text: # Rigid Object in Equilibrium Problem please
  
  1. Jan 24, 2014
  
  ### gcombina
  
  A 3.0-kg ball and a 1.0-kg ball are placed at opposite ends of a massless beam so that the system is in equilibrium as shown. Note: The drawing is not drawn to scale. What is the ratio of the lengths, b/a?
  (a) 2.0 (c) 3.0 (e) 5.0
  (b) 2.5 (d) 4.0
  
  #### Attached Files:
  
  • ###### exercise physics.png
  File size:
  1.2 KB
  Views:
  162
  2. Jan 24, 2014
  
  ### BvU
  
  Well, you formulated the problem, so we know what is asked in the exercise. What is your question, what are the relevant equations and what have you attempted to solve the problem?
  
  3. Jan 24, 2014
  
  ### BvU
  
  I see you did well in an earlier exercise which is very, similar to this one. Now there are only three forces. Pick a good point to calculate the sum of torques comfortably, and that's all!
  
  4. Jan 25, 2014
  
  ### gcombina
  
  lb = left ball
  rb = right ball
  SO I GOT 2 TORQUES
  -(Tlb) (Llb) + - (Trb) (Lrb) = 0
  -3kga - 1kgb = 0
  -3kga = 1kgb
  a = 1kgb/3kg
  a= .33 b
  
  5. Jan 25, 2014
  
  ### BvU
  
  Excellent! So b/a is in the list.
  If you really want to delight physicists, you explicitly make clear that you did not forget g:
  $\sum$ torques = 0 w.r.t. supported point $\Leftrightarrow$
  ma * g * (-La) + mb * g * (Lb) = 0 $\Leftrightarrow$ etc.
  
  6. Jan 25, 2014
  
  ### gcombina
  
  hahaha delight physicists? I am a language major trying to get into a physician assistance program and this course is just WAY TOO HARD
  
  7. Jan 25, 2014
  
  ### gcombina
  
  OHHHH OK OK Let me try with the m.g
  
  8. Jan 25, 2014
  
  ### BvU
  
  The more respect you earn by passing !
  Remember: physicists are lazy nerds. But not insensitive to this subtle way of flattering.
  
  9. Jan 25, 2014
  
  ### BvU
  
  As I said, lazy. the etc means that from there it's "only" math.
  Math is even more important for physicians assistants: if they can prevent the doctor from making a factor 10 mistake in their recipes even only once in their career, all the effort was well worth it!
  
  10. Jan 25, 2014
  
  ### gcombina
  
  prescriptions, not recipes.......language major )
explanation: The text extract contains a physics problem involving a rigid object in equilibrium. It includes a question about the ratio of lengths, b/a, and a series of posts where the user attempts to solve the problem with guidance from another user. The user eventually arrives at an answer, and the helper confirms the solution.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Rigid Object in Equilibrium Problem please
  
  1. Jan 24, 2014
  
  ### gcombina
  
  A 3.0-kg ball and a 1.0-kg ball are placed at opposite ends of a massless beam so that the system is in equilibrium as shown. Note: The drawing is not drawn to scale. What is the ratio of the lengths, b/a?
  (a) 2.0 (c) 3.0 (e) 5.0
  (b) 2.5 (d) 4.0
  
  #### Attached Files:
  
  • ###### exercise physics.png
  File size:
  1.2 KB
  Views:
  162
  2. Jan 24, 2014
  
  ### BvU
  
  Well, you formulated the problem, so we know what is asked in the exercise. What is your question, what are the relevant equations and what have you attempted to solve the problem?
  
  3. Jan 24, 2014
  
  ### BvU
  
  I see you did well in an earlier exercise which is very, similar to this one. Now there are only three forces. Pick a good point to calculate the sum of torques comfortably, and that's all!
  
  4. Jan 25, 2014
  
  ### gcombina
  
  lb = left ball
  rb = right ball
  SO I GOT 2 TORQUES
  -(Tlb) (Llb) + - (Trb) (Lrb) = 0
  -3kga - 1kgb = 0
  -3kga = 1kgb
  a = 1kgb/3kg
  a= .33 b
  
  5. Jan 25, 2014
  
  ### BvU
  
  Excellent! So b/a is in the list.
  If you really want to delight physicists, you explicitly make clear that you did not forget g:
  $\sum$ torques = 0 w.r.t. supported point $\Leftrightarrow$
  ma * g * (-La) + mb * g * (Lb) = 0 $\Leftrightarrow$ etc.
  
  6. Jan 25, 2014
  
  ### gcombina
  
  hahaha delight physicists? I am a language major trying to get into a physician assistance program and this course is just WAY TOO HARD
  
  7. Jan 25, 2014
  
  ### gcombina
  
  OHHHH OK OK Let me try with the m.g
  
  8. Jan 25, 2014
  
  ### BvU
  
  The more respect you earn by passing !
  Remember: physicists are lazy nerds. But not insensitive to this subtle way of flattering.
  
  9. Jan 25, 2014
  
  ### BvU
  
  As I said, lazy. the etc means that from there it's "only" math.
  Math is even more important for physicians assistants: if they can prevent the doctor from making a factor 10 mistake in their recipes even only once in their career, all the effort was well worth it!
  
  10. Jan 25, 2014
  
  ### gcombina
  
  prescriptions, not recipes.......language major )
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract contains a physics problem involving a rigid object in equilibrium. It includes a question about the ratio of lengths, b/a, and a series of posts where the user attempts to solve the problem with guidance from another user. The user eventually arrives at an answer, and the helper confirms the solution.
  
  Label: question-and-answer