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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.

There is no net upward force due to buoyancy

Am I missing something?

santhe
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  • 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 01 '16 at 01:04
  • @David White : Perhaps the answer does not depend on the shape of the object? – sammy gerbil May 07 '16 at 16:44
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    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 07 '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 07 '16 at 17:43

1 Answers1

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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.

stafusa
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Chet Miller
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  • Let's say you put a sticker on the side of a bathyscaphe, and then headed for the Mariana Trench. Is there some depth at which the sticker would be suddenly ripped from the side of the bathyscaphe? If not, then why would the object in the OP's problem be ripped from the bottom of the container at some critical depth? – Daniel Griscom Mar 31 '16 at 20:42
  • @DanielGriscom, I don't understand your bathyscaphe question, but, if you would like more detail, I would be glad to provide the full development for the critical depth problem. – Chet Miller Mar 31 '16 at 22:19
  • Sorry; I was just pointing out that, in truth, the object would never be pulled from the bottom. The buoyancy of an object submerged in water is independent of its depth. If you're hypothesizing that there is some inequality at the attachment point that adds an upwards force greater than there would be if surrounded by water, you have to explain where that increased force comes from. – Daniel Griscom Mar 31 '16 at 22:30
  • No. I'm saying that the object would be pulled down at the bottom by tension in the glue. You are familiar with the concept to tensile stress, correct? – Chet Miller Mar 31 '16 at 23:04
  • 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
  • @Daniel Griscom See the Added Supplement to my answer. – Chet Miller Mar 31 '16 at 23:55
  • You've got a sign wrong. Assuming no pressure is placed on the object across the adhesive contact patch and the object is completely submerged, then the net buoyancy of the object would decrease with the depth of the water. So, depending on the shape of the object and the resulting buoyancy changes as the water depth increased, the object would be torn away at different depths, or perhaps even never torn away. – Daniel Griscom Apr 01 '16 at 00:51
  • I stand by what I said. My equation does predict that the buoyancy of the object would decrease with the depth of the water, in agreement with what you said. The problem statement seems to have the sign wrong. It seems to imply that the depth has to be greater than a certain amount to cause the glue to fail. Actually, it has to be less than a certain amount to cause the glue to fail. – Chet Miller Apr 01 '16 at 01:01
  • ... but you're still assuming the object is always fully immersed, when different shapes/depths will change that. – Daniel Griscom Apr 01 '16 at 01:29
  • Yes. That is a constraint that I'm assuming. – Chet Miller Apr 01 '16 at 02:03
  • Using values given – santhe Apr 02 '16 at 02:33
  • Yes, that's correct. – Chet Miller Apr 02 '16 at 02:35
  • $$h\leq\frac{V}{A} (\frac{\rho_{w}}{\rho}-1)-\frac{T}{A\rho g}$$ $$ h\leq\frac{0.5}{10^{-2}} (\frac{1000}{500}-1)-\frac{2000}{10^{-2}5000}$$ $$h\leq 10m$$ The answer given in the book is wrong. – santhe Apr 02 '16 at 02:49
  • @santhe You made an algebraic error in solving the Eqn. 1 for h. The book answer is correct. $h\leq 5m$ – Chet Miller Apr 02 '16 at 11:46
  • Oops! $$ h \leq \frac{V}{A}(1- \frac{\rho}{\rho_{w}})-\frac{T}{Ag\rho_{w}}$$ $$h \leq \frac{0.5}{10^{-2}}(1- \frac{500}{1000})-\frac{2000}{10^{-2}10^{4}}$$ $$h \leq 5m$$ With the assumption that the whole object is submerged. – santhe Apr 04 '16 at 08:46