Member GH is axial hence; it can only transmit vertical force to the support at H.

$\Sigma M_A = 0$

$1.2R_H = 1.2(100) + 4.5(80)$

$R_H = 400 \, \text{ kN upward}$ *answer*

$\Sigma M_H = 0$

$1.2A_V = 4.5(80)$

$A_V = 300 \, \text{ kN downward}$ *answer*

$\Sigma F_H = 0$

$A_H = 80 \, \text{ kN to the left}$ *answer*

If you wish to check the reaction forces at A and H, pass a cutting plane through members AB, AG, and AH. As shown in the figure above, the cutting plane b-b cut these members. Take the section above b-b and replace the cut members by internal force to maintain equilibrium. Solve for the internal force in members AB, AG, and AH. With these forces known, you can then check the reactions through joints A and H. The free body diagram of the section above b-b is similar to the free body diagram of section above a-a. See the solution of section a-a below for easy reference of doing the section b-b.

**From section above a-a**

$\Sigma F_H = 0$

$\frac{4}{\sqrt{41}}F_{BF} = 80$

$F_{BF} = 128.06 \, \text{ kN tension}$ *answer*

## Comments

## when do you use (5/squareroot

when do you use (5/squareroot of 41)and when do you use (4/squareroot of 41)

## The slope is 4 horizontal 5

The slope is 4 horizontal 5 vertical. For $\Sigma F_x$ use the horizontal or $\frac{4}{\sqrt{41}}$ and for $\Sigma F_y$, use the vertical or $\frac{5}{\sqrt{41}}$.