Free Body Diagram of 3-Bar Linkage. Splitting the force?

[noodle951]

Homework Statement

Hi, please see the attached file. I am trying to find the force as illustrated in the diagram.

Homework Equations

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The Attempt at a Solution

I've found the force for the 4deg beam and therefore the horizontal beam however I am unsure how to take into account the third angled beam.

Thank you in advanced for any help. It is much appreciated.

 

[Nidum]

As drawn that mechanism is just floating in space with all the links loose relative to one another .

Any applied forces will distort the mechanism and accelerate the whole thing away from the starting location .

You need to have some constraints somewhere .

 

[noodle951]

Hi Nidum, does this explain it better? The long horizontal beam represents an actuator. The mechanism is to tip a container.

 

[Nidum]

That's better . We don't want our problems propelled into outer space .

The mechanism won't actually function with the 4 deg angle . Linkages like that typically need 25 deg + to work .

Do you want to alter the problem to be a bit more realistic ?

 

[noodle951]

That's better . We don't want our problems propelled into outer space .

The mechanism won't actually function with the 4 deg angle . Linkages like that typically need 25 deg + to work .

Do you want to alter the problem to be a bit more realistic ?

Alright, handy to know! :)

So if i change the 4 deg angle to 25, I get: 9/(sin(25)) = 21.3KN on the upper angled beam.

 

[Nidum]

Let's see a proper free body diagram for the linkage with all the forces acting shown clearly ..

 

[noodle951]

This ok?

 

[Nidum]

There are three missing forces ...?

 

[noodle951]

This better? Not sure what other forces I need.

 

[Nidum]

Left hand side of diagram ? Two horizontal forces + one vertical force ?

 

[noodle951]

The horizontal component of F1? at either end of the beam? Then the vertical at the end of the first angled beam?

 

[Nidum]

Reaction forces from container fixing pin and ground fixing pin . Four in total but you are given one so three to find .

The mechanism and force system is symmetric about c/l so not much work to do .

 

[noodle951]

Something like this?

 

[Nidum]

 

[noodle951]

View attachment 114047

Ah that makes more sense. How would I go about finding the X and Y components at the lower ground fixing pin? Would I find the force going along the lower angled beam (hypotenuse) and go from there?

Also, once I've found the X and Y components, how do I go about finding the force required "F ??".

Thanks

 

[Nidum]

For the complete mechanism what can you say about the sum of externally applied forces in the X direction and in the Y direction ?

 

[noodle951]

So the force required "F ??" Is taken from the the sum of the X components and the sum of the Y components for the container and linkage fixing pin?

 

[Nidum]

Not quite - sum in X direction and sum in Y direction are separate calculations .