In the figure shown blocks 'A' and 'B' are kept on a wedge 'C'. A, B and C each have mass m. All surfaces are smooth. Find the acceleration of C.

Consider free body diagram for mass A.

There is a pseudo force $m{a}_c$ acting on the masses A,B if we take reference frame fixed to C.

$N_a+m\times a_{c}cos{53}^0=mgcos{37}^0$

$N_a=mgcos{37}^0-m{a}_{c}cos53^0$.........(i)

Now consider free body diagram for mass B,

$N_B=mgcos{53}^0+m{a}_{c}cos{37}^0$.......(ii)

Consider the wedge C.

The net horizontal force on the wedge =

$N_{a}cos53^0-N_{b}cos37^0$

Substitute the values of

$N_a,N_b$

Thus,

$(mgcos37^0-m{a}_{c}cos53^0)cos53^0-(mgcos53^0+m{a}_{c}cos37^0)cos37^0=m{a}_c$

Therefore,

$m{a}_c(1+{cos}^{2}53^0+{cos}^{2}37^0)=mg(cos37^{0}cos53^0-cos53^{0}cos37^0)$

Thus,

$m{a}_c=0$

or

$a_c=0$

Thus the acc of wedge C is $0$.