Something about applied dynamics.

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The discussion centers on the differentiation of the equation 2SA + SB = l, leading to the transformation into 2aA = -aB. The key point is that l is a constant, which allows for the differentiation to eliminate it from the equation. Participants clarify that if sA is constant, then both velocity (vA) and acceleration (aA) would also be zero, implying that sA and sB cannot be constants in this context. The conversation emphasizes understanding the dynamics of the variables involved in the equation. Overall, the thread highlights the importance of recognizing variable behavior in applied dynamics.
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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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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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