Okay plus 50! What is the added pressure of the container on the desktop? 50

The buoyancy and gravity acting on the object are canceled out, but now the pressure of the container on the table is studied, and the object, water, and the container are all pressed down, so it is the sum of the three.

Archimedes tells us that the pressure of the liquid is only related to the depth of the liquid,,, and the principle of the communicator tells us that the pressure of the liquid in the connected container at the same depth is the same, so the pressure of the liquid on the top of the container on the right is equal to the pressure of the liquid at the same depth on the left, that is, it is equal to the pressure at the depth of h1-h2, and then multiplied by the area to get the pressure.

The second question: No, because the liquid can flow, if you want to calculate the pressure at the bottom of the container, you must calculate the total height h1, if you think about it, if there is no liquid in the tube, the pressure of the liquid in the branch is only the liquid pressure of the large container, if you add the liquid in the tube, you have to add h1-h2 as high...

I know you may not be able to fully understand it for a while, but the key is that you have to understand the Archimedes' principle, you can ask me if you have any questions, I will do my best, and promise not to return to the error.

Think of the container and the object as a whole, and the pressure of the container on the table is the sum of the weight of the object and the container. What increases is the weight of the object.

Since the problem outside the container is considered, it has nothing to do with buoyancy. No matter what you put in the container, the pressure of the container on the table should be g container + g object. It's like a kilogram of water and a kilogram of iron, which are 2 kilograms when weighed together.

The pressure of the container on the table top is considered the pressure of the solid, which is equal to the total weight. In other words, no matter what shape the container is, the total weight of 10n water and 5n container is 15n, but if it is a non-cylindrical or cuboid container, the pressure of water on the container is not equal to the gravity of the water, that is, the pressure generated by 10n water is not equal to 10n. Because the pressure is distributed to the side walls of the vessel.

The pressure of the container on the table = the gravity of the container + the gravity of the water.

The pressure of water on the container = the bottom area of the container * the height of the water, not necessarily equal to the gravity of the water, so the pressure of the container on the table is not equal to the pressure of the water on the container + the gravity of the container (for example, when the mouth of the container is large and the bottom is small, or the bottom is small, the pressure of water on the container and the gravity of the water are different) I hope it can help you

There's something wrong with your habitual thinking. You think that the container is cylindrical, but in fact, if you look at the data, you can know that if it is 50*10cm 3 volume of water, the mass should be, and the weight should be 5N. In the question, 6n, it can be seen that the container has a large mouth and a small bottom. Herein lies the problem.

Water and the container should be regarded as a whole, and the whole should be placed on the tabletop, so the pressure of the whole on the table is naturally the gravity of the whole. And I think the reason for your mistake is to ignore the fact that the "pressure of water on the container" is calculated by the water as the external environment, while the "pressure of the container on the tabletop" is calculated by the air for the external environment.

So, the "pressure of the container on the tabletop" should be equal to the "pressure of the container" plus the "gravity of the water".

Now the force of the table is sought, and the "pressure of the water on the container" does not act on the table, and the "pressure of the water container" cannot be used to calculate the force on the table directly.

It can't be equal. Because the pressure of water on the container is related to the depth and density, and has nothing to do with the gravitational bottom area. The pressure of the water on the table is related to the force area and pressure. Therefore, these two quantities are not the same thing and cannot be equal.

The pressure of the water is in all directions, and the pressure is also in all directions, and the pressure is generated by the pressure, so the pressure at the bottom of the cup should be used to find the pressure at the bottom of the cup.

The cup in question is not cylindrical, but a container with a small bottom and a large mouth, and the bottom of the cup only bears the vertical downward pressure of the cylindrical water column with the same area as the bottom of the cup. The pressure of the rest of the water is borne by the walls of the cup.

Only if the shape of the container is regular, can it be added directly...

The pressure of water on the bottom of the container increases the weight of water equal to the volume of the object, and the pressure of the container on the table increases the weight of the object.

It will definitely increase, how much does it increase, the conditions you give are not enough, if you give the bottom area of the container, calculate the displacement of the metal ball, you can calculate the buoyancy of the metal ball, and the increase is the buoyancy of the metal ball.

You can imagine that if the metal ball is not connected by a thin wire, but a spring, the number of the spring must decrease, and the reduced part is the part of the container that increases the pressure on the table.

Increase. If the liquid level rises HCM if it is not filled, the pressure increases Water gh*s 100 Newtons s is the bottom area of the cup.

Isn't it full of water? So how can the liquid level rise?

You are standing on the ground, what is your pressure on the ground? If you think about it this way, wouldn't the vertical gravity and support force also cancel it out?

After the block sinks to the bottom, the container, water and object can be regarded as a whole, and the horizontal desktop only receives the pressure of this whole, and is not affected by other external forces, and the pressure is equal to the gravity of this whole, so the pressure on the table is equal to the sum of the container and water and the gravity of the object.

Since it is sinking to the bottom, according to the classification of high school physics, it is in a state of equilibrium, and the block, water, and container can actually be regarded as a whole, and the pressure of the container on the desktop can be understood as the gravity of the container plus water and the object