#### Answer

Part A) The reaction at support A is 1500N downward. The reaction at support B is 2000N upward.
Part B) The reaction at support A is 1800N downward. The reaction at support B is 2600N upward.

#### Work Step by Step

Part A) To find the forces at the supports we will sum the torques about support B and then sum the forces vertically to find both reactions.
$(+ \circlearrowleft) \sum\tau=0$
$1m * F_A-52kg*9.8m/s^2*3m=0$
$F_A=\frac{52kg*9.8m/s^2*3m}{1m}$
$F_A\approx1500N$
The reaction at support A is 1500N downward.
$(+ \uparrow) \sum F_y=0$
$-52kg*9.8m/s^s-F_A+F_B=0$
$F_B=52kg*9.8m/s^s+F_A$
$F_B=52kg*9.8m/s^s+1528.8N$
$F_B\approx 2000N$
The reaction at support B is 2000N upward.
Part B) The same as part A except we will take the weight of the board into account.
$(+ \circlearrowleft) \sum\tau=0$
$1m * F_A-52kg*9.8m/s^2*3m-28kg*9.8m/s^2*3m=0$
$F_A=\frac{52kg*9.8m/s^2*3m+28kg*9.8m/s^2*3m}{1m}$
$F_A\approx1800N$
The reaction at support A is 1800N downward.
$(+ \uparrow) \sum F_y=0$
$-52kg*9.8m/s^s-F_A+F_B-28kg*9.8m/s^2=0$
$F_B=52kg*9.8m/s^s+F_A+28kg*9.8m/s^2$
$F_B=52kg*9.8m/s^s+1528.8N+28kg*9.8m/s^2$
$F_B\approx 2600N$
The reaction at support B is 2600N upward.