Part (a): P to start the wedge under block A
From the FBD of block A
$\Sigma F_H = 0$
$R_1 \cos 15^\circ = R_2 \sin 35^\circ$
$R_1 = 0.5938R_2$
$\Sigma F_V = 0$
$R_2 \cos 35^\circ = R_1 \sin 15^\circ + 100$
$R_2 \cos 35^\circ = (0.5938R_2) \sin 15^\circ + 100$
$0.6655R_2 = 100$
$R_2 = 150.27 \, \text{ kN}$
From FBD of block B
$\Sigma F_V = 0$
$R_3 \cos 15^\circ = R_2 \cos 35^\circ + 40$
$R_3 \cos 15^\circ = 150.27 \cos 35^\circ + 40$
$R_3 = 168.85 \, \text{ kN}$
$\Sigma F_H = 0$
$P = R_2 \sin 35^\circ + R_3 \sin 15^\circ$
$P = 150.27 \sin 35^\circ + 168.85 \sin 15^\circ$
$P = 129.89 \, \text{ kN}$ answer
Part (b): P to pull the wedge out from under the block
From FBD of block A
$\Sigma F_H = 0$
$R_1 \cos 15^\circ = R_2 \sin 5^\circ$
$R_1 = 0.0902R_2$
$\Sigma F_V = 0$
$R_2 \cos 5^\circ + R_1 \sin 15^\circ = 100$
$R_2 \cos 5^\circ + (0.0902R_2) \sin 15^\circ = 100$
$1.0195R_2 = 100$
$R_2 = 98.08 \, \text{ kN}$
From FBD of block B
$\Sigma F_V = 0$
$R_3 \cos 15^\circ = R_2 \cos 5^\circ + 40$
$R_3 \cos 15^\circ = 98.08 \cos 5^\circ + 40$
$R_3 = 142.57 \, \text{ kN}$
$\Sigma F_H = 0$
$P + R_2 \sin 5^\circ = R_3 \sin 15^\circ$
$P + 98.08 \sin 5^\circ = 142.57 \sin 15^\circ$
$P = 28.35 \, \text{ kN}$ answer
