Buoancy (block floating in water)

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SUMMARY

The discussion centers on calculating the fraction of a wooden block submerged in water after oil is poured over it. Initially, 90% of the block's volume is submerged in water. The buoyancy force equations used are ρ_w g(0.9)V = ρ_w gxV + ρ_o g(1-x)V, where ρ_w is the density of water and ρ_o is the density of oil (875 kg/m³). The confusion arises when the user incorrectly factors the initial submerged volume, leading to the conclusion that the submerged fraction remains 0.9, which is incorrect. The correct approach requires careful handling of the equations without premature factoring.

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etagg
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Im confused by this question:

A block of wood floats above water with 90% of its volume submerged. Oil with a density of 875kg/m^3 is then poured over the block so that it covers the entire block. Find the fraction of the block now submerged in water.

I know that the fraction of the block submerged will decrease.
I started by recognizing that the initial force of buoyancy is equal to the final force of buoyancy

ρ_w g(0.9)V=ρ_w gxV+ρ_o g(1-x)V

Where the V and g cancel out, and the x is equal to the portion of the volume submerged in the water and the oil.
However, when i use this method, x equals 0.9, meaning that the same amount of the block is submerged in the water, when it should be less of the block is submerged in the water.
Please help!
 
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etagg said:
I started by recognizing that the initial force of buoyancy is equal to the final force of buoyancy

ρ_w g(0.9)V=ρ_w gxV+ρ_o g(1-x)V
OK.

Where the V and g cancel out, and the x is equal to the portion of the volume submerged in the water and the oil.
x is the fraction of the block submerged in water.
However, when i use this method, x equals 0.9, meaning that the same amount of the block is submerged in the water, when it should be less of the block is submerged in the water.
Please show how you solved for x.
 
ohhhh, what a stupid mistake. I mistakenly factored 0.9 when i should not have. Goes to show what happens when you look at a problem for too long!
thanks for your help.
 

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