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6. The two blocks, m = 10kg and m = 50 kg are free to move as shown. The coefficient of static friction between the blocks Is 0.5 and there is no friction between M and the ground. A minimum horizontal force F is applied to hold m against M that is equal to –
plz explain it. answer is 10 N
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...
but the answer was given as 10 n i got the same answer
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