Question

The
following forces act at a point:
(
i
) 20N inclined at 30
0
towards North of east
(
ii
) 25N towards
North
(
iii
) 30N
inclined at 45
0
towards North
of
west,
(
iv
) 35N inclined at 40
0
towards south of west.
Find the magnitude and direction of the resultant force.

Answer 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.