$P_x = 500(\frac{1}{\sqrt{2}}) = 353.55 \, \text{ lb}$ to the right

$P_y = 500(\frac{1}{\sqrt{2}}) = 353.55 \, \text{ lb}$ upward

$Q_x = 361(\frac{3}{\sqrt{13}}) = 300.37 \, \text{ lb}$ to the right

$Q_y = 361(\frac{2}{\sqrt{13}}) = 200.25 \, \text{ lb}$ downward

$M_O = 5Q_y = 5(200.25)$

$M_O = 1001.25 \, \text{ lb}\cdot\text{in.}$ clockwise

$R_x = P_x + Q_x = 353.55 + 300.37$

$R_x = 653.92 \, \text{ lb}$ to the right

$R_y = P_y - Q_y = 353.55 - 200.25$

$R_y = 153.3 \, \text{ lb}$ upward

x-intercept of the resultant

$aR_y = M_O$

$a(153.3) = 1001.25$

$a = 6.53 \, \text{ in.}$ to the left of point O *answer*

y-intercept of the resultant

$bR_x = M_O$

$b(653.92) = 1001.25$

$b = 1.53 \, \text{ in.}$ above point O *answer*

can you explain how Mo = 5Qy? in solution 228, why x-component not taken in the moment?

In order to produce the value of MO (moment about O) you need to transfer the Q components to the end of the Q arrow. However, if this is inconvenient to you, you can also take the moment about O without transferring Qx and Qy and it should yield similar results. Just always remember to use the sign conventions. Try it.

MO = 5(Qy-Py) + 5Px, since Py and Px have the same magnitude they cancel each other out.

Or since P passes through point O, it doesn't contribute any moment about O at all. Hence, MO = 5Qy.

Is it possible po ba na 1.53 ft below O and 6.53 ft right of O ung magiging coordinates?