Problem 533 | Friction

$N_B(2x) = 424x + f_B(2x)$

$2N_B = 424 + 2f_B$

$2N_B = 424 + 2(0.268N_B)$
 

$1.464N_B = 424$

$N_B = 289.62 \, \text{ N}$
 

$f_B = 0.268(289.62)$

$f_B = 77.62 \, \text{ N}$
 

$\Sigma F_V = 0$

$N_A = f_B + 200 + 424$

$N_A = 77.62 + 200 + 424$

$N_A = 701.62 \, \text{ N}$
 

$f_A = 0.20(701.62)$

$f_A = 140.32 \, \text{ N}$
 

$\Sigma F_H = 0$

$P = f_A + N_B$

$P = 140.32 + 289.62$

$P = 429.94 \, \text{ N}$           answer