Question

A two member truss ABC is shown in the figure The force (in kN) transmitted in member AB is

Answer 1

Method I:
AB = 1 m
AC = 0.5 m
$BC=\sqrt{1^2+{0.5}^2}$
$=\sqrt{1.25}=1.118m$
$A_x+C_x=0$
$A_y+C_y=10$
(from force equilibrium)
$\sum M_A=0$
$C_x\times 0.5=10\times 1$
or $C_x=20kN$
and $A_x=-20kN$

$\sum M_c=0$
$\Rightarrow F_{AB}\times 0.5=20\times 0.5$
$\therefore F_{AB}=20kN$
Method II:

Free body diagram of point B,

Horizontal Reaction:
$F_{BC}\cos \theta =-F_{AB}.....\left(i\right)$
Vertical Reaction :
$F_{BC}\sin \theta =-10....\left(ii\right)$
Putting the value of $F_{BC}$ from equation (ii) to equation (i)
$\frac{-10}{\sin \theta }\times \cos \theta =-F_{AB}$
$\therefore F_{AB}=10\times \cot \theta =10\times \frac{1}{0.5}=20kN$