Aluminum cylinder in alcohol

[mikefitz]

An aluminum cylinder weighs 1.01 N. When this same cylinder is completely submerged in alcohol, the volume of the displaced alcohol is 3.88 × 10-5 m3. The apparent weight of the cylinder when completely submerged is 0.778 N. What is the specific gravity of the alcohol?

m=W/g
m(cylinder)=.10296 kg
p=m/V
.10296kg/3.88*10^5 => 2653.608kg/m^3 (avg density)
SG= 2653.608/1000=> 2.65

What did I do wrong? Thanks

 

[Doc Al]

You found the specific gravity of the cylinder, not the alcohol. Hint: Buoyant force is involved.

 

[mikefitz]

Doc Al,

So I can find the mass of the alcohol, then the avg density, divide by 1000 and get the alcohol's SG?

Also, how do I relate Buoyant force to this problem? Thanks again

 

[Office_Shredder]

The force applied upwards by the alcohol is equal to the weight of the displaced alcohol

 

[Doc Al]

So I can find the mass of the alcohol, then the avg density, divide by 1000 and get the alcohol's SG?

That'll work.

Also, how do I relate Buoyant force to this problem?

The buoyant force equals the weight of the displaced fluid (alcohol). You can find the buoyant force by comparing the actual weight of the cylinder to its apparent weight when submerged in the alcohol.