# 227 - Moment of resultant force about a point

**Problem 227**

Two forces P and Q pass through a point A which is 4 m to the right of and 3 m above a moment center O. Force P is 890 N directed up to the right at 30° with the horizontal and force Q is 445 N directed up to the left at 60° with the horizontal. Determine the moment of the resultant of these two forces with respect to O.

**Solution 227**

$R_x = 890 \cos 30^\circ - 445 \cos 60^\circ$

$R_x = 548.26 \, \text{ N}$ (to the right)

$R_y = P_y + Q_y$

$R_y = 890 \sin 30^\circ + 445 \sin 60^\circ$

$R_y = 830.38 \, \text{ N}$ (upward)

$M_O = 4R_y - 3R_x$

$M_O = 4(830.38) - 3(548.26)$

$M_O = 1676.74 \, \text{ N}\cdot\text{m}$ (counterclockwise) *answer*

The moment of resultant about O can be solved actually without the use of R_{x} and R_{y}. The moment effect of the components of R is the same as the combined moment effect of the components P and Q. Thus, $M_O = 4P_y + 4Q_y + 3Q_x - 3P_x$. Try it.

You can also find M_{o} by finding the magnitude of R and its moment arm about point O. Moment arm is the perpendicular distance between the line of action of R and point O.