Let F
4 = the fourth force and for couple resultant, R is zero.
$R_x = 0$
$110 + 150(\frac{3}{5}) + F_{4x} = 0$
$F_{4x} = -200 \, \text{ lb}$
$F_{4x} = 200 \, \text{ lb to the left}$
$R_y = 0$
$150(\frac{4}{5}) - 120 + F_{4y} = 0$
$F_{4y} = 0$
Thus, $F_4 = 200 \, \text{ lb to the left}$
Assuming F4 is above point O and clockwise rotation to be positive
$M_O = C$
$110(4) + 120(2) - F_4d = 480$
$110(4) + 120(2) - 200d = 480$
$d = 1 \, \text{ ft}$
d is positive, thus, the assumption is correct that F4 is above point O.
Therefore, the fourth force is 200 lb acting horizontally to the left at 1 ft above point O. answer