Calculate Normal Force

[Calvin Murphy]

I am hoping to settle a debate between two engineers.

I have created a simple diagram of the part of the structure we are investigating.

It can be solved in 5 minutes or less by anyone familiar with static equilibrium problems.

The goal is simple enough: find Fn.

Yet, the discussion over whose solution was correct is still unresolved. I build a rig to physically measure it, but the disagreement persisted. Then I reached out to 5 university professors to see if they would help, but I have not heard a response. So here I am. Please help.

 

[human909]

None additional requirements or caveats are necessary calculate the static equilibrium answer. The fact that it is stated that the system is in static equilibrium is enough. Of course we should all recognise that this static state is inherently unstable and hence this is a academic exercise not a real world exercise.

As I said earlier if you need extra confirmation then your favourite frame analysis software should give you an answer. Here is my result for what should be a simple statics problem. I've provided two diagram simply to show that geometric length is not relevant.

 

[GregLocock]

5/tan(30) obviously for the horizontal component. I think it is stable, it is very similar to the jammed kitchen drawer problem, and can be 'fixed' the same way, pulling up at the centre. The constraint at the LH wall is hard to achieve in practice, the reality is the upper wheel will lift off and then there will be an expensive noise.

 

[human909]

5/tan(30) obviously for the horizontal component. I think it is stable, it is very similar to the jammed kitchen drawer problem, and can be 'fixed' the same way, pulling up at the centre. The constraint at the LH wall is hard to achieve in practice, the reality is the upper wheel will lift off and then there will be an expensive noise.

No. It is not a stable structure by definition.

Even the slightest variation away from "perfect" a horizontal of the first beam will remove static equilibrium you end up with a mechanism. A perfectly vertical needle balancing on its tip is in static equilibrium, but it is unstable because even the smallest deviation will lead to a situation where static equilibrium is not possible.

Sure, if you have a real world scenario with roller static friction then you can achieve some stability. But that isn't what we have here we have zero vertical restraint on the left end of the beam.

Again if you want to check these claims try to calculate a static equilibrium for the case where the angle of the horizontal member is 0.1degree off horizontal.

 

[GregLocock]

Agreed.

 

[jec67]

If one takes a free body diagram of the horizontal member and applies a vertical load it will reveal itself unstable. Summing moments about the right end (hinge point) must equal zero so to satisfy the condition of the hinge - i.e. zero moment. For example, say there is a 2 kip vertical applied load at the midpoint between the roller and point A. Additionally, let's call the length of the horizontal member L. Then summing moments about point A yields M = (2kip)*(0.5 L) which will not equal zero for any real load. Hence unstable.

This is all assuming we are looking at rigid body mechanics.

 

[jec67]

Another clarification I should add: based on the system this represents, the skate roller is in essence constrained to a track so that Fx is always perpendicular to the wall. It can react moments and the system should be stable.

I tried to do this in a short-hand way using a perpendicularity symbol on the member that attaches to the skate. I apologize for the confusion.

So it can react moments through a couple of horizontal forces - one at each wheel? If so, this is quite a different diagram than the one you showed in your OP.

 

[Tomfh]

I think it is stable

It's unstable equilibrium.

 

[GregLocock]

If the horizontal strut angles down clockwise the lower wheel will push harder, therefore providing a restoring moment on the strut.

 

[Tomfh]

If the horizontal strut angles down clockwise the lower wheel will push harder, therefore providing a restoring moment on the strut.

Edit: If the two wheels are two separated X restraints, then you're right. One of them fights back. I was assuming they were a single X restraint....

 

[rb1957]

This is a problem, not the real world. If the three forces intercept (exactly) at pt A, then point A (pinned though it may be) is in equilibrium (the three forces are bearing on different parts of the pin [1]). no?

so, 10kN


[1] the load is pushing down on the pin, strut B is reacting this vertical component but pushing up and to the left, strut A is pushing to the right

 

[GregLocock]

If the constraint at the left resists moments and all forces perpendicular to the wall then the mechanism is bistable, it will snap through to a position where the right hand strut is at -30 degrees. This is analogous to elastic buckling, and like that real world factors will have a great impact on the maximum force that can be applied before it snaps through.

 

[jec67]

If the horizontal strut angles down clockwise the lower wheel will push harder, therefore providing a restoring moment on the strut.

Agree - the point I am making: the diagram in the original post is misleading as there is no indication the rotational restraint exists.

 

[rb1957]

but ... all the joints are pins ... A and B are defined as pins, the LH support is drawn as a "roller" which would normally be considered as a normal force only (no moment capability).

we can say that practically this is a bad/unstable design, but that isn't answering the question. As a textbook, 2nd year type problem, we can reasonably assume (and state) that the three forces are coincident (intercept at one point) and that deflections are "small". Sure, part of the "trick" might be to consider the strain deformation of the elements, but we aren't told anything about them so we'd neglect them and the answer is 10 kN.

I don't understand why people say it is not solvable because of the pin at A ... you can certainly assemble this structure. as you apply load to point A I can see the mathematical perfection of joint A becoming less perfect and I'd expect the LH support would collapse when the load becomes non-normal to the LH face.

 

[Eng16080]

It's difficult to model the roller support, but I think this is reasonably accurate:

View attachment unstable structure.MOV

 

[human909]

I don't understand why people say it is not solvable because of the pin at A ... you can certainly assemble this structure. as you apply load to point A I can see the mathematical perfection of joint A becoming less perfect and I'd expect the LH support would collapse when the load becomes non-normal to the LH face.

Most people here are not saying this is unsolvable. They are saying it is unstable. There is a huge difference.

 

[Celt83]

Just like the balancing bird toy example I imagine some of us have seen in statics class. Although the system itself is inherently unstable, under the particular loading presented it is in static equilibrium.

 

[HTURKAK]

Unbelievable.. With this one , 37 responds . This reminds me a Russian proverb ' A man may throw a stone into a well and a hundred wise men cannot pull out."

 

[Tomfh]

They are saying it is unstable.

I originally thought the roller was a single roller as you showed in your model, in which case yes it’s unstable.

However if it’s a pair of rollers, then as Greg is saying it’s stable due to the restoring moment that develops.

 

[Greenalleycat]

Jesus, people, talk about overcomplicating a basic situation
This is simply solvable by hand or computer in a variety of ways and the diagram given is obviously just a basic representation

Why are we arguing about whether the roller has one wheel or two???

 

[Just Some Nerd]

If the constraint at the left resists moments and all forces perpendicular to the wall then the mechanism is bistable, it will snap through to a position where the right hand strut is at -30 degrees. This is analogous to elastic buckling, and like that real world factors will have a great impact on the maximum force that can be applied before it snaps through.

Agree - the point I am making: the diagram in the original post is misleading as there is no indication the rotational restraint exists.

I originally thought the roller was a single roller as you showed in your model, in which case yes it’s unstable.

However if it’s a pair of rollers, then as Greg is saying it’s stable due to the restoring moment that develops.

If the horizontal strut angles down clockwise the lower wheel will push harder, therefore providing a restoring moment on the strut.

Edit: If the two wheels are two separated X restraints, then you're right. One of them fights back. I was assuming they were a single X restraint....

Why are we arguing about whether the roller has one wheel or two???

The original diagram shows a single reaction at the roller, no moment reactions anywhere. If it was two separate wheels, there would be two separate reactions. The slight deviation from diagrammatic convention shouldn't be tripping us up this badly just because they used a line instead of a triangle for the roller, the normal symbol always (edit: often, not always) has multiple wheels on the roller anyway