Something about applied dynamics.

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    Applied Dynamics
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Discussion Overview

The discussion revolves around the differentiation of a dynamic equation involving positions (SA and SB) and their relationship to a constant length (l). Participants are exploring the implications of differentiating the equation and the conditions under which certain variables are considered constants.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why the equation 2SA + SB = l leads to 2aA = -aB after differentiation, noting that l disappears in the process.
  • Another participant suggests that since l is a constant, differentiating the equation results in 2aA + aB = 0.
  • There is a challenge regarding the assumption that SA and SB are not constants, with one participant arguing that if SA is constant, then both its velocity (vA) and acceleration (aA) would be zero, and the same applies to SB.

Areas of Agreement / Disagreement

Participants express differing views on whether SA and SB can be considered constants, leading to an unresolved discussion about the implications of these assumptions on the differentiation process.

Contextual Notes

The discussion lacks clarity on the definitions of SA and SB and their dependence on the context of the problem, which may affect the differentiation and the resulting equations.

aiklone1314
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May i know actually why initially 2SA+SB=l , but after differentiate, it becomes 2 aA = -aB ?
why does the l disappear?
 
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This thread does not belong here. Please move it to the homework thread.

Matt
 
Hi aiklone1314! :smile:
aiklone1314 said:
May i know actually why initially 2SA+SB=l , but after differentiate, it becomes 2 aA = -aB ?
why does the l disappear?

2SA+SB = l, so 2aA+aB=0 (because l is a constant), so 2aA = -aB :wink:
 
but how to know SA and SB is not a constant?
 
aiklone1314 said:
but how to know SA and SB is not a constant?

If sA is constant, then vA and aA will be 0 (same for sB)
 
tiny-tim said:
If sA is constant, then vA and aA will be 0 (same for sB)

ok thank you very much...
 

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