The following vectors define the position of each of the important nodes.
The location of each node will have two variables defined by a position vector. For the articulated rods, you know the distance separating them and for the sliders, you know they move within the dotted line.
- ##A## is located at the origin so ##\overrightarrow{AO}=0##.
- ##\overrightarrow{{BA}}## is the vector starting from ##{A}## going to ##{B}## and defines the position of point ##B##. That vector is equivalent to ##\overrightarrow{BO}## because ##A## is the origin.
I know it might seem counterintuitive to say ##\overrightarrow{BA}## instead of ##\overrightarrow{AB}## but it's done like that so that vector subtraction cancels the letters and vectors point the right way. Since addition is commutative, it makes sense to use subtraction to define the order of the letters. In the end, you can choose to call things any way you want as long as they are coherent.
I'll recommend you use my notation so we can communicate more easily and because of the reason I previously showed.
- ##\overrightarrow{BA}## + ##\overrightarrow{CB}## is the vector defining the position of ##C##. That's equivalent to the vector ##\overrightarrow{CO}##.
- And so on.
To calculate the CM of any polygon (line segments or triangles) add the position vectors defining their nodes and calculate the average.
To calculate the mechanism, you see 3 loops. In each loop, the addition of the vectors is 0. That will give you the constraints. Since you have 3 loops and each of them is a 2D vector equation, you'll have 6 equations.
- A→B→C→D→A is 0 because you come back to the same point.
- A→B→C→P→F→A is 0 for the same reason.
- C→P→D is 0 for the same reason.
Create as many angles for the vectors as necessary. In the triangle case, notice that the vector's angles will define its shape so they'll be related to the angle ##\beta_y## given in the exercise. If your mechanism only has one degree of freedom, by considering ##\theta_{12}## (the angular position of the red bar) known, the rest of the mechanism will be defined.
NOTE: Mechanisms sometimes have multiple solutions.
Picture's source
In a "normal" exercise where the size of each element is defined, you'd have 6 variables and 6 equations to find the position of the mechanism.
However, in this case, besides those variables, you'll have variables defining the size of each of the elements so you'll have infinite solutions because your 6 equations are not enough to constrain the system.
You're then optimizing for the solution that reduces the maximum reaction ##R## at point ##A## for the duration of a cycle.