The forces acts at a point1 20N inclined at 30deg towards north of east2 25N towards north3 30N inclined at 45deg towards north of west4 35N inclined at 40deg towards south of westFind the magnitude and direction of the resultant force.With its diagram
Answers
Answered by
1
Let us resolve the components of forces along the positive x-axis = East... and positive y-axis ie., North.
F_x = 20 Cos 30⁰ - 30 Cos 45⁰ - 35 Cos 40⁰ Newtons
= - 30.704 Newtons in the direction of West...
F_y = 25 + 30 Cos (90-45) - 35 Cos (90-40)
= 23.715 Newtons towards North.
Resultant is = [tex]\sqrt{F_x^2+F_y^2}= 38.796 Newtons
direction is : Ф deg. North of West
Tan Ф = | F_y / F_x | = 23.715 / 30.704 = 37.68 deg.
F_x = 20 Cos 30⁰ - 30 Cos 45⁰ - 35 Cos 40⁰ Newtons
= - 30.704 Newtons in the direction of West...
F_y = 25 + 30 Cos (90-45) - 35 Cos (90-40)
= 23.715 Newtons towards North.
Resultant is = [tex]\sqrt{F_x^2+F_y^2}= 38.796 Newtons
direction is : Ф deg. North of West
Tan Ф = | F_y / F_x | = 23.715 / 30.704 = 37.68 deg.
Similar questions