This is an edited version (I added a solution and a physics concept related query) of a closed question that I'd like to see the answer to
Three identical rods are connected by hinges to each other, the outmost ones are hinged to a ceiling at points A and B. The distance between these points is twice the length of a rod. A weight of mass m is hanged onto hinge C. ***At least how strong a force onto hinge D is neces- sary to keep the system stationary with the rod CD horizontal?
My Solution
One can deduce the minimum angle for $F$ as follows: equate moments about $D$ to obtain an expression for $T_1$ – the tension in $AC$. $T_1 = \frac{mg}{cos(30)}$. It is easy to see that the angle $AC$ to the vertical is 30 degrees. $T_2$ – the tension in $BD$ can be written in terms of $F$ similarly, $Fsin(\theta)=T_2cos(30)$. We can also setup an expression for horizontal forces in equilibrium as $T_1sin(30) = Fcos(\theta) + T_2sin(30)$. There is no use resolving vertically since that equation is implied by solving for momentum in the first two equations. Substituting what we have for $T_1$ and $T_2$ gives $F=\frac{mgtan(30)}{cos(\theta)+sin(\theta)tan(30)}$. Then we can find the obvious root of it's derivative which is $tan(\theta)=tan(30)$ where $\theta$ is acute. Plugging this into the equation for $F$ gives us the minimum as $\frac{mg}{2}$.
However the question booklet gives a hint/theorem to what seems to be a much more elegant solution (problem 3).
"if forces are applied only to two endpoints of a rod and the
fixture(s) of the rod at its endpoint(s) is (are) not rigid (the rod
rests freely on its supports or is attached to a string or a hinge),
then the tension force in the rod is directed along the rod."
I cannot understand the reasoning they have used to explain to why the direction of $F$ must always be along $AC$ for any setup of this kind. It's seems to be explained very vaguely, for instance, they seem to imply that this is always the case, rather than it being a solution to the minimum force needed. Could someone clarify please? Any help is appreciated.
Best Answer
The force triangle for the hinge at $C$ is fully determined as shown in the force diagram and so force $X$ can be found in terms of $mg$.
For hinge $D$ the magnitude and direction of the horizontal force $X$ is fixed and the angle that the force $t'$ in rod $DB$ makes with the horizontal is fixed.
An external force $f'$ has to complete the force triangle $KLM$.
However, there is an extra constraint in that you have to make the force $f'$ as small as possible and that is done by making angle $MNL$ a right angle as shown in red. So force $F$ can be found.
What the booklet answer does is to make force $F$ be along $KM$ which by inspection is going to be a larger force than that shown in red.