**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*

## Good day po, ask ko lang sana

Good day po, ask ko lang sana kung bakit involved na agad yung sec(20) sa Normal Area at hindi 100 times 50 muna. If I am not mistaken, yung Normal Area is 5000mm^2 lang naman kasi yun yung area perpendicular to the force P.