Problem 538 | Friction on Wedges

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

$R_1 \cos 20^circ = R_2 \sin 10^\circ$

$R_1 = \dfrac{R_2 \sin 10^\circ}{\cos 20^\circ}$

$R_1 = 0.1848R_2$
 

$\Sigma F_V = 0$

$R_1 \sin 20^\circ + R_2 \cos 10^\circ = 40$

$(0.1848R_2) \sin 20^\circ + R_2 \cos 10^\circ = 40$

$1.048R_2 = 40$

$R_2 = 38.168 \, \text{ kN}$
 

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

$R_3 \cos 20^\circ = R_2 \cos 10^\circ$

$R_3 \cos 20^\circ = 38.168 \cos 10^\circ$

$R_3 = 40 \, \text{ kN}$
 

$\Sigma F_H = 0$

$P = R_2 \sin 10^\circ + R_3 \sin 20^\circ$

$P = 38.168 \sin 10^\circ + 40 \sin 20^\circ$

$P = 20.308 \, \text{ kN}$           answer