Find the minimum value of horizontal force P applied to the lower block that will hold the system in equilibrium.

If P= 0, then in that case A will slip down and B will move to the right. For this not to happen, force P is required to maintain the equilibrium.

Applying static conditions of equilibrium to wedge A,

$∴Σ F_x (→ +ve)=0 \\ ∴ N_1 -N_2 \cos⁡ 30+0.2N_2 \cos60=0…………(I) \\ ∴Σ F_y (↑ +ve)=0 \\ 0.3 N_1+0.2 N_2 \sin60+ N_2 \sin30=500…………(II) $

Solving equations (I) and (II) simultaneously,

$∴ N_1 =424.15 N \\ ∴ N_2 =553.70 N \\ ∴ N_2= N_3 =553.70N $

Applying static conditions of equilibrium to wedge B,

$∴Σ F_y (↑ +ve)=0 \\ ∴ -0.2N_3 \sin60- N_3 \sin30-1000+N_4=0 \\ ∴ N_4=1000+0.2×553.70 \sin60+553.70 \sin30 \\ ∴ N_4=1372.754N \\ ∴Σ F_x (→ +ve)=0\\ ∴ -0.2N_3 \cos60+ N_3 \cos30-P-0.25N_4=0 \\ ∴ P= -0.2 ×553.70 \cos60+553.70 \cos30-0.25 ×1372.754 \\ P = 80.96N$