1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is the total torque zero?

  1. Dec 20, 2011 #1
    1. The problem statement, all variables and given/known data
    When a body is weighed on an ordinary balance we demand that arm should be horizontal if the weights on two pans are equal. Suppose equal weights are put on two pans, the arm is kept at an angle with the horizontal and released. Is the torque of the two weights about the middle point (point of support) zero? Is the total torque zero? If so, why does the arm rotate and become horizontal?


    2. Relevant equations



    3. The attempt at a solution
    Here's what i think so far:-
    2zgzvw3.jpg

    (Sorry for a bad drawing :) )

    For the first part, individual torque is not zero. For each weight the torque is mgrcosθ.
    For the second part, the total torque is zero since the torques of weight are in opposite direction.
    Now i am confused in third part. Since the net torque is zero, the balance should not move but thats not observed. Then why does it come to the horizontal position? :confused:
     
  2. jcsd
  3. Dec 20, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    I think you'll find that real equal-arm balances have the pivot point located slightly above the center line of the arms. Try drawing the arm as a triangle with a wide base (the span of the arms) and the pivot at the apex.
     
  4. Dec 20, 2011 #3
    Do you mean something likr this:-
    10wruol.jpg

    Can you explain a bit more?
     
  5. Dec 20, 2011 #4

    gneill

    User Avatar

    Staff: Mentor

    Schematically something like this:

    attachment.php?attachmentid=42101&stc=1&d=1324398147.gif

    Real scales often have ornately shaped arms that tend to disguise the offset of the pivot from the horizontal line joining the pan attachment points.

    If you take the torques about the actual picot point, I think you'll see a difference in the contributions from each pan when the arm is at an angle to the horizontal.
     

    Attached Files:

  6. Dec 20, 2011 #5
    I still don't get it.
    I am having problems taking the perpendicular distances.
    Maybe i am not able to visualize when the arm is kept at an angle.
    Can you please show me a figure to help me?

    Thanks! :smile:
     
  7. Dec 20, 2011 #6

    gneill

    User Avatar

    Staff: Mentor

    Here's a diagram. I've indicated the appropriate angles for the left hand pan. You should work out the angles for the right hand pan yourself.

    attachment.php?attachmentid=42105&stc=1&d=1324410447.gif
     

    Attached Files:

    • Fig1.gif
      Fig1.gif
      File size:
      10.8 KB
      Views:
      355
  8. Dec 20, 2011 #7
    I am sorry for asking stupid questions but what does those red arrows represent. :uhh:
     
  9. Dec 20, 2011 #8

    gneill

    User Avatar

    Staff: Mentor

    They are perpendicular to the line joining the pivot point to the point of application of the force. What do you think they might represent?
     
  10. Dec 22, 2011 #9

    I like Serena

    User Avatar
    Homework Helper

    I've also made a picture. :shy:

    attachment.php?attachmentid=42150&stc=1&d=1324590764.gif
     

    Attached Files:

  11. Dec 24, 2011 #10
    I am assuming the red ones are the forces causing torque ... but the one on the right is not really perpendicular to arm as the left one ... why?
     
  12. Dec 24, 2011 #11

    gneill

    User Avatar

    Staff: Mentor

    It is perpendicular to the arm. The arms are also in red.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Is the total torque zero?
  1. Total Torque (Replies: 4)

  2. Zero total charge (Replies: 1)

Loading...