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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

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