Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 80

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Mathemat:

Are you implicitly implying that if the boxes are equal, you need 3/2*K*m*g?

Yes. But that doesn't mean that you can drag them further with the same force. Sooner or later, if you don't increase the force F, the system will stop because of friction.
 
alsu:
The formula will not work because it turns out that the factor u is infinite, so the potential energy goes there as well. But if we assume that the rod stretches the required distance according to Hooke's law (which is not the case in reality), the formula will be the same.
How many centimeters is required? I am in search of a racially correct spring.
 

Although, in principle, the process will repeat itself. It would be like a caterpillar.

You may remember that this is almost exactly what happens in practice - two bodies connected by springs move jerkily if pulled with a constant force. Although there is another effect here - the maximum force of rest friction is actually slightly greater than sliding friction, which is not usually taken into account in problems.

 
MetaDriver:
How many centimetres are needed? I'm just looking for the right spring.
OK, I'll go to the handbook and calculate it.
 
alsu:

Although, in principle, the process will repeat itself. Like a caterpillar.

No. The process will stall. (sort of) How about an interchangeable vector solution?

 

Yes, I didn't take friction into account. I'll have to think and break another pattern...

OK, let's take friction into account. Apply K(m+delta)g. The acceleration begins, the spring compresses/extends.

The balance of forces is such that due to the friction energy consumption, only K*delta*g acts on the spring, which will charge it and push the big body when the spring fully balances the small one and it stops.

It turns out that you need K(m+M)g. Again it doesn't matter which body to push.
 
Mathemat:


It turns out that you need K(m+M)g.

was on page 78. I guess it didn't work.
 
Mathemat: It turns out that you need K(m+M)g. Again it turns out it doesn't matter which body to push.

There is a simple explanation for this: the spring inside is just a distraction. This is how any solid body works.

It is a composite body, and to move it all the way somewhere, you have to apply exactly this force and not less.

 

It is clear that the force needed is less than K(m+M)g. By a positive delta. It is clear that the delta depends on how much distance (and therefore time) the baby has to spare before the spring will bounce it back. I.e. the spring stiffness is not only important, it is also the main thing in all this jostling.

Waiting for Alsou from the handbook.

 
alsu:
Okay, I'll go to the reference book, let's do the math.

Here's the deal. Let's take steel. It has a Young's modulus of 210 gigapascals. Recall, Young's modulus is a characteristic of the ability to elastic deformation, calculated as

E = F*l/(S*x), where F is the force, l is the length of the rod, S is the cross-sectional area, x is the strain.

Let the box have a weight of 1 kg and a coefficient of friction of 0.5. Then the force required for shear, k*m*g ~ 5N.

If the bar has a cross-section of 1 square millimetre and is 1 metre long, the required strain to produce this force is

x = F*l/(S*E) = 5*1/(10^(-6)*210*10^9) = 2.4 * 10^(-5) meters.

Looks like I'm wrong with the explanation: in fact, at such displacements under real-world conditions, rest friction doesn't have time to convert to sliding friction simply because sliding won't even start. The point is that the friction model we are using is a very approximate one, and will not work at such offsets, comparable to the size of the surface roughness.

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