Advice on drawing the FBD (Free Body Diagram)

[LOKI1983]

Hello.
Sorry for the foolish question but I am somehow got lost and feel like cornered in solving a simple problem.
I had attached a picture: the "boomerang" look like thing is a arm which is fixed at A (pin type connection) (allows the arm to rotate around A),in point E we have a hydraulic cylinder, and in point D we have the actual load.Actually when the hydraulic cylinder is applying force to point E, point D is moved horizontally left (in point D we have a wheel which is moved on a flat surface) , while the A is moving the Up (A is attached to a load - in order to avoid further questions point A can move up and down and rotate,right and left is constrained). I am trying to figure out if my hydraulic cylinder can lift the load.My question is: when the hydraulic cylinder is applying the force the point of action where it is: at the intersection of the direction of force (B) on the line AD or at the intersection of the perpendicular on line AD ( point C)?
Thank You.

 

[oq172]

LOKI - no worries .

Please see the attached FBD for just the little link. I am assuming you know how we arrived at the ground load at B (Rby) - Yes? This was done by doing an FBD on the entire assmebly with your weight and c.g. location.

 

[rb1957]

"A can move vertically" ? ... this was discussed (at length) previously and determined that A is pinned and not in a vertical slot.

the CG is used only to determine the ground reaction, at ptB.

the link is a "three force body", all the forces pass through a single intersection point. If we know two directions (actuator laod and ground load) then we can figure out the third (reaction at ptA), and draw a force polygon. once we know one of the forces, the triangle shows us the other two. the math way is solve sumFx and sumFy (since you know the directions of the forces, the only unknowns are the magnitude of the two other forces ... you know one force, the ground contact force at ptB; and you can use components to sum in x- and y-.

now if pt A is in a slot, then your geometry won't solve. But "all" that'll happen is the link will rotate a more to allow the actuator force to pass throu the intersection of the vertical ground force and the horizontal reaction at ptA.

but, aft 80+ posts, we should be able to state with some assurance whether point A is able to translate or not ... sigh

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