A structure is submitted to two distributed loads:

$f_{1\text{m}\text{a}\text{x}}=4\mathrm{.}4k\frac{N}{m}$

$f_2=2\mathrm{.}4k\frac{N}{m}$

The structure's dimensions are: $I_1=4m,I_2=2m$ and alpha $=22$ deg.

The ultimate goal is to replace the distributed loads with one resultant force on its line of

action.

Drawing $1$: Draw the distributed loads and distances on a drawing. Use $1cm=0\mathrm{.}5m$ and

$1cm=1k\frac{N}{m}.$ Respect the scales.

Replace the blue distributed load by one resultant force $F_{R1}$ at the right location from $B$ :

$d_1.$

Replace the green distributed load by one resultant force $F_{R2}$ at the right location from $B$ :

$d_2.$

Drawing $2$: Draw the resultant forces $F_{R1}$ and $F_{R2}$ on a new drawing. Use $1cm=0\mathrm{.}5m$ and

$1cm=1kN.$ Respect the scales.

Find the resultant force $F_{R_R}$ Its components on $x$ and $y,$ its magnitude, and its direction

from the horizontal.

Find the location of its line of action, the vertical$/$horizontal$/$minimal distance from B$.$

Explain your steps.

Note: The line of action is an infinite line, so you can find its location using any point, A$,$ B$,$ C$,$

and horizontal, vertical, or minimum distance, but the easiest way is from B$)$

Drawing $3$: Draw the resultant force $F_R,$ including Magnitude, Direction, and Location of

the line of action. Use $1cm=0\mathrm{.}5m$ and $1cm=1kN.$ Respect the scales. On the same drawing:

Add graphically Fr$1+$ Fr$2,$ using the parallelogram method. Compare with your Fr and its line

of action $($graphic answer and calculation answer$).$ Are your answers correct, why, why not?

In the box below, enter the magnitude of the resultant force $F_R$ in kN $,$ including one decimal.