A block of mass *m *sits atop a mass *M *which rests on a frictionless table. The coeffecient of friction between the two masses is $\displaystyle \mu$

The mass *M *is connected to a spring of force constant *k *attached to the wall.

(i) How far can mass *M *be pulled so that upon release, the upper mass *m *does not slip off?

(ii) Repeat if $\displaystyle \mu'$s the coeffecient of friction between *M *and the table.

im getting bit confused about the signs.

for big+small block we have upon release

$\displaystyle F=(M+m)a=-kx $

giving $\displaystyle a= \frac{-kx}{(M+m)}$

for small block

F=$\displaystyle \mu R$= $\displaystyle mg\mu$

now as big block goes back upon release then does small block move opposite way so friction acts in same direction as acc?

so

ma= F= $\displaystyle mg\mu$?

then

$\displaystyle \frac{-kx}{(M+m)}= $$\displaystyle mg\mu$?

so as we want max x we need

$\displaystyle \frac{-kx}{(M+m)}< $$\displaystyle mg\mu$?

but answers negate the Lhs. Why is this?

for part (ii) for the big block with friction the answers have

$\displaystyle (M+m)a=-xk+\mu'(M+m)g $

but once again should the friction not be in opposite direction to acc of the block?