Question

With how much force must you push down on a solid ball of aluminum of radius 9 cm in order for it to be halfway submerged in mercury? (The densities of aluminum and mercury are 2700 kg/m3 and 13,690 kg/m3 respectively.)

Answer 1

Answer:

$P = 329.117\,N$

Explanation:

According to the Archimedes Principle, the buoyancy force exerted on a body is equal to the displaced weight of fluid, an external force is needed to guarantee the equilibrium of the solid ball, since denstity of aluminium is lower than denstity of mercury. The equation of equilibrium is:

$\Sigma F = -P - \rho_{Al}\cdot V_{ball}\cdot g + \rho_{Hg}\cdot V_{ball}\cdot g = 0$

$-P+(\rho_{Hg}-\rho_{Al})\cdot g\cdot V_{ball} = 0$

$P=(\rho_{Hg}-\rho_{Al})\cdot g\cdot V_{ball}$

$P = \left(13,690\,\frac{kg}{m^{3}} - 2,700\,\frac{kg}{m^{3}} \right)\cdot \left[\frac{4}{3}\pi\cdot (0.09\,m)^{3} \right] \cdot (9.807\,\frac{m}{s^{2}} )$

$P = 329.117\,N$

Answer 2

Answer:

124.2 N

Explanation:

9 cm = 0.09 m

When the aluminum is half weight submerged in the mercury, the force you must push down must be equal to the buoyant force, which is the difference between the weight of the submerged portion and the mercury weight displaced by the ball:

Let g = 9.81 m/s2

$F = P_m - P_a$

$F = m_mg - m_ag$

$F = V_s\rho_mg - 2V_s\rho_ag$

$F = gV_s(\rho_m - 2\rho_a)$

where $V_s = 2\pi r^3/3 = 2\pi*0.09^3/3 = 0.00153 m^3$ is the volume submerged, which is half of the total spherical  volume of the aluminum ball. $\rho_m = 13690 kg/m^3, \rho_a = 2700 kg/m^3$ are the density of mercury and aluminum, respectively.

$F = 9.81*0.00153*(13690 - 2*2700) = 124.2 N$