#暑期创作大赛#
The conclusion is: the change in the pressure of the liquid to the bottom of the container = the change in the total buoyancy of the liquid to the object
We mentioned in the previous article that using a very bullish conclusion helps us solve the buoyancy problem quickly. This is equivalent to telling you a conclusion that is very close to the answer, and you can fast forward directly to the third or third from the bottom to solve the problem. Others are starting from scratch and starting from zero, and you are using the penultimate and third steps of the whole process as the starting point to solve the problem, so for you, the difficulty of solving the problem will be greatly reduced, and your solving speed will be greatly improved.
This conclusion saves you a lot of thought process, all you have to do is believe and remember this conclusion.
The conclusion is:
This article will continue to show how to use this conclusion to solve seemingly complex problems super quickly. The title is shown in the picture:
(This is the last question of the middle school entrance examination in a certain region, and the amount of calculation is relatively large,)
First, according to the combined float suspended in water, it is stated that the overall density of the float, or the average density of the water, is equal, so we can get:
When the hollow tube that makes up the float and the filling cylinder are completely detached, there are no more than three results:
The first is that both the hollow tube and the filled cylinder are suspended in water, which means that the float density of the hollow tube, the filling cylinder, and the combination of the two is equal and equal to ρ0. Obviously, it is not suitable, because the title says that the material is different, which implies a difference in density (don't raise the bar, ask is Chinesecharacteristics), so this kind of scene does not exist.
The second is that the hollow tube floats up, and the filling cylinder sinks. This means that the hollow tube density is less than ρ0 and the packed cylinder density is greater than ρ0.
The second is the sinking of the hollow tube, which floats up with the filling cylinder. This means that the hollow tube density is greater than ρ0 and the packed cylinder density is less than ρ0.
So the second and third scenarios are real and reasonable. We need to discuss the second and third scenarios. According to the stem information, the total volume of the float is 2Sh, the gravity of the hollow tube is mg, the volume of the hollow tube is Sh, the volume of the filling cylinder is Sh, the density is unknown, set to ρ, then the gravity of the filling cylinder is ρgSh.
In fact, the main thing is to compare ρ and ρ0.
Regardless of the two outcomes, the final state is one floating, one sinking bottom.
The main thing is to correctly represent the gravity of the filling material, and to introduce the density into the expression, so that the total gravity of the float can be correctly expressed.
There are two outcomes:
The hollow tube floats, and the filling material sinks to the bottom;
The filling material floats and the hollow tube sinks to the bottom.
That is, when ρ is greater than ρ0, the hollow tube floats and the filling material sinks to the bottom
That is, when ρ is less than ρ0, the filling material floats and the hollow tube sinks to the bottom
【Conclusion】:
So, the answer to this question has two when the density ρ of the filler material:
When ρ is greater than ρ0, the hollow tube floats and the filling material sinks to the bottom
When ρ is less than ρ0, the filling material floats and the hollow tube sinks to the bottom
The difficulty of this topic is that the gravity of the original material is 2mg, and once the gravity of the float is considered to be 2mg, then this topic must be wrong. In addition, we can directly use the logic from the previous article: the amount of pressure change of the liquid to the bottom of the container = the amount of buoyancy change. We don't need to think too much about the specific details of the change, we can directly express the buoyancy in different states, and then do the difference, get ΔF floating, and then directly write the conclusion of the buoyancy change or convert it into other physical quantities such as pressure, as for which one to convert, adjust it according to the problem of the topic.
Ordinary problem solving ideas, generally to derive the change of liquid surface height Δh, but also to derive the volume of displacement and the change of volume ΔV, relatively troublesome and large amount of calculation.
Our method avoids the calculation and derivation of the change amount Δh before and after the liquid and the change amount ΔV of the volume of the discharge, looks at the essence through the phenomenon, hits the core, and directly runs to the end point: the buoyancy change ΔF float before and after calculation and derivation.
This makes it faster and easier to steer ideas down the right path.
Have you learned this conclusion and method that can be called an "artifact" level?
Have you figured it out now?