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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?


enter image description here

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.

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  • $\begingroup$ Did you post this on a different account yesterday? physics.stackexchange.com/questions/333386/… $\endgroup$ – JMac May 16 '17 at 11:29
  • $\begingroup$ not my account or question $\endgroup$ – lucky-guess May 16 '17 at 12:27
  • $\begingroup$ @rpfphysics If you are not "HocusPocus", why did you make substantial edits to another user's question, without any discussion with that user? $\endgroup$ – sammy gerbil May 16 '17 at 23:50
  • $\begingroup$ idk, i suppose i thought it wouldn't be ok to steal someone else's question. But it was up to other people to have that edit accepted, so it shouldn't be a problem $\endgroup$ – lucky-guess May 17 '17 at 9:47
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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 $mg$.

enter image description here

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.

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    $\begingroup$ @rpfphysics You could also say: to be static the torque of $mg$ on $\bar{AC}$ must be compensated by $x$ which is equal and opposite to $x'$. Now we need a force to compensate the torque on $\bar{DB}$. Due to the sine properties of torque, it is clear that the smallest force doing that is perpendicular on $\bar{DB}$, which is what is shown above $\endgroup$ – mikuszefski May 16 '17 at 14:57
  • $\begingroup$ well it conveniently turns out that this does lead to the same solution. (30+60) degrees is a right angle. I basically broke down these forces into up and down rather than at right angles to the axes of rotation. I suppose it's much easier to tell where force is being 'wasted' using your method. don't know if i understood the booklet correctly, $F$ is not parallel to $AC$ because $AC$ is 60 degrees from the horizontal whereas $F$ is 30 degrees. Their answer is correct though $\endgroup$ – lucky-guess May 16 '17 at 16:41

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