**Based on maximum compressive stress:**
Normal force:

$N = P \cos 20^{\circ}$

Normal area:

$A_N = 50 (100 \sec 20^\circ)$

$A_N = 5320.89 \, \text{mm}^2$

$N = \sigma A_N$

$P \cos 20^\circ = 20 (5320.89)$

$P = 113\,247 \, \text{N}$

$P = 133.25 \, \text{kN}$

**Based on maximum shearing stress:**

Shear force:

$V = P \sin 20^\circ$

Shear area:

$A_V = A_N$

$A_V = 5320.89 \, \text{mm}^2$

$V = \tau A_V$

$P \sin 20^\circ = 5 (5320.89)$

$P = 77\,786 \, \text{N}$

$P = 77.79 \, \text{kN}$

For safe compressive force use

$P = 77.79 \, \text{ kN}$ *answer*