A block of wood floats on water. A layer of oil is now poured on top of the wate

Question

A block of wood floats on water. A layer of oil is now poured on top of the water to a depth that more than covers the block, as shown in Figure 15-30. Figure 15-30

(a) Is the volume of wood submerged in water greater than, less than, or the same as before? _____________

(b) If 87% of the wood is submerged in water before the oil is added, find the fraction submerged when oil with a density of 849 kg/m3 covers the block. (Do not neglect the buoyant force of air before the oil is added.) ___________ %

Explanation / Answer

(a) Is the volume of wood submerged in water greater than, less than, or the same as before?

Volume of wood in water is LESS THAN before, since now thereis a buoyant force from the oil.

(b) If 87% of the wood is submerged in water before the oil is added, find the fraction submerged when oil with a density of 849 kg/m3 covers the block. (Do not neglect the buoyant force of air before the oil is added.) ___________ %

The density of water

dw = 1000 kg/m^3   
The density of air

da = 1.225 kg/m^3   

Then before the oil is poured,

weight of the block = buoyant force due to water + buoyant force due to air

m*g=0.87*V*dw*g+(1-0.87)*V*da*g equation 1

Suppose after the oil is added the new fraction of the block submerged in water is given by P. Then again,

m*g=PV*dw*g+(1-P)*V*doli*g equation 2

equating equations

P*V*dw*g+(1-P)V*doli*g=0.87*V*dw*g+(1-0.87)*V*da*g

simplifying g and V

P*dw+(1-P)*doli=0.87*dw+(0.13)*da

Expression for P

P*dw+doli-P*doli=0.87*dw+(0.13)*da

P*dw-P*doli=0.87*dw+(0.13)*da-doli

P*(dw-doli)=0.87*dw+(0.13)*da-doli

finally

P=(0.87*dw+0.13*da-doli )/(dw-doli)=(0.87*1000 kg/m^3   +0.13*1.225 kg/m^3-849 kg/m3 )/(1000 kg/m3-849 kg/m3)

P=0.1401*100%=14.01%

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