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 F_{4} 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 F_{4} is above point O.

Therefore, the fourth force is 200 lb acting horizontally to the left at 1 ft above point O. *answer*

i have a correction sir,,, my answer is, the fourth force is 200 lb, 1 ft above O.

The four forces in the force system produces a couple of 480 lb-ft clockwise, meaning the resulatant of the four forces is 0 but it has a resultant moment of 480 lb-ft which is a couple.

Considering the figure P-268 taking moments at O clockwise positive, and the equation should be written in this way,

-200x+120(2)+110(4)=480. (assuming the missing fourth force which is 200lb left, is above O.)

solving for x.

x= 1ft positive, meaning the assumption is correct, therefore 200 lb force is 1 ft above O.

The solution is incorrect. Assuming the missing fourth force which is 200 lb left is above O, taking moments at O and taking clockwise moments as positive, the equation should be written in this way

M = -200d + 120(2) + 110(4) = 480

solving for d,,, we get d= 1ft above O. The answer is positive because the assumptions are correct

Yes you are correct. The solution above will yield a 480 ft·lb couple in "counterclockwise" direction. This page is noted for revision. Thank you.

This page has been corrected. Thank you very much kainur for calling our attention.