Question

A hollow plastic sphere is held below the surface of a fresh-water lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.650 cubic meters and the tension in the cord is 900 N. a) Calculate the buoyant force exerted by the water on the sphere. b) what is the mass of the sphere? c) the cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

Answer 1

Answer: a) B = 6811N

              b) m = 603.2kg

              c) 86.8%

Explanation: Buoyant force is a force a fluid exerts on a submerged object.

It can be calculated as:

$B=\rho_{fluid}.V_{obj}.g$

where:

$\rho_{fluid}$ is density of the fluid the object is in;

$V_{obj}$ is volume of the object;

g is acceleration due to gravity, is constant and equals 9.8m/s²

a) For the hollow plastic sphere, density of water is 1000kg/m³:

$B=10^{3}.0.695.9.8$

B = 6811N

b) Anchored to the bottom, the forces acting on the sphere are Buoyant, Tension and Force due to gravity:

B = T + $F_{g}$

B = T + mg

mg = B - T

$m=\frac{B - T}{g}$

Calculating:

$m=\frac{6811 - 900}{9.8}$

m = 603.2kg

c) When the shpere comes to rest on the surface of the water, there are only buoyant and gravity acting on it:

B = m.g

$\rho_{w}.V_{sub}.g=m.g$

$V_{sub}=\frac{m}{\rho_{w}}$

$V_{sub}=\frac{603.2}{1000}$

$V_{sub}$ = 0.6032m³

Fraction of the submerged volume is:

$\frac{V_{sub}}{V_{obj}}$ = $\frac{0.6932}{0.695}$ = 0.868