Why did my book consider the ratio $7:5$ when it also could've considered $5:7$? Was it just a matter of choice?

Question

Question:

Four forces $P$, $2P$, $3P$, and $4P$ act respectively along the sides of a square taken in order. Find the magnitude, direction, and line of action of their resultant.

My book's attempt:

(All of the following can be ignored. Just skip to the concluding paragraph: look for "Start reading from here" and read from there onwards)

The resultant of unlike parallel forces $3P$ and $P$ is $2P$, which acts along $EF||CD$ cutting $AD$ produced to $E$ such that,

$$P\cdot AE=3P\cdot DE$$

$$AD+DE=3DE$$

$$DE=\frac{1}{2}AD$$

$$DE=\frac{1}{2}a,\ \text{where $a=$ each side of the square}$$

Again the resultant of unlike parallel forces $4P$ and $2P$ is $2P$, which acts along $OG||DA$ cutting $BA$ produced to $G$ such that,

$$2P\cdot BG=4P\cdot AG$$

$$AB+AG=2AG$$

$$AG=AB=a$$

Let $EF$ and $OG$ meet at $O$. Then the resultant of the four forces is

$$\sqrt{(2P)^2+(2P)^2}$$

$$=2\sqrt{2}P$$

which acts along $OH$ and bisects $\angle GOF$. If $AG$ (produced) meets $OH$ at $H$,

$$AH=AG+GH$$

$$=a+OG$$

$$=a+AE$$

$$=a+a+0.5a$$

$$=\frac{5}{2}a$$

and,

$$BH=AH+AB$$

$$=\frac{5}{2}a+a$$

$$=\frac{7}{2}a$$

Now,

$$BH:AH=\frac{7}{2}a:\frac{5}{2}a$$

$$=7:5$$

(Start reading from here)

The magnitude of the resultant of the given forces is $2\sqrt{2}P$, and the line of action of the resultant is parallel to the diagonal that is drawn through the point of intersection of the lines of action of the forces $2P$ and $3P$, and the line of action of the resultant divides the side of the square along which $P$ amount of force acts externally from the opposite direction in the ratio $7:5$ (Ans.)

My comments:

The line of action of the resultant divides the side of the square along which $P$ amount of force acts externally from the opposite direction in the ratio $7:5$. This just means that $OH$ divides $BA$ externally in the ratio $7:5$. My question is, why did the book divide $BA$ by $OH$ externally? Why couldn't the book just divide $AB$ by $OH$ externally? Then the ratio would've been $5:7$. I saw another book, and they also divided $BA$ by $OH$. Why did the book have to divide externally from the opposite direction?

My question:

1. Why did $OH$ divide $BA$ externally instead of $OH$ dividing $AB$ externally? If I divided $AB$ by $OH$ would my answer be acceptable?

Answer 1

Clearly, as you say, we have the following equivalent sentences:

• The segment $OH$ divides externally the segment $BA$ in a $BH$:$HA$=$7$:$5$ ratio,
• The segment $OH$ divides externally the segment $AB$ in a $AH$:$HB$=$5$:$7$ ratio.

Some minos comments

• Ratios are for segments magnitudes, not vectors. Hence, some authors will write the external ratio $BH$:$HA$=$7$:$5$ as $BH$:$AH$=$7$:$5$, $HB$:$HA$=$7$:$5$ or even using negative signs to reflect the external ratio $BH$:$HA$=$-7$:$5$. All of them are ratios of magnitudes. Remembering that ratios are for segment and not for vectors should help. If you really wish to take vectors ratios, just go into $\mathbb{C}$omplex numbers.

Some following observations:

• The question "find the line of action of their resultant" is bad formulated, since there are four different lines of action, which will lead to one net force plus a net momentum, so depending the point of reference you choose to express the net force, the corresponding net momentum will be calculated accordingly. Hence the concept "resultant line of action" is absolutely wrong.
• A completely different issue is there is a mechanical reason to consider a third object reacting at point $O$. Hence, taking the point $O$ as reference, you indeed have a net force, and a momentum. But again, you do not have a "resultant line of action". As always, you have a net force plus a net momentum.