Can someone help me with my Physics (asap)?

Question

I know the answers but not how to get them. Here is the question:

A 40kg plank 6.0m long is supported by sawhorses A and B. A 60kg crate is placed on the plank, 2.0m from B.
(a) Find the reaction forces at A and B
(b) If the box is moved 1.0m closer to A and samhorse B is moved in 1.5m, find the reaction forces at A and B.

Answer 1

See reference below

Sum of forces in y equal to 0

Fa+Fb - 60g - 40g=0
Fa= 100g-Fb
Moment about say point A is 0.

6 Fb - (6-2) 60g - 3(40)g=0
Fb=(4(60) + 3 (40))g/ 6= 588 N

so Fa= 100g-Fb=980 - 588=392N

Test

W box + W plank= Fa + Fb
(60+40) x 9.8=392 + 588N
980N = 980N check

for b)
Same thing
Fa+Fb= W box + W plank
Fa + Fb = (60 + 40) 9.8=980N
Fa=980-Fb
Moment about A
Ma= (6 -1.5)Fb - (3)60g - (3)40g=0
Fb=3(980)/4.5=653N
Fa= 980-653=327N

Answer 2

(a) To balance the moments with reference to point B:
6 Ra - 3 * 40 - 2 * 60 = 0
so Ra = 40 kg (considering positive an upward reaction)
The reaction in B is the difference between the weights (the plank and the crate) and the reaction on A:
Rb = 40 + 60 - 40 = 60 kg.
(b) The box is now in the middle of the plank, distant 1.5 m from B. To balance the moments with reference to point B:
4.5 Ra - 1.5 (40 + 60) = 0
so Ra = 33.33 kg.
To balance forces:
Rb = 40 + 60 - 33.33 = 66.67 kg.