Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.4 - Equations of Motion: Rectangular Coordinates - Problems - Page 137: 46

$P=2mg\space tan\theta$

We can determine the largest horizontal force $P$ as follows: For Block $A$ $\Sigma F_y=0$ $\implies Ncos\theta-mg=0$ $\implies Ncos\theta=mg~~~$eq(1) and $\Sigma F_x=ma$ $\implies Nsin\theta=ma~~~$eq(2) Dividing eq(2) by eq(1), we obtain: $a=gtan\theta$ Now for block $B$ $\Sigma F_x=ma$ $\implies P-Nsin\theta=ma$ $\implies P-mgtan\theta=mgtan\theta$ $\implies P=2mgtan\theta$