Sum of horizontal and vertical components:

The two forces shown in the figure, are to be replaced by an equivalent force R applied at P. Locate point P by finding its distance x from AB and specify magnitude of R and the angle O it makes with horizontal. 

Sol.: Let us assume equivalent force R (Resultant force) is acting at an angle of θ with the horizontal. And  ∑H and ∑V be the sum of horizontal and vertical components.

∑H = -1500 cos 30º = -1299N                                   ...(i)

∑V = 1000 -1500 sin 30º = 250 N                            ...(ii)

R = { ∑H²  + ∑V²  }^1/2

For the direction of resultant

R = 1322.87 N                                                                        .......ANS

tanθ  =  ΣV/ ΣH

= 250/-1299

= -10.89º                                                                              .......ANS

Now for finding position of resultant, we make use of Varignon's theorem, that is R × d =∑M , Take moment about point 'O'

1322.878 × x = 1500 cos 30º × 180 + 1500 sin 30º × 50 -1000 × 200

By solving x = 53.92 mm                                                                                                       .......ANS