Determine the resultant of the following force system as shown in figure. Also locate its position w.r.t. O. Radius of circle is 2 m.[Figure: circle of radius 2 m, centre O, with horizontal and vertical centre lines. Four horizontal forces act at poi…

Coordinates of points (O origin, x right, y up, r = 2 m):

150 N at \((-2\cos60^\circ, 2\sin60^\circ) = (-1, 1.732)\); 90 N at \((1.732, 1)\); 200 N at \((1.732, -1)\); 100 N at \((-1.414, -1.414)\).

Resultant (all forces horizontal, so \(\sum F_y = 0\)):

\(\sum F_x = -150 - 90 + 200 - 100 = -140\ \text{N}\) \(\Rightarrow\) R = 140 N acting horizontally to the left (←).

Moments about O (anticlockwise +; a horizontal force F at height y gives moment \(-yF_x\)):

150 N: \(150 \times 1.732 = +259.81\) N·m; 90 N: \(90 \times 1 = +90\) N·m; 200 N: \(200 \times 1 = +200\) N·m; 100 N: \(-100 \times 1.414 = -141.42\) N·m

\(\sum M_O = 408.39\) N·m (anticlockwise)

Position (Varignon): \(R \times d = \sum M_O \Rightarrow d = \dfrac{408.39}{140} = 2.917\ \text{m}\). A leftward force gives an anticlockwise moment about O when it lies above O.

Answer: R = 140 N, horizontal, acting to the left, along a line 2.917 m above O (it cuts the Y-axis at (0, 2.917) m, outside the circle).