## Question:

Replace the loading by an equivalent resultant force and specify its location on the beam,measured from point $B$.

Units Used:

$kip$ = ${ 10 }^{ 3 }$ $Ib$

Given:

${ w }_{ 1 }$ = $800$ $\frac { Ib }{ ft }$

${ w }_{ 2 }$ = $500$ $\frac { Ib }{ ft }$

$a$ = $12$ $ft$

$b$ = $9$ $ft$

## Step-by-step

${ F }_{ R }$ = $\frac { 1 }{ 2 } a{ w }_{ 1 }+\frac { 1 }{ 2 } ({ w }_{ 1 }-{ w }_{ 2 })b+{ w }_{ 2 }b\quad \quad \quad\quad\quad\quad$ ${ F }_{ R }$ = $10.65$ $kip$

${ F }_{ R }x$ = $-\frac { 1 }{ 2 } a{ w }_{ 1 }\frac { a }{ 3 } +\frac { 1 }{ 2 } \left( { w }_{ 1 }-{ w }_{ 2 } \right) b\frac { b }{ 3 } +{ w }_{ 2 }b\frac { b }{ 2 }$

$x$ = $\frac { -\frac { 1 }{ 2 } a{ w }_{ 1 }\frac { a }{ 3 } +\frac { 1 }{ 2 } \left( { w }_{ 1 }-{ w }_{ 2 } \right) b\frac { b }{ 3 } +{ w }_{ 2 }b\frac { b }{ 2 } }{ { F }_{ R } } \quad \quad \quad x$ = $0.479$ $ft$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x$ = $0.479$ $ft$

( $to$ $the$ $right$ $of$ $B$ )