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Homework Help: Physics help please Torque Questions

  1. Apr 16, 2005 #1
    Physics help please :( Torque Questions...

    umm okay i have 4 physics questions that i was hoping someone could help me through it..but it's on a sheet of paper and i scanned it through the computer..i dunno if there's anyway i can put the question on here? or if someone is willing to help me, it deals with torque..if you want, can you plz private message me with ur email address and i'll send you the file personally..lemmi know if you private msged me also, so i can check :P
    thanks alot you guys..
    it's part of a lab and im loost 'cuz i missed a couple days :( :frown:
  2. jcsd
  3. Apr 16, 2005 #2


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    there's an option when you post a message that lets you upload a file.

    See "Additional option" --> "Attach Files" Right under the emoticons and "submit reply" etc buttons

    Attached Files:

  4. Apr 16, 2005 #3
    ahh ijust tried to upload it that way..it says my file is too big though..my file is 238kb... :frown:
  5. Apr 16, 2005 #4
    okaay so i made it smaller! so can anyone help guide me thru it...

    Attached Files:

  6. Apr 16, 2005 #5


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    Note that they are currently "pending approval". I'm not sure what this means, but other people won't have acess to it until they get approved. (By who?)
  7. Apr 16, 2005 #6
    hmm....so you can't check it out yet? cuz i can..lol maybe it's cuz i put it up though...ahhh!
  8. Apr 17, 2005 #7
    can you guys see the pics yet? if not..can someone msg me and i'll email it..cuz i need it by monday.. :( thanks you guys
  9. Apr 17, 2005 #8
    Its still pending. Itll be up by tomorrow since youve apparently gone to bed already.
  10. Apr 17, 2005 #9
    hmm so how is it now?
  11. Apr 17, 2005 #10
    For problem 1 (first picture),

    you have a bar balanced at the center, and a torque applied on the right side of force 12 N at a distance of 80cm. Calculate the torque about the axis ofrotation. Since the bar would rotate about the triangle in the center, your radius is 80cm.

    Now the question is asking, what torque when applied at 30cm to the left of the axis of rotation will keep the bar balanced?

    remember the bar will balance if both torques are equal:

    [tex] \tau_r = \tau_l [/tex]
  12. Apr 17, 2005 #11
    No. 2 is the pretty much the same thing

    3 is a bit tougher, you want all the torques to sum up to zero, so equilibrium can be achieved. I would add up all the torque vectors starting fom the left, and usingthe left end (0cm) as your axis of rotation. Add allthe torques together, and then find the torque needed on the 'x' vector to make it 0.
  13. Apr 18, 2005 #12
    okay..so for 1a.) asking for the magnitude of the force,i did

    .3m x 12N = 3.6 N/M
    X= 18...im not sure what's the unit for this though? N?
  14. Apr 18, 2005 #13
    I'm still having problems with 1b though..
  15. Apr 18, 2005 #14
    The units for torque are Newton meters, (Nm).

    You are applying a force of 18N at .2m away, so the units would be N.
  16. Apr 18, 2005 #15
    1b is just asking the same question if the force was applied at 10cm point. Also I just noticed, in your solution to 1b, you want the distance from the fulcrum (the triangle), not the distance from the edge. The '30cm' labelled is the distance from the edge, and the fulcrum is at 50cm, so the distance is 20cm. This will give you a different answer for 1a. Always pay attention to things like this.
  17. Apr 18, 2005 #16
    okay..so for 1a, in what i did, i put in .2m where i put .3m before..and i got an answer of 12N, is that correct?

    And for 1b, is the answer 6N?
    soo sorry, i just have so much trouble understanding this.. :(
  18. Apr 18, 2005 #17
    The torque on the right side is 0.3m * 12N = 3.6N

    The torque on the left side is 0.2m * x N

    For equilibrium they will be equal:

    3.6Nm = 0.2x Nm

    Divide both sides by 0.2m (equal to multiplying by 5)

    18N = x N

    Sorry you did the first one correctly. I don't know how I thought otherwise.


    Torque on the right = 3.6Nm
    Torque on the left = .4m*xN

    3.6Nm = .4x Nm

    Divide by .4m

    9N = x N
  19. Apr 18, 2005 #18
    okay..ahh thanks so much for that! im gonna try number 2 now..ahh! lol
  20. Apr 18, 2005 #19
    for number 2, i got 1m...is that right?

    Torque on the right:
    2.5m * 20N = 50Nm
    to be equilibrium...
    X50N = 50Nm
    divide by 50 N and i get 1m...
  21. Apr 18, 2005 #20
    You got it.
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