Calculating Buoyancy Force & Length of Submerged Cube

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Homework Help Overview

The problem involves a wooden cube submerged in a layer of oil floating on water. The cube has a density of 960 kg/m³ and a length of 4 cm. Participants are tasked with calculating the buoyancy force acting on the cube and determining the length of the cube submerged in water, considering the densities of both the oil (930 kg/m³) and water (1000 kg/m³).

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the buoyancy force and the relationship between the submerged volume in water and oil. There are attempts to equate forces and clarify the setup of the problem, particularly regarding the distribution of the cube in the two fluids.

Discussion Status

Some participants have offered insights on the need to account for the buoyant force from both the water and the oil. There is an ongoing exploration of the equilibrium condition for the cube, with various interpretations of the submerged length being discussed.

Contextual Notes

There is confusion regarding the translation of the problem statement, specifically whether the cube is evenly submerged in both fluids or if it is only partially submerged in each. This has led to differing interpretations of the submerged length and the calculations involved.

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Homework Statement


A layer of oil that has a density of 930kg/m^3 is floating on the surface of water in a container if a wooden cube with a length of 4 cm becomes submerged where it's lower half is in water and it's upper half is in oil
the cube's density is 960kg/m^3 find
A) the buoyancy force affecting the cube
B) the length of the cube that is submerged in water

Homework Equations


density=m/v
buoyancy=density*volume*gravity
cubic volume=L^3

The Attempt at a Solution


so I know the cube's density is less than water so it floats
so I use FB=Fg. 960=m/0.04^3. m=0.06144
FB = 0.06144*9.81=0.6027N is that correct?
B)FB=Fg
1000*Vsubmegred*9.81=0.6027
V=0.000061m^3
0.04*0.04*L=V
L=0.038 now this is my problem the answer sheet says that the submerged part's length is 1.71*10^-2
I can't seem to find the significance of oil and it's density what am I doing wrong?
 
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vinamas said:

Homework Statement


A layer of oil that has a density of 930kg/m^3 is floating on the surface of water in a container if a wooden cube with a length of 4 cm becomes submerged where it's lower half is in water and it's upper half is in oil
the cube's density is 960kg/m^3 find
A) the buoyancy force affecting the cube
B) the length of the cube that is submerged in water

Homework Equations


density=m/v
buoyancy=density*volume*gravity
cubic volume=L^3

The Attempt at a Solution


so I know the cube's density is less than water so it floats
so I use FB=Fg. 960=m/0.04^3. m=0.06144
FB = 0.06144*9.81=0.6027N is that correct?
B)FB=Fg
1000*Vsubmegred*9.81=0.6027
V=0.000061m^3
0.04*0.04*L=V
L=0.038 now this is my problem the answer sheet says that the submerged part's length is 1.71*10^-2
I can't seem to find the significance of oil and it's density what am I doing wrong?
Remember, the submerged part of the cube is said to be half in the water and half in the oil.

Treat the total submergence as a variable T and work out what T must be so that the cube is floating in equilibrium with the oil and water.
 
SteamKing said:
Remember, the submerged part of the cube is said to be half in the water and half in the oil.

Treat the total submergence as a variable T and work out what T must be so that the cube is floating in equilibrium with the oil and water.
well I have seen the problem again and am sorry because I translated it wrong it doesn't say half it just says the lower surface is in water and the upper surface is in oil
 
vinamas said:
well I have seen the problem again and am sorry because I translated it wrong it doesn't say half it just says the lower surface is in water and the upper surface is in oil
OK, so T may not be evenly split between the water and the oil. You still should be able to find a relationship between the two parts given the densities of each fluid and the fact that the cube is floating in equilibrium.
 
SteamKing said:
OK, so T may not be evenly split between the water and the oil. You still should be able to find a relationship between the two parts given the densities of each fluid and the fact that the cube is floating in equilibrium.
The closest thing on my mins is
Fnet=FBwater+FBoil-Fgcube=0 but that still doesn't get me the answer
 
vinamas said:
The closest thing on my mins is
Fnet=FBwater+FBoil-Fgcube=0 but that still doesn't get me the answer
Have you adjusted your calculations from what you posted in the OP?
 
SteamKing said:
Have you adjusted your calculations from what you posted in the OP?
nope
 
vinamas said:
nope
Why not?
 
SteamKing said:
Why not?
What am I required to change?I think that FBwater is correct and the mass of the cube is correct too I can also obtain the volume of of the part of the cube in oil but I don't know what to do with it
 
  • #10
vinamas said:
What am I required to change?

Your original post does not include the buoyant force exerted by the oil. Add that according to your equation in #5.
 
  • #11
CrazyNinja said:
Your original post does not include the buoyant force exerted by the oil. Add that according to your equation in #5.
well after hours of crying and thinking why did I go into the advanced high school program I think I've found the answer and it goes like this
FBwater+FBoil-Fg=0
1000*0.04*0.04*L*9.81+930*(0.04-h)*0.04*0.04*9.81-0.602=0
L=0.0171 m.
I feel good now btw
 
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  • #12
Always feels nice to help. We can feel a little bit better @SteamKing #peaceout
 

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