Buoyancy and Fluid Statics

Question

Hello! I need help with this question please. I solved it and found that Fb in water is 4.704 Newtons. Doing the same to find teh apparent weight in alcohol, I found that that equaled 0.98784 Newtons. However, the answer key that my professor made says it should be around 0.89. What could I be doing wrong? I found the Volume of the block to be 4.8*10^-4 while the key says it should be 5.5*10^-4.

Answer 1

Hi Yogita S.,

Yes, there are some problems in your calculations. You should know that by eyeballing your values. Your value for Fb in water looks good : 0.48 kg x 9.81 m s^-2 . The apparent weight in alcohol should be 0.047 kg x 9.81 m s^-2, nowhere near your value. That 0.89 refers to the density of the wood NOT apparent weight (since the alcohol neutralizes all but 0.047 kg of the 0.48 kg initial mass, the density of the wood is correspondingly slightly proportionately greater than that of alcohol). Do you see how that works? You are substituting in the density of alcohol, and proportionating it up to the total mass-equivalent of the wood. Proportions still work, even though the quality used (density) is a stand-in for another quality (mass). So the calculation would be: 0.79 * (0.48)/(0.48-0.047) = (conceptually:) the part * (the whole/the part) .

So for volume you must divide mass by density. Your numbers will match then.

Cheers, -- Mr. d.