# No buoyancy inside liquid [closed]

The answer to this question is given as (a) 5m.

Due to the difference in pressure at the curved parts there will be a net upward force of buoyancy. And how can that be calculated without knowing the shape?

The reasoning is further illustrated in the image below.

Am I missing something?

## closed as off-topic by ACuriousMind♦, user36790, Gert, CuriousOne, Daniel GriscomMar 31 '16 at 20:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Community, Gert, CuriousOne, Daniel Griscom
If this question can be reworded to fit the rules in the help center, please edit the question.

• Yes, you're missing something. If you don't have information regarding the shape of the object, the question can't be answered. – David White Apr 1 '16 at 1:04
• @David White : Perhaps the answer does not depend on the shape of the object? – sammy gerbil May 7 '16 at 16:44
• I disagree that this question is off-topic. It contains a conceptual difficulty which the OP has identified. The OP has demonstrated effort to solve the problem, providing additional diagrams. An answer has been given, discussed at length, and accepted by the OP. The problem and its solution are instructive to other members of Physics SE. In what way could it be re-worded to 'fit the rules'? I cannot see any way in which it does not 'fit the rules' alluded to here. – sammy gerbil May 7 '16 at 16:56
• @sammygerbil, if the object is less then 5m tall, it will NOT be torn off the bottom of the container. Additional information in the question would be very helpful. – David White May 7 '16 at 17:43

The idea is to find the force that the water exerts on the object. If the object were not glued to the bottom, but were instead surrounded by water, the upward force would be given by Archimedes principle. However, since there is no water under the contact area, the Archimedes force of the water is reduced by the contact area times $\rho g h$, where h is the depth to the bottom. So this takes care of the force of the water. In addition, there is a tensile force on the bottom of the object equal to 2000 N. This is enough information to solve the problem.
Added Supplement The weight of the object is $W=\rho g V$, where $\rho$ is the density of the object. If the object were fully surrounded by water, according to Archimedes principle, the buoyant force of the surrounding water on the object would be $\rho_wgV$, where $\rho_w$ is the density of the water. However, since there is no water under the contact patch, the upward buoyant force exerted by the water on the object would only be $$F=\rho_wgV-\rho_wghA$$where A is the area of the contact patch. If $T$ is the tension in the glue, the net upward force on the object is given by: $$F_{net}=(\rho_w-\rho)gV-\rho_wghA-T=0\tag{1}$$ The glue fails when T=2000N. This gives one an equation for calculating the depth h at which the glue fails. Try a value of $h = 5\,$m in this equation, and see whether the equation is satisfied.
• I'm sorry; I miscalculated: the object would be pulled from the bottom, as the buoyancy of the object is greater than the given tensile force. However, the shape of the object determines how that buoyancy increases as the water depth increases. If it were a flat plate (on a $100cm^2$ base) this would happen quickly; if it were a tall rod (on the same base) it would happen slowly. Thus there is indeed insufficient information to answer. – Daniel Griscom Mar 31 '16 at 23:25