A block of mass `m_(1) = 20 kg` as placed on a wedge of mass `m_(2)= 30 kg` rest on a smooth flooras shpwn in Fig (a) . Friction exists between block and wedge.
a. Find the acceleration of each block if (i) `F = 180 N` and (ii) `F = 400 N`
b. Solve the problem if force `F = 180 N` is direction upwards as shown in Fig (b)

Explanation

a. Assume that the system move together and friction between `m_(1) and m_(2)` is static nature
Common acceleration
` a = (F)/((m_(1) + m_(2))) = (F)/((20 + 30))= (F)/(50)ms^(-2)`
For block ` F - f = 20a`
For wedge `f + F - F = 30a`
`or f = 30a`
From (i) and (ii) ` f = 30((F)/(50)) rArr f = (3)/(5)F`
If friction is of static nature `f le mu_(s) N`
` (2)/(3) F le 0.6 xx 20 xx 10 rArr F le 200 N`
`or F_(max) = 200 N`
For case (i) , `F = 180 N lt F_(max)`
Hence friction will be of static nature and will move with common acceleration
`a_(1) = a_(2) = (180)/((20 + 30)) = (18)/(5) ms^(-2)`
For case (ii) ` F = 400 N gt F_(max)`
The friction will be of kinetic nature both the block and the wedge will move with different acceleration .The friction between the block and wedges

`f_(k)= mu N = 0.4 xx 20 xx 10 = 80 N`
Acceleration of block, `a_(1) = (400 - 80)/(20)= 16 ms^(-2)`
Acceleration of wedge, `a_(2) = (80)/(30) (8)/(3)= 16 ms^(-2)`
b. First we will determine the force for which block and wedge move in combination .From Newton's acceond law the equation are
Block `F - f = m_(1) a`
`f - F = m_(2)a`
On solving the above equation simulaneously , we obtain
`a = 0 F = f`
For no sliding `f le mu_(s)N`
` f le 0.6 xx 20 xx 10`
or `f_(max) = 120 N`
Here `F` is greater than `120 N` therefore occure New equation are
Block : `- mu_(s) m_(1) g = m_(1)a_(1)`
`a_(1) = 5 ms^(-2)`
Wedge: `mu_(k) m_(1) g - F = m_(2) a_(2)`
`a_(2) = - 3.33 ms^(-2)`