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HW Lab PHYS101: Net torque of pulley system and meterstick position balance

  1. Oct 31, 2012 #1
    THANKS, I ONLY NEED HELP #2 NOW, PLEASE
    10/31/12 @ 9:10 pm
    bump


    1. The problem statement, all variables and given/known data
    1)If the fulcrum of a 0.1 kg meter stick was placed at the 40 cm mark and a 200 g mass at the 0 cm mark, would it be possible to balance the meter stick with a 200 g mass on the long side of the meter stick? What would the position of the mass have to be to balance the meter stick?

    2) What is the torque acting on the pulley system below

    10-31-2012040104PM.jpg

    2. Relevant equations

    LA1=LA2

    Torque net= radius x force




    3. The attempt at a solution

    FROM DIAGRAM
    #2, this is my first time working on this type of problem. how should I combine the other "counterclockwise" 2 pulleys? value correct? 10N x radius #2 of 1m= 10Nm

    and the other T1=15N x radius#1 of 2m= 30Nm??

    How do i combine these 2 forces? Are they negative?


    I understand 60Nm is correct from the clockwise but not the other. Please help!!

    Please help -- Greatly Needed
     
    Last edited: Oct 31, 2012
  2. jcsd
  3. Oct 31, 2012 #2

    tiny-tim

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    welcome to pf!

    hi pmalayavech! welcome to pf! :smile:
    yes, technically torque is a vector

    but in a 2D case like this, all you need to say is "clockwise" or "anti-clockwise" :wink:

    (I'm not clear … are you asking about question 1) also? :confused:)
     
  4. Oct 31, 2012 #3
    Re: welcome to pf!

    Tim!, hey thanks


    that is correct, so it would be

    (20N)(2m)+(15N)(1M)cc=(30N)(2m)cw


    (45Joules)cc = (60Joules)cw

    there for the net force is a Fnet of 15 Joules counterclockwise

    YES, i need help on the first one as well, any idea, please.
     
  5. Oct 31, 2012 #4

    tiny-tim

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    hey pmalayavech! :smile:
    i] are we using the same diagram? :confused:

    ii] this shouldn't be an equation!! :rolleyes:

    you should simply be adding the torques (some positive, some negative)!
    what have you done so far? :smile:
     
  6. Oct 31, 2012 #5
    i) yes, same diagram

    ii) ok, no equations, I was thinking about conservation, where net force will equal to zero.

    ok, cc is negative and cw is positive? -40Nm-15Nm+60Nm= 15Nm

    from the pulley, I am looking at the left side as negatives and the right side positives. I hope I had multiplied the radius to the right "T"
     
  7. Oct 31, 2012 #6

    tiny-tim

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    +60 is right, but i don't see how you got the other two :confused:
     
  8. Oct 31, 2012 #7
    Alright, for #1

    for Larm 1
    I did r(1)=0-40
    =-40cm
    Tcc= -40 x 200
    =-800 gcm for m1La2

    now for m2LA2
    i am stuck
     
  9. Oct 31, 2012 #8

    tiny-tim

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    let's see …
    ok, now you need the torque for the 500g stick, at the 50 cm mark, and for the other 200g mass, at x cm :wink:
     
  10. Oct 31, 2012 #9
    #1,

    for Tcc= -8000gcm

    the fact that fulcrum only moved to the left 10cm (at 40cm), where it ought to be balance at 50cm (center), if we had 2 equal weights on both side, it would balance "cc" at 0cm and "cw" at 100cm, agree?

    so, since fulcrum moved to the left 10cm(40cm), weight at 0cm, on the "cw" side, the weight of that should moved 10cm to the left as well to balance. so the "cw" 200g mass should be at 90cm!!

    yes agree?

    also for #2, this is my first time working on this type of problem. how should I combine the other 2 pulleys? value correct? 10N x radius #2 of 1m= 10Nm

    and the other T1=15N x radius#1 of 2m= 60Nm??

    do i have atleast these values correct?

    Thanks again
     
  11. Oct 31, 2012 #10
    need help with the second one. please
     
  12. Oct 31, 2012 #11
    Just the last one

    for #2, this is my first time working on this type of problem. how should I combine the other 2 pulleys? value correct? 10N x radius #2 of 1m= 10Nm

    and the other T1=15N x radius#1 of 2m= 60Nm??

    do i have atleast these values correct?
     
  13. Oct 31, 2012 #12
    still trying, and no references to check answers with
     
  14. Nov 1, 2012 #13

    tiny-tim

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    hi pmalayavech! :smile:

    (just got up :zzz:)
    yes :smile:
    no

    i] it would have to be 20 cm, wouldn't it?

    ii] anyway, you're not taking the weight of the ruler itself into account :redface:

    write the equation out properly, or you'll never learn how to do these problems​
    yes, that's correct now! :smile:
     
  15. Nov 1, 2012 #14
    Thank you, I understand now. after working out the problem, it was common sense to have to have it at 20cm, to evenly distribute the weight.
    8000cc=8000cw, duhh

    And yes the "net" force is adding it all up i presume
    130 Joules

    I was able to turn this in complete now, even though late, its fine. I understand thanks
    Timmy
     
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