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Newton, Find T

  1. Oct 6, 2012 #1
    knight_Figure_08_36.jpg

    It's the same diag. but the blocks are A and B and the angle is 30°

    Theta=30°
    Block A=15kg
    Block B=12kg
    assume the rope is massless
    us=.25=coefficient of static friction
    What is T?

    My Attempt at a solution
    I made FBD for both block A and B

    For A sum the forces
    Fx=t1-m1gsin30°-f=ma
    fy=N1-m1gcos30°=0

    Block B
    F=m2g-T2=m2a

    So I'm looking for the Tension in the rope. Is it just 1 rope or 2 ropes? I really don't know what to do now. Do I find t2 before i find t1? Do I find the acceleration before I find the Tension?

    Thank you!
     
  2. jcsd
  3. Oct 6, 2012 #2

    tiny-tim

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    Hi SherBear! :smile:

    (try using the X2 button just above the Reply box :wink:)
    from the diagram, it's one continuous rope, so (if the pulley is smooth) the tension is the same all the way along, and you can just call it T :wink:
    fine so far :smile:

    now write f in terms of N, and you'll have two equations with two unknowns (a and T) …

    eliminate one and find the other :smile:
     
  4. Oct 6, 2012 #3
    In terms of N, Would that be.... n1=m1gcos30° ?
     
  5. Oct 6, 2012 #4

    tiny-tim

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    no, f = µN1 :wink:

    (btw, did you get an email notification of this answer? i didn't get one for your answer, and only noticed it by accident :redface: some members are complaining that the notifications aren't operating properly)
     
  6. Oct 6, 2012 #5
    Yes, I did get email notifications both times for your answer :)
     
  7. Oct 6, 2012 #6
    u=.25 and n1 is? I have 2 unknowns, f and n1? I am sorry I'm really not that smart
     
  8. Oct 6, 2012 #7
    My Professor says solve for N then plug into the Fx=T-mgsin20-f=ma or is it -ma, ?
     
  9. Oct 6, 2012 #8

    tiny-tim

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    (hmm … i got both of those notifications! :frown:)
    yes, f = .25 n1 is the extra equation :smile:

    (except, i've just noticed, that's the coefficient of static friction, so it only applies if a = 0 … perhaps he means dynamic friction?)

    (and yes n1 is one of your unknowns, but you can get it from your n1 - m1gcos30° = 0 :wink:)

    (btw, best to stick capital letters for forces, small letters for distances speeds accelerations etc)
    depends which way you're measuring a …

    you must measure a for both masses in the same direction (along the rope)
     
  10. Oct 6, 2012 #9
    usf is the only coefficient for friction he gave us so for Fx Fy and F for B all = 0 in the equations if it is static friction?

    using N1-m1gcos30=0 i got
    m1gcos30=N1?
    (15kg)(9.80m/s^2)(cos30)=N1
    127.30N=n1

    then use f=uN1
    f=(.25)(127.30N)
    f=31.825N?
     
  11. Oct 6, 2012 #10

    tiny-tim

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    looks ok
     
  12. Oct 6, 2012 #11
    Thank you for your time and patience Tim. Is the final answer, f=31.825N? the Tension or?
     
  13. Oct 6, 2012 #12

    tiny-tim

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    uhh? :confused:

    f is the friction
     
  14. Oct 6, 2012 #13
    How do I find tension? I really don't know what to do next?
     
  15. Oct 6, 2012 #14

    tiny-tim

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    use your original equation …
     
  16. Oct 6, 2012 #15
    If I use my original equation I get.. T=m1gsin30-f / ma
    T = (15kg)(9.80m/s^2)(sin30)-31.825N / (15kg) (a) <---and I don't know a?
     
  17. Oct 6, 2012 #16

    tiny-tim

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    block B ? :wink:
     
  18. Oct 6, 2012 #17
    Could this finally be correct?
    T = (15kg)(9.80m/s^2)(sin30)-31.825N / (15kg) (12kg)
    T= .637 N ?
     
  19. Oct 6, 2012 #18

    tiny-tim

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    i'm confused :confused:

    you had T - m1gsin30° - 31.825 = m1a

    and m2g - T = m2a

    to find T, you need to eliminate a, so multiply the first equation by m2 and the second equation by m1, and then subtract
     
  20. Oct 6, 2012 #19
    Sorry Tim =/

    T=m1g sin 30° - 31.825 (m2)

    T=(15 kg) (9.80 m/s^2) (sin 30°) - 31.825 (12 kg)

    T=1382.1 N?

    Then...2nd Equation

    T = m2g (m1)

    T=(12 kg) (9.80 m/s^2) (15 kg)

    T=1764 N?

    Then Subtract the 2....

    1764 - 1382.1 = 381.9 N

    T= 381.9 N?
     
  21. Oct 7, 2012 #20
    Here's a different way, can you tell me if it's right or wrong?

    forces on A
    u*m*g*Cos30 -T = ma
    T - 0.25*15*9.8*cos30 = 15*a
    T - 31.82N = 15a .........................................i

    Forces on B
    m*g-T = ma
    12*9.8 -T = 12a
    117.6-T = 12a...........................................ii

    adding i and ii
    -31.82 + 117.6 = 27a
    85.78 = 27a
    a = 3.17 m/s/s
    Now T = 117.6 - 12*9.8 = 0 No motion
     
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