Why does F_3a = F_2(a+b) for forces on a stick?

[Grand]

Homework Statement

I'm reading Classical mechanics by Morin and I can't understand the following reasoning. There is a horizontal stick at the ends of which 2 forces $$F_1$$ and $$F_2$$ are applied perpendicular and upward to the stick, and another force $$F_3$$ applied downward and again perpendicular to the stick. The distances from the point where $$F_3$$ is applied to the ends of the stick are $$a$$ and $$b$$. Now the author proves that $$F_3a=F_2(a+b)$$ by saying that since the force is linear, then the following must hold:

$$F_3f(a)=F_2f(a+b)$$
where f is a function to be determined. I can't fully understand this reasoning, so could anyone give me some more detail. I'll be very grateful.

Greetings,
Grand

Homework Equations

The Attempt at a Solution

 

[hikaru1221]

I luckily have Morin's book, so I can interpret his idea in the way I understand. At the point of this problem, Morin doesn't even introduce the formulas of torque and angular momentum. What I understand is this: For this particular problem, from the condition of force balance that the author introduces before, we have F3 = F1 + F2. But that's not enough and not so intuitive, since F3 must be exerted at a certain point on the stick. There must be some other condition (and most of us know that is torque balance; but let's pretend that we don't know it). And the author's aim is, after finding that condition, he will introduce the concept of torque balance.

So his argument is this: Because this condition has something to do with both magnitude of the force and the position where the force is exerted, it can be expressed in terms of the force (F3, for example) and the position (a, or even a+some length, for example). But he chooses a, as the position is relative. That is, he chooses to consider the relative positions of the forces to the point where F1 is exerted, and therefore F1 is no longer considered. There is a real case that matches this choice: suppose that at the end where F1 is exerted, the stick is fixed to a pivot. So F1 is the force due to the pivot. We know that if a force F2 is exerted at the free end, then F3 must be exerted with one certain magnitude and one certain position: F2 determines F3, regardless of F1. Therefore the equation should contain F3, accompanied by a (relative position of F3 to F1), and F2, accompanied by a+b (relative position of F2 to F1).

Intuitively when F3 is doubled, F2 must be doubled, provided that their positions maintain the same. Therefore the equation must show the linearity of the force. Thus, the most likely equation is: $$F_3f(a) = F_2f'(a+b)$$ (the two function f and f' may be different!).

However there is interesting symmetry: F2 and F3 have the same role. If we choose to consider F2 as the applied force, F3 must be the force which is here to balance out F2. On the other hand, if F3 is the applied force, F2 must be the balancing force.
Out of topic: Such symmetry actually comes from an intuitive and interesting characteristic of mechanical problems: A mechanical problem should have one and only one solution (or show one and only one resultant phenomenon) corresponding to certain initial conditions. Here, F2 and F3 (and their positions) are initial conditions, and these initial conditions lead to that the stick stays at rest. To put it another way, if the stick stays at rest, then with one F2 (one certain magnitude and one certain position), we have one and only one F3. Notice: The last sentence "To put it another way..." is consistent with the above characteristic only under certain circumstances!

So due to that F2 and F3 shares the same role, f and f' must be the same, or the rule of the position of each force in statics is the same. Later on, Morin shows that this rule is the linearity of position (f(a) = a), an intuitive result.

Just my 2 cents