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Author Topic:   A test of your common sense
xongsmith
Member
Posts: 2578
From: massachusetts US
Joined: 01-01-2009
Member Rating: 6.8


Message 3 of 137 (665740)
06-16-2012 6:27 PM
Reply to: Message 1 by Taz
06-16-2012 5:48 PM


Your description and diagram are incomplete. I presume you meant 2 forces located L units away from the endpoints of beam, at points marked B and C, equal in strength P, pressing downward. One end of the beam, marked D, rests on a cylinder of circular cross section such that all right side endpoints of the beam rest on the top of the cylinder. The other end of the beam, marked A, rests on a triangular parallelepiped with a right triangular cross section, with the right angle edge on the ground directly below all left side endpoints of the beam. The beam is also presumed to be a rectangular parallelepiped. I also assume that the limitations of drawing the diagram prevented making the top of the triangular parallelepiped exactly meet the bottom of the beam, rather than penetrate into it as shown.
Is this correct?
BTW,
THe reason I chose this example is because lately I've been noticing that most people have misconceptions about this. I'd argue that this example represents normal everyday thing that most people deal with. And yet, most people appear to be oblivious when it comes to this simple problem.
I have never seen this problem come up anywhere in my life. When you say "most", how many is that? What "normal everyday thing" is this a representation of?
Edited by xongsmith, : further issues
Edited by xongsmith, : No reason given.
Edited by xongsmith, : No reason given.

- xongsmith, 5.7d

This message is a reply to:
 Message 1 by Taz, posted 06-16-2012 5:48 PM Taz has not replied

  
xongsmith
Member
Posts: 2578
From: massachusetts US
Joined: 01-01-2009
Member Rating: 6.8


(1)
Message 6 of 137 (665743)
06-16-2012 7:23 PM
Reply to: Message 5 by Minnemooseus
06-16-2012 7:04 PM


Again, there are many things left out. Is the triangular support made out of carbon steel, the beam made out styrofoam and the circular support a big soap bubble? Are the 2 forces P carrying the weight of dustmotes or asteroids the size of Ceres? Are there lateral perturbations distributed around the application of each force P at points B and C? Are they big enough to make the triangular support fall to the left as the right end rolls off the cylinder? Can the cylinder even roll freely? When Taz says "fail" does he mean fall down or snap apart? Or does he mean Either One, whichever comes first? All of the relative tensile strengths and characteristics of the static stability of the elements have to be specified in a far greater detail than one would need if it were to be able to be merely categorized as a "simple" problem. The diagram leaves out a presumed solid ground underneath, but what if it is correctly drawn as only thin air?
But most of all, what in the world is this "common every day thing"?

- xongsmith, 5.7d

This message is a reply to:
 Message 5 by Minnemooseus, posted 06-16-2012 7:04 PM Minnemooseus has seen this message but not replied

  
xongsmith
Member
Posts: 2578
From: massachusetts US
Joined: 01-01-2009
Member Rating: 6.8


Message 10 of 137 (665748)
06-16-2012 8:04 PM
Reply to: Message 8 by Taz
06-16-2012 7:53 PM


Sorry you see it as BS, Taz. But you can't just have one end be a triangle about to tip over to the left and other be some metal tubing or something. Make your example identical at each end. Make all of the materials the same. Make it simple so that there is a simple solution that can be mis-intuited through "common sense".
Just trying to help.
You say:
There's a reason why I said use your common sense rather than real engineering knowledge.
Too which I reply:
There's a reason the real world is built with good sound engineering knowledge rather than relying on common sense - engineering knowledge actually works.
Edited by xongsmith, : soapbox utterance
Edited by xongsmith, : whoops i meant mis-intuited

- xongsmith, 5.7d

This message is a reply to:
 Message 8 by Taz, posted 06-16-2012 7:53 PM Taz has not replied

  
xongsmith
Member
Posts: 2578
From: massachusetts US
Joined: 01-01-2009
Member Rating: 6.8


(1)
Message 12 of 137 (665750)
06-16-2012 8:37 PM
Reply to: Message 9 by Taz
06-16-2012 7:59 PM


Hi again Taz...
Not to be troublesome, but could you give me some kind of a clue about these questions I asked in Message 3:
I have never seen this problem come up anywhere in my life. When you say "most", how many is that? What "normal everyday thing" is this a representation of?
"Most" in my neck of the woods usually means "more than half".
This may actually be off-topic, but you claimed that "lately I've been noticing that most people have misconceptions about this. I'd argue that this example represents normal everyday thing that most people deal with." For example, most people do not have any everyday experience carrying a load of bricks across a plank.
Maybe you meant to say something along the lines of "this represents a simplified abstraction of certain events in ordinary life." This is similar to first studies of the quantum dynamic wave function of an atom where they first begin by assuming the atom is a spherically symmetric object, then add in the other terms.

- xongsmith, 5.7d

This message is a reply to:
 Message 9 by Taz, posted 06-16-2012 7:59 PM Taz has replied

Replies to this message:
 Message 16 by Taz, posted 06-16-2012 10:03 PM xongsmith has replied

  
xongsmith
Member
Posts: 2578
From: massachusetts US
Joined: 01-01-2009
Member Rating: 6.8


Message 14 of 137 (665753)
06-16-2012 9:35 PM
Reply to: Message 1 by Taz
06-16-2012 5:48 PM


general reply
OKAY.......my hipshot, common sense guess, without the needed information Taz left us to assume......
assuming the drawing is as i assumed in Message 3, and that the whole experiment apparatus is perhaps on a carpet-ish unyeilding surface where the triangular support cannot slide on it, but the circular support can roll on it, and that all parts of the experiment are unattached in any way, whether by glue or magnetism or molecular attraction otherwise and that the beam, being of this world and subject to the visual recognition of common sense eyes, is ordinary and will bend some before it breaks, the "Fail" here must be a fall first, not a break - unless he defines it as a break, in which case we follow the fall down. Here is why:
The beam will begin to bend before it breaks, making the endpoints A and B on it get closer together by an infinitesimally small amount at first. At this infinitely small step in time, the 2 ends move towards the middle, meaning that the end on the left drops precipitously down the triangle while the right end still rides pretty high on the circle. The next infinitesimally small increment in time finds the left end splaying out the triangular side, being a right angle triangle positioned so badly for the beam, lifting it's right corner off the rug, while the left side begins to slide off the circle, rolling it away to the right. Because Taz drew the right triangle not isosceles but 3-4-5-ish with the 5-ish pointing up, the left side is quicker to descend overall and quickly the hypotenuse meets the beam with an audible snapping sound. Then the circular descending rolling-away motion at the right completes and the right end touches the rug first. It is at this point that the relative power of the 2 forces, P, come into play. The one on the right initially got to bear down harder as the left side fell faster, but then the hypotenuse ended that and the one on left got a jolt of stubborn resistence. I suspect that, now, the common sense world beam, if "failing" meant breaking before falling, would break on the left side just fractions of a second before the right side gave way. By the time the P on the right gets to hit the beam full on, it is already on the ground, too late, and might even bounce.
That would be my hipshot, common sense angle on it, assuming all the things I had to assume about Taz's ill-formed statement of the problem. These assumptions could be way off, yes.

- xongsmith, 5.7d

This message is a reply to:
 Message 1 by Taz, posted 06-16-2012 5:48 PM Taz has not replied

  
xongsmith
Member
Posts: 2578
From: massachusetts US
Joined: 01-01-2009
Member Rating: 6.8


Message 17 of 137 (665756)
06-16-2012 10:04 PM
Reply to: Message 15 by purpledawn
06-16-2012 9:42 PM


Re: Not The Center
PD writes:
Assuming whatever the beam is sitting on is stable and won't move, I would say it will fail at B and C.
Yes, those are good assumptions, given the dearth of info Taz gave us. But I suspect he wanted us to think they would move. Why draw them the way he did if not to emphasize the differences in each supporting side? A little reverse detective work........

- xongsmith, 5.7d

This message is a reply to:
 Message 15 by purpledawn, posted 06-16-2012 9:42 PM purpledawn has not replied

Replies to this message:
 Message 19 by Taz, posted 06-16-2012 10:16 PM xongsmith has replied

  
xongsmith
Member
Posts: 2578
From: massachusetts US
Joined: 01-01-2009
Member Rating: 6.8


Message 18 of 137 (665757)
06-16-2012 10:09 PM
Reply to: Message 16 by Taz
06-16-2012 10:03 PM


Lol thanks for that!

- xongsmith, 5.7d

This message is a reply to:
 Message 16 by Taz, posted 06-16-2012 10:03 PM Taz has not replied

  
xongsmith
Member
Posts: 2578
From: massachusetts US
Joined: 01-01-2009
Member Rating: 6.8


(1)
Message 20 of 137 (665760)
06-16-2012 11:00 PM
Reply to: Message 19 by Taz
06-16-2012 10:16 PM


Re: Not The Center
If it were even remotely possible to assign myself with anyone in that funny snippet (thanks again for that!!), it would be the Peter Sellers character. But - actually the snippet was not really relevant to this thread. Sorry.
Try replying to Message 14 first, if you are set on blowing off Message 3 and Message 4 by NWR.
You say:
Actually, there are 2 reasons why I put a roller in there. (1) It's out of habit. (2) I didn't want people to be concerned with the x axis.
Well, see? there: you are now calling it a "roller"! That is information that I had to assume with probabilities less than desirable - you could have just simply said that at the outset. Habit? Weird you. Thumbs up! And if you think the x-axis is nothing to be concerned about, maybe you should restart your algebraic physics problem knowledge at this time. Is the roller going to only move in the Y-axis??? Read Message 14. See? I think you thought you had a cool simple little thing here, only it was way more complicated than you first thought. Clue: if I had all the info, it would still be WAY TOO HARD FOR ME!
You say:
Just look at the drawing. Don't over think it.
The drawing is very bad. No way to sugar-coat it for you. It's bad.
Edited by xongsmith, : Forgot
Edited by xongsmith, : No reason given.

- xongsmith, 5.7d

This message is a reply to:
 Message 19 by Taz, posted 06-16-2012 10:16 PM Taz has not replied

  
xongsmith
Member
Posts: 2578
From: massachusetts US
Joined: 01-01-2009
Member Rating: 6.8


Message 66 of 137 (665862)
06-18-2012 4:14 PM
Reply to: Message 54 by Taz
06-18-2012 1:38 PM


Taz restates his question:
We have a long thing made from a hard material that is designed to only withstand 500 pounds. The long thing is held up at the ends by two 10-96's. 2 insanely overweight women, each weighing 500 pounds, decide to sit on two points L distance from the ends and the women are also L distance from each other.
Where on the long thing will the crack or breaking occur?
This is much better!
My inferior common sense tells me it breaks simultaneously under each woman.
P.S. - what's a 10-96? Is that some kind of cinder block?

- xongsmith, 5.7d

This message is a reply to:
 Message 54 by Taz, posted 06-18-2012 1:38 PM Taz has replied

Replies to this message:
 Message 68 by Taz, posted 06-18-2012 5:00 PM xongsmith has replied

  
xongsmith
Member
Posts: 2578
From: massachusetts US
Joined: 01-01-2009
Member Rating: 6.8


(1)
Message 69 of 137 (665869)
06-18-2012 10:57 PM
Reply to: Message 68 by Taz
06-18-2012 5:00 PM


Taz replies:
xongsmith writes:
P.S. - what's a 10-96? Is that some kind of cinder block?
It's cop talk.
How do you know the two 10-96's won't immediately drop the long thing upon sight of the two fat women approaching it, run over to them and then start furiously smearing their secret stash of Boysenberry jam all over them, singing "Oh yes, we're going to Heaven now, we're going to Heaven now! Oh yes, we're going to Heaven now!" ?

- xongsmith, 5.7d

This message is a reply to:
 Message 68 by Taz, posted 06-18-2012 5:00 PM Taz has not replied

  
xongsmith
Member
Posts: 2578
From: massachusetts US
Joined: 01-01-2009
Member Rating: 6.8


Message 88 of 137 (665970)
06-20-2012 11:43 AM
Reply to: Message 84 by Taz
06-20-2012 10:52 AM


Taz explains with more detail:
On the left side, the beam is held by a pin to prevent it from moving horizontally but allow it to rotate. On the right side, it is held up by a roller.
And presumably the pin doesn't allow it to move vertically as well. See, I had no idea the triangle was an engineering shorthand symbol for a pin. I still don't have any idea why it was drawn as a Right Triangle facing the way it is instead of an isosceles triangle - what is engineer's fleshing out of that shorthand symbolization? Does it make a difference if the symbol is drawn facing the other way?
Now I am reminded of those elasticity/viscosity diagrams with short symbols for various kinds of forces, like a piston symbol for viscosity, arranged in a drawing similar to a software flow diagram.
So now I would revise my hipshot answer, since only the roller can move away, once the deformation is large enough. But I need to know if the P on the right is fast enough in acceleration to keep up pace of applying force P on the beam after it slips off the roller. If so then the fat lady on the right will be the first to experience the break, followed by it breaking under the fat lady on the left, only because the radius arm on the right hits the ground first, at a faster speed, providing a bigger jolt to the load P on the right.
But if the beam breaks before it slips off the roller, then you have the beam breaking simultaneously under each fat lady.
Still not convinced of any thing, though, because this is me trying to use the well-known error-prone common sense to make a hipshot, rather than some more careful scientific analysis.

- xongsmith, 5.7d

This message is a reply to:
 Message 84 by Taz, posted 06-20-2012 10:52 AM Taz has not replied

Replies to this message:
 Message 133 by Stile, posted 07-10-2012 3:42 PM xongsmith has not replied

  
xongsmith
Member
Posts: 2578
From: massachusetts US
Joined: 01-01-2009
Member Rating: 6.8


Message 135 of 137 (667725)
07-11-2012 4:16 PM
Reply to: Message 134 by RAZD
07-11-2012 7:17 AM


Re: Engineering Symbols
Well then: What I get here is that all of the drawing is fiction, but not only that, secretive codes...symbols! that have coded meanings only for card-carrying members of the Busily Esteemed Engineering Society (BEES). Nothing about this is actually at all commonplace, as posited in the OP.
So basically the triangle "pin" represents a roller of 0.00 radius and the roller is a fat pin (r>0) and they are both anchored to a fixed, practically infinitely strong background. One of the answers was that the pin breaks. ??? We weren't given the info that the pin might be weaker than the beam. WTF? WTF??
Why not just put a board over 2 cinder blocks? There's your commonplace example. Jeez...eyes roll....

- xongsmith, 5.7d

This message is a reply to:
 Message 134 by RAZD, posted 07-11-2012 7:17 AM RAZD has replied

Replies to this message:
 Message 136 by Taz, posted 07-12-2012 12:05 AM xongsmith has not replied
 Message 137 by RAZD, posted 07-12-2012 7:06 AM xongsmith has seen this message but not replied

  
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