Problem 537 | Friction on Wedges

From the FBD of 40 kN block
$\Sigma F_H = 0$

$R_1 \sin 80^\circ = R_2 \sin 30^\circ$

$R_1 = \dfrac{R_2 \sin 30^\circ}{\sin 80^\circ}$

$R_1 = 0.5077R_2$
 

$\Sigma F_V = 0$

$R_2 \cos 30^\circ + R_1 \cos 80^\circ = 40$

$R_2 \cos 30^\circ + (0.5077R_2) \cos 80^\circ = 40$

$0.9542R_2 = 40$

$R_2 = 41.92 \, \text{ kN}$
 

From the FBD of lower block
$\Sigma F_V = 0$

$R_3 \cos 20^\circ = R_2 \cos 30^\circ$

$R_3 \cos 20^\circ = 41.92 \cos 30^\circ$

$R_3 = 38.634 \, \text{ kN}$
 

$\Sigma F_H = 0$

$P = R_2 \sin 30^\circ + R_3 \sin 20^\circ$

$P = 41.92 \sin 30^\circ + 38.634 \sin 20^\circ$

$P = 34.174 \, \text{ kN}$           answer