$\Sigma M_D = 0$

$2xR_A = 450x$

$R_A = 225 \, \text{N}$

$\Sigma M_A = 0$

$2xV_D = 450x + (450 \sin 30^\circ)(2x)$

$V_D = 450 \, \text{N}$

$\Sigma F_H = 0$

$H_D = 450 \cos 30^\circ = 389.71 \, \text{N}$

**At Joint A**

$\Sigma F_V = 0$

$F_{AB} \sin 30^\circ = 225$

$F_{AB} = 450 \, \text{N}$

$\Sigma F_H = 0$

$F_{AC} = F_{AB} \cos 30^\circ = 450 \cos 30^\circ$

$F_{AC} = 389.71 \, \text{N}$

**At Joint C**

$\Sigma F_V = 0$

$F_{BC} = 450 \, \text{N}$

$\Sigma F_H = 0$

$F_{CD} = 389.71 \, \text{N}$

**At Joint B**

$\Sigma F_H = 0$

$F_{BD} \cos 30^\circ = 450 \cos 30^\circ + 450 \cos 30^\circ$

$F_{BD} = 900 \, \text{N}$

$\Sigma F_V = 0$

$F_{BD} \sin 30^\circ + 450 \sin 30^\circ = 450 + 450 \sin 30^\circ$

$F_{BD} = 900 \, \text{N}$ *Check!*

**At Joint D**

$\Sigma F_V = 0$

$450 = 900 \sin 30^\circ$

$450 = 450$ *Check!*

$\Sigma F_H = 0$

$900 \cos 30^\circ = 389.71 + 389.71$

$779.42 = 779.42$ *Check!*

**Summary**

AB = 450 N compression

AC = 389.71 N tension

BC = 450 N tension

BD = 900 N compression

CD = 389.71 N tension

## In the solution, second line

In the solution, second line under the first sketch (of reaction forces at supports):

the sum of the moments about A = 2xVD=450x+(450sin30∘)(2x)

How is the last term (2x) determined?

## The 450 N at B was resolved

In reply to In the solution, second line by daveB

The 450 N at

Bwas resolved into components atD. The vertical downward component is at distance 2xfromA.## I tried solving this part but

In reply to In the solution, second line by daveB

I tried solving this part but this time, considering all forces with respect to point A, here's what I got:

VD(2x) = 450(x)+(450sin30)(x)+(450cos30)(xtan30)

VD=450

I think what they've shown is a simplified equation already.