Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 83

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Mischek:
Yes, and the train cars should be connected by springs, what an economy. It's a waste of time.

There will be no economy - less force means less acceleration of the whole system, which means the train will go slower.

 
MetaDriver:
Well, if all you had to do was move them, then yeah. :)
don't tell anyone, don't (c))
 
MetaDriver:

Well, well.

Imagine a couple hundred small weights of 1 kg connected by springs. A good kick in the end of this pyramid will easily knock it down from start to finish.

Wobbly, of course.

It is unlikely that just a "good kick" will be enough for one 200 kg weight.

You'll get in a lot of trouble with the kicker, MD. Friction has to be taken into account anyway. Each successive shift will be weaker due to energy dissipation due to friction.
 
Hooke's "necessary distance" still rules.
 
alsu:
The concept of "coefficient of friction of rest" does not exist in school problems at all, because the force of friction of rest depends on the applied force (and is equal to it), and not on the weight of the body, in relation to which the coefficient is calculated at all.
Are you sure? Try looking on the web for problems related to the calculation of the required force in a bolted joint (two plates with a bolted joint).
 
MetaDriver:
And yet it's the Hooke's "necessary distance" that rules.

Nothing can be done here, we can talk about friction only after the body has moved a decent distance, until then, strictly speaking, there is no friction at all, the box just rests against the jagged surface
 
Mathemat:
You'll get tired of the kicking, MD. Friction still has to be taken into account. Each successive displacement will be weaker due to the dissipation of energy by friction.

I'll call you. If it's just the two of us, that'll be enough. But we won't kick 200 kg at once.

// Without calculations, it's just a matter of intuition.

--

Where the hell did Hooke go? I'm no friend of him, but I'm still against him!

 
joo:
Are you sure? Try looking on the web for problems relating to the calculation of the required force in a bolted joint.
The school problems do not take into account that the maximum resting friction force is greater than the sliding friction, so it is simplistically assumed that these coefficients are equal and the first one is not entered at all.
 
Man, I wondered where people had gone from the four. Everyone's been here all day.
 
alsu:

For the second box to move, the spring has to pull it with a force k*M*g. On the other hand, the same force is equal to u*X, where u is the coefficient from Hooke's law (spring stiffness), and X is the distance the first box travelled. Note that throughout this distance it was subjected to friction force k*m*g and force F external to the system. Their total work is equal to (F-k*m*g)*X. The spring tension force is internal to the system and moreover potential (not dissipative), so all its work flows into the potential energy of spring tension. At the moment of detachment this energy according to our conditions is equal to u*(X^2)/2.

So, minimum force F can be obtained from the condition that the total work of external forces must be equal to the potential energy accumulated inside the system. We obtain a system of equations:

k*M*g = u*X

(F-k*m*g)*X = u*(X^2)/2

Let's substitute u*X from the first equation into the second one and after reducing X we get F = k*(m+M/2)*g.

Yes, formally there is no error. But who says that only the big box will be touched? The small one will get it too, the spring doesn't give a shit where to act...

P.S. It seems to be a forum virus like the plane problem.

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