Tried...
Code:
Mass Number Mass (g); Angular Position
m1 200 ----------------------> 12
m2 200 ---------------------> 134
m3 200 ----------------------> 252
mass = 200 gm
g = 9.8 m/s*s
Fi = 200*9.8 = 1960.0 dynes = 1.96 Newtons
You have to use trigonometric calculations
Any force F is anglar at T units ... it has
X Component F1*Cos(T)
Y Component F1*Sin(T)
So accordingly
Code:
Force (N) Fx(n) Fy(N)
F1 1.96*Cos(12) 1.96*Sin(12)
f2 1.96*Cos(134) 1.96*Sin(134)
f3 1.96*Cos(252) 1.96*Sin(252)
---------------------------------------------------------------------
Resultant F(N) -0.05 -0.047
|F| = SQRT((-0.05)^2 + (-0.047)^2) = SQRT(0.0047) = 0.069 N
T = tan^-1 (Fy/Fx) = tan^-1 (-0.047/-0.05) = 43.23 degrees
So resultant force of 0.069 N is applied at 43.23 degrees
calculations can be erroneous