Force Needed for Lid to Tip Assembly

[resjsu]

Hello,
I read few threads about tipping hazards on here and have used some of the info to attempt a solution.
I have attached a PDF with all of the info and work.

To summarize, we have a container that has a heavy lid and I want to determine if and at what opening velocity the lid will tip the container.
The lid has considerable mass so I do not want to exclude it from the final "system" so I am assuming when the lid reaches it final open position it "locks" in place and then becomes apart of the system; the system COM reflects this assumption.

The system COM will follow an arc with radius 28.53 in (see PDF). Once the system COM begins to move it will be acted on by gravity until it returns to rest. In order for system to tip, the COM has to pass the pinned corner (see PDF). I do not know how to relate the force the system experiences caused by the opening of the lid and the distance the COM travels along the arc. All of my calcs involve constant velocities but I know this is incorrect and its been a long time since I took any of these classes.

Any help is appreciated.

Thank you.

EDIT: I thought some wording was unclear so I added a few more words to where the lid rotates and where the container should tip. USE REV2 PDF FOR REF

 

[SlideRuleEra]

...determine if and at what opening velocity the lid will tip the container.

Velocity will not provide a force to tip the container, acceleration does that (Newton's Second Law of Motion).

If the force needed to accelerate the lid (open, closed, or stop the rotation) is applied from an external source, the reaction to acceleration (or deceleration) will not affect the container... container will not tip.

That may not always be the case, look at how the direction and magnitude of the force needed to cause acceleration (deceleration) is applied in each of four cases:

1) Lid accelerates to open.

2) Lid decelerates when approaching fully open.

3) Lid accelerates to close.

4) Lid decelerates when approaching fully closed.

Case 2 is probably of most interest.

http://www.SlideRuleEra.net

 

[resjsu]

SlideRuleEra,
Thank you for the response but I am not quite understanding your comment.

I may have used the wrong language but my thought process was as follows:
I cannot measure the acceleration of the lid as easy as I can the angular velocity; we achieved this through slow motion video
Once I knew the velocity of the lid I could then calculate the force at impact and the impulse.
Once I know the resultant force of the lid I would apply this to the entire system assuming the lid remained locked into its final open position and then becomes apart of the system. I could use this force for moment calculations. But, the force is not constant, it only lasts for a split second, so simply equating moments isnt really helpful either.

I am missing the need to find the acceleration of the lid? The user is the external force opening the lid. So we took video of a few scenarios where we would open the lid slowly and then worst case scenario where we threw it open as fast as possible; this is the scenario in the included example.

Thank you

PS - I like the website! I can see it being very helpful in many situations, ive run across similar issues in my career with out of print ref material.

 

[SlideRuleEra]

resjsu - You may have enough information to solve the problem:

1) For my "Case 2", the force needed to tip the container can be calculated using statics.

2) The mass of the (moving) lid is known.

3)

If this assumption is true when the fully open lid decelerates to a stop (Case 2), you can calculate the deceleration need to produce that force.

4) Knowing the deceleration (Step 3) and distance needed for that deceleration (0.010 inches), the velocity can be found.

All this assumes the needed force is not applied externally.

Thank you for the comment on the website.

http://www.SlideRuleEra.net

 

[resjsu]

1) For my "Case 2", the force needed to tip the container can be calculated using statics.

You are talking about summing moments correct? Again, this is if the force of the lid is constant but in this scenario the force is not constant, it is an impulse.

4) Knowing the deceleration (Step 3) and distance needed for that deceleration (0.010 inches), the velocity can be found.

Yes I understand this but we got this from the video/testing which should be pretty accurate.

How do I find the angular displacement of the COM of the system after a known force is applied for say 1 second? This is what I need to know. I can calculate how far around the partial arc the COM of the system must travel for it to pass the fulcrum and result in the container tipping over. I just dont know how to relate the force and the angular displacement of the COM because the COM will have gravity acting on it the entire time.

Thank you

 

[rb1957]

I'd suggest putting weight scales under the legs of the base, so you can see, with different experiments, the load approach zero on the legs that are trying to lift.

If you Really want angular motion of the CoM, paint some lines on the side of the base.

another day in paradise, or is paradise one day closer ?

 

[SlideRuleEra]

If the force needed to accelerate the lid (open, closed, or stop the rotation) is applied from an external source, the reaction to acceleration (or deceleration) will not affect the container... container will not tip.

http://www.SlideRuleEra.net

 

[racookpe1978]

You have an advantage in that the lid is limited to a 90 degree position, and the hinge is not on the edge of the lid. The lid is also very "thick", but I cannot tell if it is hollow or solid.

Assume some over-enthusiastic worker is not violently or vigorously pushing open the lid. He (or she) pushes up the lid from the hinge. At the mid-position, his/her effort is enough to move the lid CG up to a maximum elevation above the hinge line, then it will fall on its own towards the back of the box.

It is not clear what you calculated for the maximum distance the CG of the lid can "fall" from it highest position during the swing down to its final position - That distance defines the potential energy of the lid that will be converting to kinetic energy of the lid only (just before impact) that will converted to kinetic energy (tipping energy) of the combined lid and box bottom.

 

[racookpe1978]

That 25.468 degree angle is not correct.

 

[resjsu]

What does the lid impact? Without knowing this, I don't believe a meaningful answer can be obtained.

The lid impacts a stop connected to the base, I left it out of the simplified model. When the lid reaches the max opening 90 degrees, for simplicity, I am assuming it hits the hard stop and remains "locked" in that position.

To help move things along lets forget all about the lid and focus on really the main question here about what impact force is needed to get the base to tip. Its not as easy as summing moments, I think its more along the lines of kinetic energy of rotation and angular momentum but I just cant make the connection.

Thank you

 

[rb1957]

I think you've assumed the parameters necessary to determine the acceleration of the lid. This is needed to determine the force that the lid places on the body, to calculate a moment that'll cause the front supports to lift (ie the upload from the moment exceeds the static download).

I still think the easiest way is placing weight scales under the feet and running some experiments.

another day in paradise, or is paradise one day closer ?

 

[resjsu]

rb1957,
In the included PDF I have a value ( I dont know if its right) for the calculated Force and the impulse; aside from these being correct or not was the method I used not correct?

We do not have scales available but if working this by hand does not produce anything useful I will have to resort to some form of testing as the primary means of analysis. I like working things by hand first and testing as a verification.

Thank you.

 

[rb1957]

in your original attachment, I don't think your "lid velocity" (based on a distance of 0.347m) is correct. I think you need the CG distance, 13.655m.

the assumption of a time impact is Critical to the resulting answer, is it 0.1sec (quite long) or 0.001sec ??

Saying the impact energy is equal to the initial KE is reasonable … you are missing a couple of small factors (the loss of PE as the CM rotates around the axis, the rotational energy of the lid, …)

But you can work backwards from the static load on the front feet, how much force at the lid CM to equal this (moments about the rear feet) ?

another day in paradise, or is paradise one day closer ?

 

[SlideRuleEra]

resjsu - Here are the equations to solve by Force-Mass-Acceleration:

1) Sum moments to find "F[sub]R[/sub]"
2) Solve for "A"
3) Solve for "t"
4) Solve for "V"... the answer.

You will find that the answer is a low velocity... most likely because the equations are very sensitive to the value of "S"... a small error with "S" will make a significant difference in the velocity.

Since you want to pursue your own solution, here is another error in your assumptions:

Good Luck.

http://www.SlideRuleEra.net

 

[Compositepro]

The only way you can solve this problem is is by using energy equations. When the energy of the moving lid exceeds the energy of lifting the center of mass of the system over the tipping point is when the system will tip over.

Assuming the impact time or stopping distance will simply get you an assumed answer.

 

[resjsu]

The only way you can solve this problem is is by using energy equations. When the energy of the moving lid exceeds the energy of lifting the center of mass of the system over the tipping point is when the system will tip over.

I had a few in the attached PDF but several questions for each scenario, can you explain a little further?

Assuming the impact time or stopping distance will simply get you an assumed answer.

We have 2 pieces of metal colliding, I dont know how else to get a good measure of the impact time or stopping distance. Like I said we recorded on video and watch in slow motion and estimated as best we could from there.