Answer to Question #256555 in Classical Mechanics for Lemoj

Question

A stone is found to weigh 6.2 in air, 4.5 in water, and 5.3 N in an unknown liquid. Find a.) the density of the stone and the unknown liquid, b.) the buoyant force of the liquid on the stone, c.) the buoyant force of water on the stone, and d) the volume of the stone.

Answer 1

$P_{air} = 6.2N$

$P_{water}= 4.5N$

$P_{unliquid}= 5.3N$

$g= 9.8\frac{m}{s^2}$

$\text{in water:}$

$\rho_{water} = 1000\frac{kg}{m^3}$

$BF_{water} = P_{air}-P{water}= 6.2-4.5= 1.7N$

$BF_{water} = \rho_{water}gV_{stone}$

$V_{stone} = \frac{BF_{water}}{ \rho_{water}g}=\frac{1.7}{1000*9.8}\approx 1.73*10^{-4}m^3$

$\text{in air : }$

$P_{air} = m_{stone}g$

$\rho_{stone}= \frac{m_{stone}}{V_{stone}}=\frac{P_{air}}{gV_{stone}}$

$\rho_{stone}= \frac{6.2}{9.8*1.73*10^{-4}}=3.66*10^{3}\frac{kg }{m^3}$

$\text{in an unknown liquid:}$

$BF_{unliquid}= P_{air} -P_{unliquid}= 6.2-5.3 = 0.9N$

$BF_{unliquid} = \rho_{unliquid}gV_{stone}$

$\rho_{unliquid} =\frac{ BF_{unliquid}} {gV_{stone}}=\frac{0.9}{9.8*1.73*10^{-4}}=0.53*10^3\frac{kg}{m^3}$

$\text{Answer:}$

$a)\rho_{stone}=3.66*10^{3}\frac{kg }{m^3}$

$\rho_{unliquid} =0.53*10^3\frac{kg}{m^3}$

$b)BF_{unliquid}= 0.9N$

$c)BF_{water} = 1.7N$

$d)V_{stone} = 1.73*10^{-4}m^3$