Problem 542 | Friction on Wedges

$2(R_1 \cos 25^\circ) = 1000$

$R_1 = 551.69 \, \text{ lb}$
 

From the FBD of any of the wedges
$\Sigma F_V = 0$

$R_2 \cos 10^\circ = R_1 \cos 25^\circ$

$R_2 \cos 10^\circ = 551.69 \cos 25^\circ$

$R_2 = 507.71 \, \text{ lb}$
 

$\Sigma F_H = 0$

$P = R_1 \sin 25^\circ + R_2 \sin 10^\circ$

$P = 551.69 \sin 25^\circ + 507.71 \sin 10^\circ$

$P = 321.32 \, \text{ lb}$           answer