**Problem 420**

Determine the force in members DF, DG, and EG of the Howe truss shown in Fig. P-420.

**Solution 420**

$\Sigma M_G = 0$

$3F_{DF} = 4(210)$

$F_{DF} = 280 ~ \text{kN}$ compression *answer*

$\Sigma F_V = 0$

$\frac{3}{5}F_{DG} + 120 = 210$

$F_{DG} = 150 ~ \text{kN}$ compression *answer*

$\Sigma M_D = 0$

$3F_{EG} + 4(120) = 8(210)$

$F_{EG} = 400 ~ \text{kN}$ tension *answer*