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ChunkymonkeyI Nov13-11 04:07 PM

Pressure, Buoyant Force problem
 
1. The problem statement, all variables and given/known data
A cube of metal(density=6.00 kg/dm^3) has a cavity inside it. It weighs 2.40 times as much in air as it does when completely submerged in water. What fraction of the cube's volume is the cavity?
2. Relevant equations
Density=m/v
Fb=Density of fluid times volume of fluid times g
F=mg
3. The attempt at a solution
I first made two equations:
Density=m/v (for water) Density=m/v
6.00 kg/dm^3=m/1000 kg 6.00 kg/dm^3/2.40m/V

Idk what 2 do from there because I think I set the equations up wrong could someone please explain 2 me what I should do cuz what's really bugging me is the amount of information that is given to me?

BruceW Nov14-11 07:11 AM

Re: Pressure, Buoyant Force problem
 
The information is enough to work out the answer (as long as you also know the density of water). And you could either assume the density of air is approximately zero or you could use the actual value, to get a more accurate answer.

To start this question, you should use what they give you. They are saying that the apparent weight of the cube is 2.4 times as much in air than in water. And you know the equation for the apparent weight. So you can use this to find the actual weight of the cube in terms of the volume of the cube. And then you can use this along with the equation for the actual weight of the cube due to the masses contained, to find the fraction of the cube that is hollow.

ChunkymonkeyI Nov14-11 08:48 PM

Re: Pressure, Buoyant Force problem
 
I still kinda dont get what ur saying can u show me ur work because I dervived Fw=Density of object times g times V and I made this equation equal to Fa plus 9800 times volume of the fluid and I really need help please

BruceW Nov15-11 09:07 AM

Re: Pressure, Buoyant Force problem
 
So you wrote:
[tex]Weight_{object} = F_{apparent} + 9800 V[/tex]
Is this the apparent weight when its underwater? what does the 9800 mean?

Maybe I should go through it step-by-step, since there are a few steps which could get confusing when written all in one paragraph.

So first, you need the equation for the actual weight of the object. the object has two parts. an air cavity and a metal shell. I think you should write an equation for the total weight of the object in terms of the volumes of the metal part and cavity part and the density of air and metal.

ChunkymonkeyI Nov16-11 05:12 PM

Re: Pressure, Buoyant Force problem
 
Quote:

Quote by BruceW (Post 3617095)
So you wrote:
[tex]Weight_{object} = F_{apparent} + 9800 V[/tex]
Is this the apparent weight when its underwater? what does the 9800 mean?

Maybe I should go through it step-by-step, since there are a few steps which could get confusing when written all in one paragraph.

So first, you need the equation for the actual weight of the object. the object has two parts. an air cavity and a metal shell. I think you should write an equation for the total weight of the object in terms of the volumes of the metal part and cavity part and the density of air and metal.

Could u show each step if u dont mind because I'm getting the wrong answer for 3 times now and I think it would be helpful 2 c each steps

BruceW Nov16-11 07:30 PM

Re: Pressure, Buoyant Force problem
 
Quote:

Quote by BruceW (Post 3617095)
So first, you need the equation for the actual weight of the object. the object has two parts. an air cavity and a metal shell. I think you should write an equation for the total weight of the object in terms of the volumes of the metal part and cavity part and the density of air and metal.

This is the first step. Write out this equation. It is essentially using the principle that total mass is sum of the constituent masses. In this case, the constituent masses are that of the cavity and the metal bit.
And just use symbols for now. (I find it is easier to use symbols in calculations until you get to the end of the problem). So you could use something like [itex]V_m[/itex] for the volume taken up by the metal (for example).

ChunkymonkeyI Nov19-11 12:58 PM

Re: Pressure, Buoyant Force problem
 
Quote:

Quote by BruceW (Post 3619932)
This is the first step. Write out this equation. It is essentially using the principle that total mass is sum of the constituent masses. In this case, the constituent masses are that of the cavity and the metal bit.
And just use symbols for now. (I find it is easier to use symbols in calculations until you get to the end of the problem). So you could use something like [itex]V_m[/itex] for the volume taken up by the metal (for example).

This is what I did:
Fa=Fw-Fb
Fa=m(object)g-(density of fluid)(volume of fluid)(g)
Fa=(Density of the object)(Volume of the object)(g)-(Density of fluid)(volume of the fluid)(g)
Fa=g((density of object)(volume of the object)-(density of fluid)(volume of fluid))
m(a)=(density of object)(volume of object) minus (density of fluid)(volume of fluid)
m(a)=(6000 kg/m^3)(V of object) minus (1000 kg/m^3)(Volume of fluid)
Idk what 2 do from there but the only other thing Ik is that since its submerged the volume of the fluid is equal to the volume of the object and the apparent mass=m/2.40 but idk what 2 do from there please show me the steps and the math out :)

BruceW Nov21-11 05:42 PM

Re: Pressure, Buoyant Force problem
 
Quote:

Quote by ChunkymonkeyI (Post 3624395)
This is what I did:
Fa=Fw-Fb
Fa=m(object)g-(density of fluid)(volume of fluid)(g)
Fa=(Density of the object)(Volume of the object)(g)-(Density of fluid)(volume of the fluid)(g)
Fa=g((density of object)(volume of the object)-(density of fluid)(volume of fluid))
m(a)=(density of object)(volume of object) minus (density of fluid)(volume of fluid)
m(a)=(6000 kg/m^3)(V of object) minus (1000 kg/m^3)(Volume of fluid)

Almost right. The object has an air cavity inside it, so if we assume the air density is negligible, the mass of the object equals the density of metal times the volume which the metal actually takes up. In other words, the volume of the cavity isn't contributing to the mass of the object. If you change this, you will then have the apparent weight in water. To get the apparent weight in air, its just the same equation, but the fluid is now air.


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