## Question:

Determine the magnitudes of ${ F }_{ A }$ and ${ F }_{ B }$ so that the resultant force passes through point O of the column.

## Step-by-step

Equivalent Resultant Force: By equating the sum of the forces in Fig. a along the z
axis to the resultant force ${ F }_{ R }$ , Fig. b,

$+\uparrow$  ${ F }_{ R }$   =  $\sum { F}_{ z }$ ;            $-{ F }_{ R }$  = $-{ F }_{ A }$  $-{ F }_{ B }$ – 8 – 6

${ F }_{ R }$= ${ F }_{ A }$ + ${ F }_{ B }$ + 14

Point of Application: Since ${ F }_{ R }$ is required to pass through point O, the moment of ${ F }_{ R }$
about the x and y axes are equal to zero.Thus,

${ ({ M }_{ R }) }_{ x }$  =  $\sum { M}_{ x }$ ;         0=  $-{ F }_{ B }$(750) + 6(650) –  $-{ F }_{ A }$(600) – 8(700)

750${ F }_{ B }$ – 600${ F }_{ A }$ – 1700 = 0

${ ({ M }_{ R }) }_{ x }$  =  $\sum { M}_{ y }$ ;    0= ${ F }_{ A }$(150) + 6(100) – ${ F }_{ B }$(150) – 8(100)

159${ F }_{ A }$ – 150 ${ F }_{ B }$ + 200 = 0

Solving Eqs. (1) through (3) yields

${ F }_{ A }$ = 18.0 kN                            ${ F }_{ B }$ = 16.7 kN                ${ F }_{ R }$ = 48.7 kN