Question 9.10: Floating in Two Fluids GOAL Apply Archimedes’ principle to a...
Floating in Two Fluids
GOAL Apply Archimedes’ principle to an object floating in a fluid having two layers with different densities.
PROBLEM A 1.00 × 10³-kg cube of aluminum is placed in a tank. Water is then added to the tank until half the cube is immersed. (a) What is the normal force on the cube? (See Fig. 9.26a.) (b) Mercury is now slowly poured into the tank until the normal force on the cube goes to zero. (See Fig. 9.26b.) How deep is the layer of mercury?
STRATEGY Both parts of this problem involve applications of Newton’s second law for a body in equilibrium, together with the concept of a buoyant force. In part (a) the normal, gravitational, and buoyant force of water act on the cube. In part (b) there is an additional buoyant force of mercury, while the normal force goes to zero. Using V_{ Hg }=A h, solve for the height of mercury, h.
(a) Find the normal force on the cube when half-immersed in water.
Calculate the volume V of the cube and the length d of one side, for future reference (both quantities will be needed for what follows):
V_{ Al }=\frac{M_{ Al }}{\rho_{ Al }}=\frac{1.00 \times 10^3 kg }{2.70 \times 10^3 kg / m ^3}=0.370 m ^3
d=V_{ Al }^{1 / 3}=0.718 m
Write Newton’s second law for the cube, and solve for the normal force. The buoyant force is equal to the weight of the displaced water (half the volume of the cube).
n-M_{ A 1} g+B_{\text {water }}=0
n=M_{ A 1} g-B_{\text {water }}=M_{ A I} g-\rho_{\text {water }}(V / 2) g
=\left(1.00 \times 10^3 kg \right)\left(9.80 m / s ^2\right)
-\left(1.00 \times 10^3 kg / m ^3\right)\left(0.370 m ^3 / 2.00\right)\left(9.80 m / s ^2\right)
n=9.80 \times 10^3 N -1.81 \times 10^3 N =7.99 \times 10^3 N
(b) Calculate the level h of added mercury.
Apply Newton’s second law to the cube:
n-M_{ A 1} g+B_{\text {water }}+B_{ Hg }=0
Set n = 0 and solve for the buoyant force of mercury:
B_{ Hg }=\left(\rho_{ Hg } A h\right) g=M_{ A 1} g-B_{\text {water }}=7.99 \times 10^3 N
Solve for h, noting that A = d²:
h=\frac{M_{ Al } g – B_{\text {water }}}{\rho_{ Hg _g} A g}=\frac{7.99 \times 10^3 N }{\left(13.6 \times 10^3 kg / m ^3\right)(0.718 m )^2\left(9.80 m / s ^2\right)}
h = 0.116 m
REMARKS Notice that the buoyant force of mercury calculated in part (b) is the same as the normal force in part (a). This is naturally the case, because enough mercury was added to exactly cancel out the normal force. We could have used this fact to take a shortcut, simply writing B_{ Hg }=7.99 \times 10^3 N immediately, solving for h, and avoiding a second use of Newton’s law. Most of the time, however, we wouldn’t be so lucky! Try calculating the normal force when the level of mercury is 4.00 cm.