$\Sigma M_A = 0$
$8R_H = 2(55) + 4(90) + 6(45)$
$R_H = 92.5 \, \text{ kN}$
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
we did pythagores theorem to
we did pythagores theorem to the triangle the multiplied the sides by 2
Is there a reason why we
Is there a reason why we needed to find FDG 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.❤️❤️
value of a force is not equal
value of a force is not equal to the length so we put a relation between them since they are dependent to each other : F1/F2 = L1/L2
F: force L: length
we found root 41 by applying pythagores theorem and we multiplied everything by 2 since the ratio stays the seem