jojosg's latest activity

Note that this is only justified because section DG is in equlibrium with only 2 forces acting on it. So: a) the 2 forces must have...

I used basic trigo and drew a triangle with adjacent and opposite sides both a to get 45 degrees

Ok I think I got it. ##\sum F_x = 0 : \quad A_x - B_x = -F \tag{1}## ##\sum F_y = 0 : \quad A_y + B_y = 0 \tag{2}## ##\sum M_A = 0 ...

Been trying for a While. All I know after these is to take moment about O in the sectioned FBD but I keep getting redundant equation...

\sum M_0 = 0 -F(2a) + A_y(2a) = 0 A_y = F \sum F_x = 0 F + A_x = O_x \sum F_y = 0 O_y + A_y = 2F \sum M_A = 0 -F(2a) + 2F(2a) -...

Member BE: \sum F_x = 0 \quad \Rightarrow \quad \frac{F_G}{\sqrt{2}} = F + E_x \sum F_y = 0 \quad \Rightarrow \quad E_y =...

so how should I draw the FBD if I ignore the internal forces? I tried just the L for A and C and took Cx & Cy into account. then ACDG as...

on second maybe this 2a(By + Ey - F - Gx) + a(Gy + Cy - Ex - Cx) = 0.

Like this?

ok for Fy=0; Oy + Ay = 2F, Fx=0; Ox + F = 0, Ma=0; 2F(2a) - Oy(2a) - F(2a) = 0; which gives Oy = F & Ay = F & Ox = -F.

2a(By + Ey +F) + a (Gy + Gx + Cy + Cx + Ex) like this?

what if I take method of sections and use the left side?

I took moment about A and the distance from A for Ex & Bx are 2a.

Like this?

TL;DR: determine the internal force of member OA Hi I'm struggling with this problem. I have calculated the angle to be 45 degrees...