Question

if 30kn inclined as 35 degrees towards north of east and 25kn towards north and 30kn inclined at 30 degrees towards north of west and 35kn inclined at 25 degrees towards south of west find the magnitude and direction of resultant force​

Answer 1

Answer:

With East as reference, the orientation of the given force-vectors are

# 30kN inclined at 35degrees

# 22kN inclined at 90d

# 30kN inclined at 150d

# 35kN inclined at 205d

The resultant sum of horizontal components is given by

30*cos(35d)+22*cos(90d)+30*cos(150d)+35*cos(205d)

=30(0.8192)+22(0)+30*[-sqrt(3)/2]+35*(-0.90631)=24.575-25.981-31.72085=-23.12685

The resultant sum of vertical components is given by

30*sin(35d)+22*sin(90d)+30*sin(150d)+35*sin(205d)

=30(-0.2589)+22+30*[0.5]+35*(-0.4226)=-7.2575+22+15–14.792=4.9515

Resultant vector=sqrt[(-23.12685)^2+(4.9515)^2]=7.003kN

Amplitude=arctan(4.9515/(-23.12685))=167.915degrees