$\Sigma M_A = 0$

$8R_H = 2(55) + 4(90) + 6(45)$

$R_H = 92.5 \, \text{ kN}$

From section to the right of a-a

$\dfrac{x + 2}{1.5} = \dfrac{x + 4}{2.5}$

$2.5x + 5 = 1.5x + 6$

$x = 1 \, \text{ m}$

$\Sigma M_O = 0$

$(x + 2)\left( \frac{5}{\sqrt{41}}F_{DG} \right) + xR_H = (x + 2)(45)$

$(1 + 2)\left( \frac{5}{\sqrt{41}}F_{DG} \right) + 1(92.5) = (1 + 2)(45)$

$\frac{15}{\sqrt{41}}F_{DG} + 92.5 = 135$

$\frac{15}{\sqrt{41}}F_{DG} = 42.5$

$F_{DG} = 18.14 \, \text{ kN tension}$ *answer*

## Comments

## where did that (√41) came

where did that (√41) came from ?

## Pythagorean Theorem: $\sqrt{4

Pythagorean Theorem: $\sqrt{4^2 + 5^2}$

## bakit po 4 and 5 ginamit at

bakit po 4 and 5 ginamit at hindi 2 and 2.5?

## nagmultiply lng ng 2 sa 2 at

nagmultiply lng ng 2 sa 2 at 2.5 para mag whole number yung 2.5 pero same pa din makukuhang sagot pag 2 at 2.5

ginamit

## Is there a reason why we

Is there a reason why we needed to find

F_{DG}with moments? Why can't we use summation of y-forces instead? They seem to give different answers.## What's the principle behind

What's the principle behind plotting an imaginary right triangle to solve for the forces at member DG? Can't we just take the moment at H

Advance thank you to whoever replies.❤️❤️