vikas askiitian expert
Last Activity: 14 Years ago
F = (M+M)a
a=F/m+M ............1
bw m and M there is normal reaction let it be N,
then at m
F-N=ma
N=F-ma or
=F(1 - m/(m+M) )
N=F( M/m+M)................2
if both blocks are at rest wrt each other then it can be said that friction is sufficient to balance the weight of block of mass m..
(friction force) f =mg
f =u(N) (u is cofficient of friction and N is normal reaction)
uN=mg
u(F)(M/m+M) =mg
F=mg(m+M)/uM
on solving
F=240N
this is the minimum force required for which block remmains in equilibrium...
if force if more then friction is more and then also block remains in equilibrium,so 240N is the value is minimum possible value of force...