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The retardation force on the horizontal portion is not the only reason the tension reduces. Consider a massless string but with a mass on each end. The tension is reduced when accelerating even though there is no retardation force. This corresponds to the ##-\lambda\frac{dv}{dt}## term in the vertical.NTesla said:No retardation condition is fulfilled when no motion is happening. So, yes, when the chain is not moving at all, then Tension is more than when it is moving. So, this statement is true.
I'm still trying to understand this.
Let's say that before the chain starts to move, the tension at the rightmost point of the purely horizontal portion of the chain is ##T_{when\,static}##.
Now, let's say, the chain is in moving condition, then the tension at the same location is ##T_{when\,moving}##.
Let ##F_{on\,element}## be the leftward force on the element which is just at the bend.
then ##T_{when\,moving}## < ##T_{when\,static}## as long as there is a horizontal chain to talk about.
When you say, that
It means that equation wise, it would be like this:
##T_{when\,moving}## = ##T_{when\,static}## - ## F_{on\,element} ## . ...........................................(i)
And from earlier posts, we have established that magnitude of ## F_{on\,element} ## = ##\lambda v^{2}##.
But, ##T_{when\,static}## = ##\lambda hg##, and ##T_{when\,moving}## = ##\lambda h(g - \frac {dv}{dt})##
So, ##T_{when\,moving}## = ##T_{when\,static}## - ##\lambda h \frac {dv}{dt}## ...........................(ii)
Clearly, equations (i) and (ii) are not same, the last term that is.
At this point I'm certain that I'm misunderstanding the critical point of view/ information about what you've written. But I've read your above post multiple times, and have spent much time thinking about it, but I still don't understand it.
The correct comparison is with the massless string with two point masses model, though the vertical equation may be complicated by table impact.