I actually did the same thing. For rth block and blocks above it I can write
Fr required <= F*(mr + m(r+1)+ … + mn)/(m1 + m2 +…+ mn)
So ur * g should be >= F/(m1 + m2 +…+ mn) as Fr= Normal reaction *g = (mr + m(r+1)+ … + mn)*g
I think I have found it. It involves using the concept of finding max. friction required at any surface which can be done by using an FBD, considering a system of either the upper of lower blocks depending upon net forces on this system and taking its acc to be Ac that is (net force on all...
edited the question
Uk stands for friction coefficient of kth block with the lower one, and I reached that eqn by considering common acc as if no sliding was there in whole system because that is the maximum possible acc in any case for a combined system of blocks above kth block and it.
So far I have reached a pretty promising equation that if uk * g >= F/(m1+m2+…+mn) then kth block will atleast surely not slide wrt the lower one. But the converse to this that if uk * g < … then it will “surely” slide is wrong!!
Stuck on this from many days!
Thanks (:
Sorry to bother but do you mean the acceleration of blocks as if m1 and m2 only didn’t slide relative to each other in first line because if I consider acceleration such that all three blocks don’t slide relative to each other and then compare the max frictional value, answer is coming wrong in...