Problem 268 | Resultant of Non-Concurrent Force System

Let F₄ = 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