Analyzing Forces in Three Rods and a Block Problem

[IamVector]

Homework Statement

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 necessary to keep the system
stationary with the rod CD horizontal?

Relevant Equations

consider the system “rod CD + the
mass m” as a whole; there are four forces acting on it: m⃗g, F⃗ ,
and the tension forces of the rods, T⃗AC and T⃗BD. The tension
forces are the ones which we don’t know and don’t want to
know. so how can I cancel them out? rest of the solution is easy I think.

 

[wrobel]

Actually it is quite advanced problem. You need D'Alembert's principle.
Imagine the point D has a velocity $\boldsymbol v_D$ then the point C has a velocity $\boldsymbol v_C$.
D'Alembert says $(\boldsymbol v_D,\boldsymbol F)+(\boldsymbol v_C,m\boldsymbol g)=0$

Take into account that $\boldsymbol v_D\perp DB,\quad \boldsymbol v_C\perp AC$
and $(\boldsymbol v_D,\boldsymbol{CD})=(\boldsymbol v_C,\boldsymbol{CD})$

 

[IamVector]

Actually it is quite advanced problem. You need D'Alembert's principle.
Imagine the point D has a velocity $\boldsymbol v_D$ then the point C has a velocity $\boldsymbol v_C$.
D'Alembert says $(\boldsymbol v_D,\boldsymbol F)+(\boldsymbol v_C,m\boldsymbol g)=0$

Take into account that $\boldsymbol v_D\perp DB,\quad \boldsymbol v_C\perp AC$
and $(\boldsymbol v_D,\boldsymbol{CD})=(\boldsymbol v_C,\boldsymbol{CD})$

well I can try but scope of the problem was to teach me how to cancel out unwanted forces and torques so I think I will have to stick with torques and forces .

 

[haruspex]

how to cancel out unwanted forces and torques

To avoid involving a force in a linear force equation you consider forces normal to it, obviously. But you can't do that for both these tensions at once.
How can you avoid a force featuring in a torque equation? How can you do that for both at once?

 

[IamVector]

To avoid involving a force in a linear force equation you consider forces normal to it, obviously. But you can't do that for both these tensions at once.
How can you avoid a force featuring in a torque equation? How can you do that for both at once?

I will come on thus question later first I think I need to learn more about torque so I will just brush up my theory a bit :) thank you. And I need an advice if you don't mind is reading feynman lectures nice idea prep for olympiads?

 

[kuruman]

The direction of $\vec F$ is not explicitly specified in the statement of the problem or in the diagram. I assume that one has to find the direction for which the magnitude $F$ is least. Answering the two questions in post #4 is the key.

 

[wrobel]

By the way it is exactly D'Alembert's equation written above does not contain the "unwanted forces and torques" I believe that the Topic starter should notice that for the future

 

[haruspex]

By the way it is exactly D'Alembert's equation written above does not contain the "unwanted forces and torques" I believe that the Topic starter should notice that for the future

Yes, the two methods are more similar than first appears. They both argue that any component of F parallel to BD is irrelevant, and any component of mg parallel to AC is irrelevant. In each case, a change in the applied force only leads to a change in tension/compression in the corresponding strut.

 

[Lnewqban]

...At least how strong a force onto hinge D is necessary to keep the system stationary with the rod CD horizontal?View attachment 258829

If you apply onto D a vertical force equivalent to the weight on C, the system will be balanced and CD should be horizontal.
They are asking about the magnitude and direction of the smallest force you could apply onto C.
Creating a free body diagram, you could see that the force would be horizontally towards the right.
Having the proportions of the mechanism, you could calculate the angles among bars and the magnitude of that horizontal force.

 

[haruspex]

Creating a free body diagram, you could see that the force would be horizontally towards the right.

That's not what I get.
Besides, the OP states an objective of avoiding involving the other tensions.