Cube suspended in liquid - find tension

[Saladsamurai]

a cube with edge length L=.600m and mass 450kg is suspended by a rope in an open tank of liquid of density 1030 kg/m^3.

The top of the cube is L/2 deep in the tank. Find the tension in the rope.

Vol=.600^3=.216
Area of top side of cube=.6^2=.36

$$\sum F=0$$
$$\Rightarrow T+F_b-w-F_{press}=0$$
$$\Rightarrow T=F_{press}+w-F_b$$
$$\Rightarrow T=p*A+mg-\rho Vg$$
$$\Rightarrow T=(p_{atm}+\rho_lgh)*A+mg-\rho Vg$$
$$\Rightarrow T=[1*10^5+1030(9.8)(.3)]*(.36)+450(9.8)-1030(9.8)(.216)=39.4kN$$

I don't know what my error is, but the text answer is different. Is it the text or me?
BTW it says use 1 atm for p_atm

Casey

 

[Saladsamurai]

Any ideas? I am jammed.

Casey

 

[dynamicsolo]

I don't think that pressure term $$F_{press}$$ should be there. While it is true that atmospheric pressure is being applied to the top of the fluid, the effect on the fluid is to raise the pressure everywhere by 1 atm. So it contributes no pressure differential to the fluid and so no net force.

I get just 2230 N for the tension in the rope.

 

[Saladsamurai]

I don't think that pressure term $$F_{press}$$ should be there. While it is true that atmospheric pressure is being applied to the top of the fluid, the effect on the fluid is to raise the pressure everywhere by 1 atm. So it contributes no pressure differential to the fluid and so no net force.

I get just 2230 N for the tension in the rope.

Wow, that is the correct answer too...so is this--- While it is true that atmospheric pressure is being applied to the top of the fluid, the effect on the fluid is to raise the pressure everywhere by 1 atm. So it contributes no pressure differential to the fluid and so no net force<---basically Pascal's principle?

I am just trying to figure out why sometimes in class we need P_atm and sometimes we don't...

 

[stewartcs]

Sounds like it to me. The pressure is undiminished in all directions, so if you increase the pressure on top of the submerged object, then the pressure on the bottom is increased proportionally.

Do you have some examples of the problems involving p_atm?

 

[dynamicsolo]

Wow, that is the correct answer too...so is this--- While it is true that atmospheric pressure is being applied to the top of the fluid, the effect on the fluid is to raise the pressure everywhere by 1 atm. So it contributes no pressure differential to the fluid and so no net force<---basically Pascal's principle?

Yes, that's what's at work here.

I am just trying to figure out why sometimes in class we need P_atm and sometimes we don't...

Can you give an example of a problem where the atmospheric pressure had to be considered? (It generally doesn't contribute in buoyancy problems.)

 

[Saladsamurai]

Yes, that's what's at work here.
Can you give an example of a problem where the atmospheric pressure had to be considered? (It generally doesn't contribute in buoyancy problems.)

Yeah...but it does not involve buoyancy... PROBLEM: Crew members attempt to escape from a damaged submarine 100m below the surface, what force must be applie to the pop-out hatch (1.2 x .6)m^2.

Now this is clearly much different, but I need to know why. Is it because in this case, the force due to water pressure does not "cancel out" because the air pressure inside the sub is 1 atm?

So would this statement be true: with buoyancy problems, we don't consider pressure because we are not considering what is happening inside the object...just on it.?

Casey

 

[stewartcs]

Now this is clearly much different, but I need to know why. Is it because in this case, the force due to water pressure does not "cancel out" because the air pressure inside the sub is 1 atm?

Yes. The hydrostatic pressure pushing down on the hatch is greater than the air pressure pushing up on the hatch.

So would this statement be true: with buoyancy problems, we don't consider pressure because we are not considering what is happening inside the object...just on it.?

Casey

Not really.

The reason an object floats is due to the net hydrostatic pressure at the bottom of the object being greater than the downward force of the object due to gravity...i.e. it's weight is less than or equal to the buoyant force. So you need to consider the pressure. But don't forget, as pointed out before, the air pressure pushing down on the fluid in a buoyancy problem, is transmitted through the water as well, so the differential due to the atmosphere is still zero.

 

[Saladsamurai]

Yes. The hydrostatic pressure pushing down on the hatch is greater than the air pressure pushing up on the hatch.



Not really.

The reason an object floats is due to the net hydrostatic pressure at the bottom of the object being greater than the downward force of the object due to gravity...i.e. it's weight is less than or equal to the buoyant force. So you need to consider the pressure. But don't forget, as pointed out before, the air pressure pushing down on the fluid in a buoyancy problem, is transmitted through the water as well, so the differential due to the atmosphere is still zero.

Seems like I am just changing the wording now, but it looks like what you are saying is still the same as what I wrote. If it is being applied everywhere, then there is no net pressure. So with regards to bouyancy, it needs to be considered, but only to be disregarded as a force. ?

Casey

 

[stewartcs]

If the object is completely submerged, then any applied pressure on top of the fluid in which the object is submerged whether it is another liquid or gas (atmosphere) will be applied to the bottom of the object as well. This results in a net differential pressure of 0. But only if it is completely submereged.

That is why the Fpress you had above in your equation should not have been there. It was already accounted for in the buoyant force.

 

[Saladsamurai]

If the object is completely submerged, then any applied pressure on top of the fluid in which the object is submerged whether it is another liquid or gas (atmosphere) will be applied to the bottom of the object as well. This results in a net differential pressure of 0. But only if it is completely submereged.

That is why the Fpress you had above in your equation should not have been there. It was already accounted for in the buoyant force.

Hate to be a nuisance, but in this problem the object is only half submerged and we still did not need to consider p*A as a force.

What gives? I just trying to understand when we do and do not consider p*A to be a force...

Casey

 

[stewartcs]

In buoyancy problems, you usually just need to consider the weight of the object and the weight of the displaced fluid per Archimedes' principle.

The air pressure is normally disregarded in buoyancy problems. However, I think technically, there would be a slight buoyancy increase on an object if it were only partially submerged due to the atmospheric pressure. But for a completely submerged object, like the one in your original post, it does not add any net buoyancy because it is applied equally throughout the water (Pascal's principle).

I think that textbook problems generally ignore the atmospheric pressure since it may complicate the concept of Archimedes' principle. Like I mentioned before, the buoyant force exists because the pressure in the surrounding fluid increases with depth below the surface. Thus, the pressure near the bottom of the object will be greater than the pressure near the top. If you wanted the exact buoyant force on any object, you would have to find the vector sum of all of the forces on the object (the forces would be perpendicular to the surface of the object).

 

[stewartcs]

By the way, the vector sum of the force componets (the buoyant force) is normally equal to the weight of the displaced fluid per Archimedes' principle. This is generally accepted as the buoyant force. My point about the vector sum was in regards to the possibility of a slightly more buoyant force due to the atmospheric pressure imbalance on the top of the object as opposed to the bottom.

 

[Saladsamurai]

By the way, the vector sum of the force componets (the buoyant force) is normally equal to the weight of the displaced fluid per Archimedes' principle. This is generally accepted as the buoyant force. My point about the vector sum was in regards to the possibility of a slightly more buoyant force due to the atmospheric pressure imbalance on the top of the object as opposed to the bottom.

Cool. So it really is just a differential difference,,,like the force of gravity difference at my head and at my feet (that is a little more extreme though).

thanks stewartcs!

Casey