
     
  9. Oct 11, 2015 #8

    sophiecentaur

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    Looks ok to me.
    The 60N F is acting on all the masses - gives you the acceleration. All three masses are accelerating at a.
    The second piece of string T2 is accelerating masses two and one - gives you the force / tension in that string.
    The last string T1 is just accelerating the m1.
    I don't see your reasoning here. It looks OK to me. Oh I see - he has mixed up the numbering of the masses (here "Tension of the rope between m1 and m2" when he means m2 and m3 - but still gets the right answer.
    It just shows how much better it is to use the notation (T1, T2, T3) given in any question - to avoid that sort of confusion
     
  10. Oct 11, 2015 #9
    I'm sorta confused now.. so T1=(10kg+20kg)1m/s which gives me 30N and the second part would be T2=(20kg+30kg)1m/s, right or this is formula completely messed up?
     
  11. Oct 11, 2015 #10

    JBA

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    No, I said that your first equation is correct all of the tension forces are horizontal and additive. Gravity has no effect since the surface under the blocks is frictionless. When I said what would be the tension with the m1 mremoved I meant: If you move m1 to the left side of the equation, then what is the result. At the same time a/gc is the correct term because m1/gc, m2/gc, m3/gc are the masses of each block.
     
  12. Oct 11, 2015 #11

    JBA

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    F total = 10 + 20 + 30 so what is left if you subtract m1=10 and then subtract m1 +m2 = 30 + 20

    Sorry I made a typo see this edit
     
  13. Oct 11, 2015 #12

    haruspex

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    if you read exactly what is in the OP, it would appear that Kingyou has involved the masses of both m1 and 2 in finding the tension in the rope between them. This is regardless of how the tensions are numbered.
     
  14. Oct 11, 2015 #13

    JBA

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    No if I subtract both m1 and m2 then that leaves the tension between m2 and m3.

    Lets look at from a different perspective. If you are holding a rope with m1 + m2 + m3 hanging down and you grab the rope below m1 and remove it how much weight will you be holding? and then do the same thing with m2 then how much weight will you be holding.
     
  15. Oct 11, 2015 #14
    so the tension between m1 and m2 is just t1=(10kg)1m/s? and then the tension between t2=(10kg+20kg)1m/s?
     
  16. Oct 11, 2015 #15

    JBA

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    No the tension between m1 and m2 = m2 + m3.
    To get the correct solutions use your first equation and start subtracting the masses starting at m1 not m3.
     
  17. Oct 11, 2015 #16
    Thank you, so the first part would be t1=(20kg+30kg)1m/s 50 N and then t2=(30kg)1m/2
     
  18. Oct 11, 2015 #17

    JBA

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    Yes, good job, even if it took us a bit of time to get there. But you need to eliminate the "1 m/s" (that is the unit for velocity).

    t1=(20kg+30kg)= 50 Kg and then t2 = 30 Kg
     
  19. Oct 11, 2015 #18
    This is correct if you change as shown in red.
     
  20. Oct 11, 2015 #19

    JBA

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    Alright, at this point I realize that I have been adding to the confusion all along by missing the fact that T1 is between m2 & m3 and T2 is between m1 and m2. rather than the reverse.
    I apologize for this oversight when viewing the figure.

    Kingyou123, I wish you would have named me directly when referring to the error. I saw your note but didn't realize it was my error you were referring to.
     
  21. Oct 11, 2015 #20
    It's okay, I figured it out thank you for the help
     
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