my mistake in m1 and m2............corrected it...........
and @ sankara........rethink.....[12]
Two bars f masses m1and m2 connected by a non deformed light spring rest on a horizontal plane. The coefficient of friction between surface and the bodies is μ. What minimum constant force has to be applied in the horizontal direction to the bar of mass m1 in order to shift the bar m2?
(i think question is complete if the diagram is needed then tell )
and do post ur method with ur answer
my mistake in m1 and m2............corrected it...........
and @ sankara........rethink.....[12]
Eureka... u are right...in ur soln... :)
in the mean time i will think how to xplain it to u... abhi tak nahi soche kaise xplain karna hai.. [3]
now how i am gonna explain u this????[7][7][7]......
i think if u rethink on this one....u will get the concept urself..........in the mean time i will think how to xplain it to u.[3][3][3]
the extension of the spring will always be less than tht of the displacement of the block
but how will the extension of the spring will be equal to the displacement of the block . i think constraint cant be applied.eureka pls reply for my dbt in ur revisons pbms post
yeah u r rite eureka
answer perfect
philip it is just different language in the question i think
they mentioned bars and just gave the usual blocks [3]
n philip ur method is wrong i dont think this one can be solved without the work energy equation written by eureka
i dont have hcv dear.......so i dont know about that.............but someone else may know.
i guess if ma memory wrks well
this is an hcv sum ....frm chap center of mass u'll get d sol der may b in solved eg
eureka i think ur rong in the energy conservation. i found it rong becos of the x term can u explain
ans should be in terms of μ m1, m2, and g
@eureka can u explain ur second equation with works im not getting it
x ki value apne aap daal lena yaar.............concept aur ans yehi hai.....[1][1]
@philip i should have given diag earlier sorry
to have caused the confusion with bars
for the second block
μmg≤kx
for the first block
F-μmg=kx1
but x is < x1
so 2μg<F
know this is rong. but this is the thing dat i got.pls correct me guys.
hmmm...
why have u used rods even two blocks wud give same result.....
if no then plz post diagram
@eureka quickly check my answer in your questions thread