Forces of F1=975 N,F2= 745 N, f3= 1175 N and are applied at the same point. The angle between f1 and f2 is 60.0 degree and the angle between f2 and f3 is 30.0 degree. F2 is between f1 and f3 find the resultant force.

Answer 1

Fr = 975N[0o] + 745N[60o] + 1175N[90o]

X=975 + 745*cos60 + 1175*cos90=1347.5 N
Y = 745*sin60 + 1175*sin90 = 1820.2 N.

tan A = Y/X = 1820.2/1347.5 = 1.31079
A = 53.5o

Fr = Y/sin A = 1820.2/sin53.5=2264.3 N.
[53.5o].