the resultant force due to action of four forces is f1 which is 100N, along the negative y-axis.three of the force are 250N,320 above the x-axis,100N,60 above the x-axis, 200N, 140 above the x-axis find the fourth force

Answer 1

Resultant = F1 = 100N[270o].
F2 = 250N[320o].
F3 = 100N[60o].
F4 = 200N[140o].
F5 = ? Add to get F1.

X=250*cos320+100*cos60+200*cos140=88.3N
Y=250*sin320+100*sin60+200*sin140=54.5N
tan A = Y/X = 54.5/88.3 = 0.61680
A = 31.7o
F2+F3+F4=X/cos A = 88.3/cos 31.7 = 103.8N[31.7o] = 88.3 + 54.5i

F1 = 103.8[31.7o] + F5 = 100N[270o]
(88.3+54.5i) + F5 = 100i
F5 = 100i-54.5i-88.3 = -88.3 + 45.5i
Tan Ar = 45.5/-88.3 = -0.51529.
Ar = -27.3o
A = -27.3 + 180 = 152.7o
F5=X/cos A=-88.3/cos152.7=99.4N[152.7o]

Answer 2

Correction: F1 = -100i

F5 = -100i-54.5i-88.3 = -88.3-154.5i
Tan Ar = -154.5/-88.3 = 1.74972
Ar = 60.3o = Reference angle.
A = 60.3 + 180 = 240.3o
F5 = X/cos A = -88.3/cos240.3 = 178.2 N.
[240.3o]