Calculating Buoyancy Force & Length of Submerged Cube

In summary: I go into the advanced high school program I think I've found the answer and it goes like this FBwater+FBoil-Fg=01000*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 btwAlways feels nice to help. We can feel a little bit better @SteamKing #peaceout
  • #1
vinamas
43
1

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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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
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
 
  • #6
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?
 
  • #7
SteamKing said:
Have you adjusted your calculations from what you posted in the OP?
nope
 
  • #8
vinamas said:
nope
Why not?
 
  • #9
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
 
  • Like
Likes CrazyNinja
  • #12
Always feels nice to help. We can feel a little bit better @SteamKing #peaceout
 

1. What is buoyancy force?

Buoyancy force is the upward force exerted on an object by a fluid (such as water) that is equal to the weight of the fluid displaced by the object.

2. How is buoyancy force calculated?

Buoyancy force can be calculated using the formula Fb = ρVg, where Fb is the buoyancy force, ρ is the density of the fluid, V is the volume of the object submerged, and g is the acceleration due to gravity.

3. How does the shape and size of an object affect its buoyancy force?

The shape and size of an object do not affect its buoyancy force, as long as the object's volume and density remain the same. However, the shape and size can affect the object's stability and ability to float.

4. How can the length of a submerged cube be calculated using buoyancy force?

The length of a submerged cube can be calculated using the formula L = (Fb/ρg)^1/2, where L is the length of the cube, Fb is the buoyancy force, ρ is the density of the fluid, and g is the acceleration due to gravity.

5. Can buoyancy force be greater than the weight of an object?

Yes, if the object is less dense than the fluid it is submerged in, the buoyancy force can be greater than the weight of the object, causing it to float.

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