Forces / Free-Body-Diagram question (not school related)...

[SteveAatE]

Greetings,

It has been a long time since I used FBDs in school.

I do remember that for the system to be in static equilibrium (in XY plane) that:

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The sum of the forces must equal zero

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Sum of forces in X must equal zero

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Sum of forces in Y must equal zero

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The sum of the moments (about a chosen point) must equal zero

I do remember how:

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1. (when taking the X & Y forces) how to calculate the reaction forces

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2. To obtain a moment from the force magnitude, line of action and perpendicular distance from the chosen point about which moments are being summed

What I am unsure about is whether/how the single applied force in the attached diagram is used both for summing the forces (in X-Y) and also for calculating the CW moment.

I do not need reaction force values. I need to determine whether the depicted work-piece, cam-followers and air cylinder will result in the applied force rotating the work-piece (in contact with the cam-followers) until it contacts the fixed stop as shown. I am assuming that the force-direction-arrows will be as shown (direction only, not magnitude).

Any guidance concerning where my understanding is incorrect is greatly appreciated!

NOTE: What may look look like small open arrowheads on the force direction arrows are small triangles indicating the X-Y direction components (not to scale) for the forces...

Thanks,
Steve

 

[3DDave]

What you need to look at is the amount of work that is done by the air cylinder. If there is a force that can act through a distance (neglecting friction) then the cylinder will extend and the part will rotate. When the part reaches the stop, no amount of force, within the strength of the parts and neglecting non-rigid deformations, will allow the cylinder to extend and therefore it will no longer rotate.

A typical FBD only looks to see that the sum of all forces is equal, not whether the item will move.

 

[SteveAatE]

Thanks 3DDave,

I wanted to be sure that the part will move until it contacts the stop.

What I am also unsure of is whether there will be some forces (of undetermined magnitude) as shown by the force directions shown in the "Final" diagram - such that there are reaction forces holding the workpiece in place. The operation that occurs, after the part is located in the "Final" position, produces (theoretically) no forces or moments in the XY plane.

Steve

 

[hydtools]

In order for the cam to move the forces and moments can not sum to zero. There must be some kinetics involved or work-energy concideration.

Ted

 

[SteveAatE]

Thanks folks. My initial post was not as clear as it could have been. I realize that the layout shown in the "Initial" image is a dynamic and not a static situation (and that the work-piece should rotate in a CW direction until it contacts the fixed stop).

I believe that the layout shown in the "Final" image should be a static condition. What I'm wanting to determine is whether the "Final" condition will have reaction forces as shown (direction only [not magnitude]).

Any guidance about the direction of the static forces being as shown in the "Final" image is surely appreciated.

Thanks,
Steve

 

[hydtools]

What keeps the plane of the cam aligned with that of the rollers and stop?

Ted

 

[SteveAatE]

Ted,

The views in the images are from above the work-piece. Gravity holds the work-piece in the fixture. A vector parallel to the center of either cam-follower is the Z-Axis (assuming the X-Y plane is parallel to the floor.

Although, in the images, it appears that the work-piece is a cam it is actually a part that will be fixtured as shown. We will be locating an insert into a c-bore in the face of the part (the c-bore centerline is parallel to the centerline of either cam-follower).

Thanks for your time looking at this.
Steve

 

[hydtools]

Do you have any concern about the sideward force on the follower?

Ted

 

[SteveAatE]

Attached is an isometric-type view of the components:

 

[SteveAatE]

Ted,

I realize that there will initially be some side-force acting on the (red) actuating follower (along the Y-Axis, in latest image). Once the (purple) work-piece rotates in-contact with the 2x (green) cam-followers - until the flat on the work-piece contacts the cyan hard-stop - the line of action of the (red) follower will align with a radius in the work-piece and the side-force (along the Y-axis) should theoretically be zero at that point (although part tolerance deviations will allow a very small amount of side-force). A guided air-cylinder will probably be used to resist cylinder damage from any side loads.

Steve

 

[BrianE22]

As shown, the problem is indeterminate. You have 4 unknowns and only 3 equations. You would need to look at relative stiffnesses to come up with an answer. I'd guess the lower right follower is significantly less stiff than the other 3 reaction points.

 

[SteveAatE]

Thanks BrianE22.