test job interview

[exxon]

was asked to answer simply yes or no, or were they asked to justify their answer?

Yes or no. asking not to give other answers.

the motivation is also to see the reaction to the need to give a dry answer. Surely someone imagines some kind of patibula where the answer "no" would have left the mannaia to truncate the candidate's fingers, but things never go in this direction.

apart from what he responded with the third option (when he was explicitly asked to respond with one of the other two), which in doing so was played almost all possibilities in his favor (who chooses the escape routes in the moments of binary choice is a danger in the company...), with all others we discussed their answer.

another has pretended to have understood the error (thinking to have before someone who would not have noticed...), putting in this way end the discussion on the topic (and also its possibilities...).

one insisted on the correctness of his answer (which I liked), and also said that he would deepen the thing regardless of the outcome of the interview (and still on the list...).

one of them did not understand the answer (the graduate), but in general proved a good preparation and (of course) the best experience and is the second candidate to be past.

I prefer to forget that I saw them.

 

[Fulvio Romano]

no: reasoning is wrong.

if you apply this reasoning to the second drawing, where the tension is generated by the spring, and apply a mass of 0.1 kg, even in this case the tension of the rope should increase, but since the spring does not stretch, it means that the tension in the wire has remained unchanged.

deduction must be the one that the two tensions do not sum up.

the fact is that 2 n should not be added to 10 kg to get the load. once the load is applied (which must exceed 2 n to create a detachment, and therefore a further stretch of the wire) it is only the latter to stretch the wire, for 98 n. this because the support is fixed and rigid. between an applied load of 0 n and 2 n the wire does not see any load difference
has been one of the hardest rocks to understand at times of construction of machines, is the same case of the tightened screw

I'm sorry, but I'm sorry. Premute that indeformability does not exist in nature, so it is necessary to choose a model and apply it, and if the result appears absurd, this is due to the absurdity of the premises.

if it is all indeformable the final condition is hyperstatic and the two tensions should be added. It is equivalent to applying two forces to the wire and there is no reason why the application of the second, the first should be discharged. free body diagram gives this answer.

Don't tell me about preloaded screws. It has nothing to do with it, the phrase "the preload of the bolt does not vary its ultimate resistance" is based on the elongation of the screw that detaches the plate. hypothesis that in the text has been excluded.

 

[Fulvio Romano]

However, "unextensible and negligible mass wire" is an ideal model of something that does not suffer from temperature, magnetic fields, neutrinos or other. It is a means that flexiblely transports the point of application of one force into another.

I do not doubt that "unextensible thread" is in principle a simplification. I do not doubt that the simplifications are the basis of scientific knowledge. What I'm saying is that if you're studying a phenomenon based on elasticity (the detachment of the capital), putting the inextensibility between the starting hypotheses is a bit like shooting in a foot.

 

[exxon]

if it is all indeformable the final condition is hyperstatic and the two tensions should be added.

no, because binding reactions do not depend on the move (see below). the binding reaction is always equal "to what remains" after having summed (vettorially) all forces agents. in the case in question, the tension of the wire is applied to the plate that "draws up", the weight attached that "gets down", and what remains to hold everything still is the binding reaction of the support. adding everything, you see that the tension of the wire does not change until the weight force does not exceed the tension set in the wire.

regarding the fact that binding reactions exist and change even in the case of rigid bodies and without any movement, I carry the example of my previous post.

example: on an infinitely rigid horizontal plane an object is also infinitely rigid; the binding reaction of the plan is equal to the force-weight of the object; if I apply to that object a vertical force upwards, equal to half of its weight, the object does not move and does not deform, but the binding reaction of the plane halves.

 

[Fulvio Romano]

no, because binding reactions do not depend on the move (see below). the binding reaction is always equal "to what remains" after having summed (vettorially) all forces agents.

Where did you read this law? under what assumption is valid?
I say that it is valid only in the case of "deformable hazards in the hypothesis of small shifts", i.e. the binding reaction continues not to do work when applying the principle of virtual works, but is originated from a deformation of the support, which is the measure of "what remains" after adding the forces.
if you deny deformability, deny the law you have just exposed, except in isostatic contexts.

example: on an infinitely rigid horizontal plane an object is also infinitely rigid; the binding reaction of the plan is equal to the force-weight of the object; if I apply to that object a vertical force upwards, equal to half of its weight, the object does not move and does not deform, but the binding reaction of the plane halves.

easy so, the system remains isstatic. we try with an example more similar to that of the wire. you have an infinitely rigid horizontal plane with supported an object too infinitely rigid. above the object and in contact with it there is another infinitely rigid horizontal plane.

How do you solve it?

 

[exxon]

Where did you read this law? under what assumption is valid?

third principle of dynamics (newton).
valid in any case (otherwise it would not be a principle... )

come lo risolvi?

the demonstration (mathematic) of tension in an inextensible medium between two fixed points is in #62.

 

[exxon]

Let us see if we can put an end to the question of that tension that you should sum. . .

I try to do it with a slightly different problem that serves as a support to better delimite some details of the initial problem placed in #1

a f, inextensible and of negligible mass, is kept in tension between a p, indeformable, and also of negligible mass, and a suspended mass with the postponement of an ideal carrucola (indeformable and negligible mass). the plate supports the fixed support s, held in position from the tension generated by the weighted mass. to the plate are then hung two masses from 0.1 kg and 0.2 kg respectively.

1) the tensions of the inextensible wire are established in the two cases listed.
2) tell if in the cases listed or in the absence of any weighted mass, you have movement in any part of the system.

note (important): questa is not a review of the first exercise. It is a different exercise that sets itself different purposes. it will be useful to understand some errors that many commit when systems are not immediately intuitive as instead most times others are.

 

[tanticapelli]

You are right in the example you carry, but we get out of the conditions of the question as a place. in many ways.
the pillar is free to move and is not confined by the two walls, so you will not have mai the condition of switching from a traction to a compression, in the examination rod.
then it is not an auction but a thread for which, by definition, in a wire you cannot have compression.
born as traction and traction remains also after the application of the load.
If I want to follow your example, I should say cool the element until I have 2n of tension, then I cool it again to add another 98.06n, what would you get?
the core of the matter lies in less 2, condition guaranteed excluding the presence of the plate s and the type of bond of the plate (unidirectional).
I repeat, it is correct to say that at the piattello there will be 98-2 newton, but it is equally correct to say that just after the bat there will be 98+2 newton.

It almost seems like a guy looking in the sky puts the fingertips in front of the moon, whispers and thinks: I've got a lot of fat, I've got my thumb as much as the moon now.
But I remembered that these menades were doing in the middle to joke, I didn't think they could be the subject of intake tests. ...

I thank God (seen that it is also Easter) not having been subjected to a test like this with a taste of talent show but to a normal test (I would say quite long since it lasted 16 or 18 hours I do not remember) that evaluated my practical attitudes in good and evil.
And I won on an engineer, and I'm just a gemetra.

I think sometimes it's better to stop a moment and cut it off.

Good Easter to all

 

[Fulvio Romano]

Let us see if we can put an end to the question of that tension that you should sum. . .

I try to do it with a slightly different problem that serves as a support to better delimite some details of the initial problem placed in #1


View attachment 53236note (important): questa is not a review of the first exercise. It is a different exercise that sets itself different purposes. it will be useful to understand some errors that many commit when systems are not immediately intuitive as instead most times others are.

known with pleasure that here the exercise has become isostatic and therefore its solution returns unique, regardless of the constitutive bond of a hypothetical indeformable material.
I'm sorry exxon, you made a mistake judging professionals (both your candidates, and us on the forum) based on a bad exercise. and it is useless that you dare to say that it is trivial and who does not know how to solve it is an ignorant. This thread shows you the opposite. Unless we're almost all ignorant, and then considering where you put the asticella maybe you were making the nasa selections.

is a bit like asking if it weighs more than one kg of feathers or a kg of lead. The elementary boy tells you they're the same, an engineer can only point out that if you're talking inertial mass, the lead kg weighs a little more.

 

[exxon]

@meccanicamg what you have reported and what is written are two totally different things, just read.
@fulvio RomanI wrote clearly that this was not a revisitation of the exercise in #1!

However, since you wrote that it is now simple, answer the points I have listed: they serve for the next step. . .

 

[Fulvio Romano]

However, since you wrote that it is now simple, answer the points I have listed: they serve for the next step. . .[/QUOTE]a couple of clarifications:
1. I'm not interviewed by you
2. in the interviews that I do I prefer the relational aspect. a candidate who is technically better than you do, but as you risk breaking my office, it makes me more harm than anything.
3. I'm sure of what I say more or less than you're sure of what you say. I don't think this discussion has a lot of chances to end up with a winner. You said a series of chestnuts, like that of the "third pricipio that as a principle always applies" that godel is turning in the grave. I'd end it with the auditions.

 

[exxon]

convenient to accuse others of saying chestnuts instead of giving two answers that would demonstrate their own (see isostatic systems and company).

However, no competition and no search for winners: until you withdraw into your towers, avoiding confrontation, every discussion is constructive. I'm always open to technical discussion if you call yourself out... sin.

 

[PaoloColombani]

I put myself in your exercise, so we unlock the discussion.
definitions:τ : load providing voltage [N]q : hanging load on the plate [N]τ : rope tension [N]...the τ symbol could cause confusion for its common use as a specific internal voltage having the size of a pressure.
known values:Applause = -2 p.
g = -9.81 m/s2
unknown:And?
the equation that identifies τ:τ = |t| (only if -q ≤ -τ)

the condition -q ≤ -τ meets the static balance of the system. when -q > -τ the balance is violated and you enter the dynamic field.

dunqe, cases 1 and 2:

== sync, corrected by elderman ==
== sync, corrected by elderman ==

in both cases is satisfied the condition -q ≤ -τ (quesite 2) for which:

τ1 = τ2 = |t| = 2n (quesite 1)

 

[exxon]

I would like to point out that even if I happen to examine candidates as part of my business, it is certainly not my intention to do so with colleagues I can find on the net. the answers I asked are to create a logical process that leads to the clarification of the technical analysis of a system.

well, we have established that in systems in #107 (up to break-even point, 2 n of force weight applied to the plate) nothing moves.

remaining within the break-even limit, is there any difference if we change the system as follows?
Note: I remember that the thread is flexible but inextensible.

 

[PaoloColombani]

It seems that no one wants to intervene anymore. How come?
in the static equilibrium condition the bond in 'o' is null, does not generate any action, so the scheme is equivalent to the previous one.

 

[cacciatorino]

It seems that no one wants to intervene anymore. How come?
in the static equilibrium condition the bond in 'o' is null, does not generate any action, so the scheme is equivalent to the previous one.

The author of the discussion has been "feeled" for a while until he is able to write again we could avoid feeding the discussion given the impossibility of the contradictory.

 

[Zac69]

I didn't make high squoles but 10kg are 98.06n, so adding the 2n for me there would be (approximately)

 

[stevie]

I didn't make high squoles but 10kg are 98.06n, so adding the 2n for me there would be (approximately)

This seems to me the best answer. . .

 

[exxon]

It seems that no one wants to intervene anymore. How come?

because when it comes to making controversy there is much more willingness than to discuss technical issues.

in the static equilibrium condition the bond in 'o' is null, does not generate any action, so the scheme is equivalent to the previous one.

perfect (and correct).

we have established that the wire connected to the plate does not see any difference in case there is the bond in 'o', or not. in both cases, for weight forces applied to the plate variables from zero to 2 n, the tension of the wire remains unchanged. It's still 2 n.

the first consideration (all but trivial) is that the weight force (up to 2 n) not sum to the tension of the wire. there are those who had supported the opposite, and the initial test (now perfectly superimposed to the system described in #114) aimed precisely at evaluating the ability to realize it.

lacks little to the complete definition of the system:
"What happens in #114 when the weight force is greater than 2 n? "

note: as said back, with "tension" I always meant to indicate a force. in Italian the term tension is used both for force, and for pressure (but in this case with the symbol σ, not τ). Since in other technical languages it uses "tension" for strength and "specific pressure" for pressure, I prefer to use these terms (admitted) also in Italian.

 

[+forte]

seems someone posted it on a foreign site

doubt and dismay o_o