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 = { ∑H^{2} + ∑V^{2} }^{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