A rubber ball is plugging a whole in bottom of tank. When will it fail

[DylanW]

Homework Statement

A 3 cm radius rubber ball weighing 5 g is used to plug a 4cm hole in the base of a tank. The tank is in use and is gradually emptying. At approximately what depth will the plug fail by floating up and out of the hole?

Homework Equations

Density

The Attempt at a Solution

Ok so I don't have a numbers and all attempt at a solution :(
But I do have a logical thought process as to how the solution should be solved, but I'm stumped on finding the volume of a partial sphere when given radius and the radius of the base circle - I know it is enough information I just cannot work out how to calculate it.

The way I see it the upward force acting on the sphere will be proportional to the mass of the water displaced, whilst the downward force acting on the sphere will be proportional to the whole sphere so theoretically the plug should fail when the water level is at the height corresponding to the radius of the base circle on the opposite side. I am not sure about this though, as according to the figures given the density of the sphere is only 44 kg/m3 so all my common sense is telling me the plug wouldn't even work

Am I taking the right approach?

 

[Andrew Mason]

Homework Statement

A 3 cm radius rubber ball weighing 5 g is used to plug a 4cm hole in the base of a tank. The tank is in use and is gradually emptying. At approximately what depth will the plug fail by floating up and out of the hole?

Homework Equations

Density

The Attempt at a Solution

Ok so I don't have a numbers and all attempt at a solution :(
But I do have a logical thought process as to how the solution should be solved, but I'm stumped on finding the volume of a partial sphere when given radius and the radius of the base circle - I know it is enough information I just cannot work out how to calculate it.

The way I see it the upward force acting on the sphere will be proportional to the mass of the water displaced, whilst the downward force acting on the sphere will be proportional to the whole sphere so theoretically the plug should fail when the water level is at the height corresponding to the radius of the base circle on the opposite side. I am not sure about this though, as according to the figures given the density of the sphere is only 44 kg/m3 so all my common sense is telling me the plug wouldn't even work

Am I taking the right approach?

If the ball was just submerged, you would be right because the water at any depth would provide almost equal pressure all around (slightly more on the bottom than top - the difference being the buoyant force/cross-secitonal area of the ball), so there would always be a net buoyant force. But here the ball is not subject to water pressure all around because there is only atmospheric pressure on the part that is covering the hole. You have to find the downward water force on the ball (the downward water pressure over the 4 cm diameter hole x area) as a function of depth and then compare that to the upward buoyant force.

AM

 

[DylanW]

Thanks very much for the information AM :) I really appreciate it.
I'm just about to go to sleep after I watch the West Ham game, but I'll be working on this first thing tomorrow so I will report back my calculations - I am stoked that I don't need to do that partial sphere calculation as my head may have exploded, and how would I ever complete my assignment then!

But seriously thanks, a nudge in the right direction makes the world of difference.

DylanW

 

[DylanW]

Okay so for this one I've done F(UP)=1000.9.8.0.000113 = 1.1N
F(DOWN)=pghA + mg = 1000.9.8.h.(0.02^2).Pi+0.005.9.8 = 12.315h + 0.049 N
When F(Down)=F(Up) plug will fail
h = (1.1-0.049)/12.315 = 8.5 cm.

Feedback please?

 

[Nickie]

As a scientist, it is important to first gather all the necessary information and data before attempting to solve a problem. In this case, we are given the radius and weight of the rubber ball, as well as the size of the hole and the fact that the tank is gradually emptying. However, we do not have the density of the water or the tank, which are important factors in determining the depth at which the plug will fail.

One possible approach to solving this problem would be to use Archimedes' principle, which states that the upward buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. In this case, the fluid is water and the object is the rubber ball.

To calculate the volume of the partial sphere, we can use the formula for the volume of a sphere, V = (4/3)πr^3, and then subtract the volume of the cylinder that is formed by the water level at the base of the tank. This will give us the volume of the sphere that is submerged in the water.

Then, using the density of water, we can calculate the weight of the water displaced by the submerged portion of the sphere. This weight will be equal to the upward buoyant force acting on the sphere.

Next, we can use the weight of the rubber ball to calculate the downward force acting on the sphere. If this downward force is greater than the upward buoyant force, the plug will fail and the ball will float up and out of the hole.

However, as mentioned earlier, the density of the water and tank are important factors in this calculation and without that information, it is difficult to accurately determine the depth at which the plug will fail. It is also worth noting that the density of the rubber ball is very low, which may also affect the outcome of the calculation.

In conclusion, while your approach of considering the upward and downward forces acting on the sphere is correct, more information is needed to accurately determine the depth at which the plug will fail. It is important to gather all the necessary data before attempting to solve a problem in order to ensure a accurate and reliable solution.