I.184 A chain AB of length l is located in a smooth horizontal tube so that its fraction of length h hangs freeely and toches the surface of the table with End B. At a certain moment the end A of the chain is set free. With what velocity will the chain slip out of the tube.
My doubt is this in the offical solutions it is written that at moment t the only force acting on the chain left in tube is due to the tension force(which is obvoiusly mhg/l). However it is also written that force due to removing mass which is F=v rel dm/dt =0 since v rel =0. i dont undersand this as the mass which is falling down is falling down verticall with velocity v(at that instant ) so obviusly vrel !=0 ...
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1 Answers
71You need to see the direction in which that force is considered. If the velocity in that direction is zero, then certainly Ft = 0