The order of the free-body diagrams (fbds) according to the magnitudes of their net forces are FBD 2, FBD1, FBD 4, and FBD 3.
This is evident based on the analysis described below:
FBD 1:
A = -30i + 40j;
B = 20j;
C = 20i + 10j;
D = -50j;
R = A + B + C + D;
R = -10i + 20j;
|R| = √ 10² + 20² => 22.36 N
FBD 2:
E = -40i;
F = 40i + 50j;
G = -50j;
R = E + F +G => 0;
|R| = 0 N
FBD 3:
H = -20i + 50j;
I = 60i + 10j;
R = H + I => 40i + 60j;
|R| = √ 40² + 60² = 72.11 N.
FBD 4:
J = -50i +30j;
K = 10i + 40j;
L = 50i + 10j;
M = -10i - 10j;
N = 10i - 50j;
R = J + K + L + M + N
R = 10i + 20j;
|R| = √ 10² + 20² = 22.36 N
Hence, in this case, it is concluded rust FBD 1 and FBD 4 are the same since they have a net force of 22.36N.
Therefore the lowest net force is FDB 2 and the highest net force is FDB 3.
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