Reaction Point

Beam AB, which is simply supported at Points A & B, is subjected to a distributed load as shown. If the beam’s weight is negligible, what is the reaction force at Point A?

Hint
Draw the free body diagram:
Hint 2
Take the moment about Point B to reduce the amount of unknown forces.
$$$\sum M_B=0=Force \times Distance$$$
First, draw the free body diagram:
The triangular force distribution can be replaced with a concentrated force $$F$$ , which is located through the triangle’s centroid. The force’s magnitude is equal to the triangle’s area:
$$$F=\frac{1}{2}bh=\frac{1}{2}(12m)(200N/m)=1,200\:N$$$
where $$b$$ is the triangle’s base, and $$h$$ is the triangle’s height.
We can take the moment about Point B to reduce our unknown variables down to 1 ( $$A_x$$ , $$B_x$$ and $$R_B$$ will zero out since their radius vector goes through Point B). Remember, $$Moment=Force \times Distance$$ :
$$$\sum M_B=0=(18m)R_A-(8m)F$$$
$$$(18m)R_A=(8m)(1,200N)$$$
$$$R_A=\frac{(8m)(1,200N)}{18m}=\frac{9,600N}{18}=533.3\:N$$$
533.3 N