Near the surface of a planet like Earth but with no atmosphere a small solid
sphere of mass m and density ρ = ½ρ_{water} is held near the center of a
large box filled with water on top of a high tower. The box and the sphere are
released simultaneously.

(a) What are the force of gravity **F**g acting on the sphere, the net force
**F**_{net }acting on the sphere, and the acceleration **a** of
the sphere?

(b) If only the sphere is released, but the box is held fixed, what, initially,
are the force of gravity **F**_{g} acting on the sphere, the net force **F**_{net
}acting on the sphere, and the acceleration **a** of the sphere?

(c) In case (b), if we set **F**_{g} = m***a**, what is the
effective mass m* of the sphere?

Solution:

- Concepts:

Newton's 2nd law, gravity. the buoyant force - Reasoning:

Only gravity is acting on the freely falling sphere, while gravity and the buoyant force act on the sphere in the fixed pool. - Details of the calculation:

(a)**F**_{g}=**F**_{net}= m**g**, where**g**points towards the surface,**a**=**g**= 9.8 m/s^{2}downward.

(b)**F**_{g}= m**g**,**F**_{net}= m**g**+**B**= m**g**– 2m**g**= - m**g**. Here**B**is the buoyant force,**B**= -2m**g**,

**B**points upward.**a**=**F**_{net}/m = -**g**, upward.

(c)**F**_{g}= m**g**= m***a**= -m***g**. m* = -m, the effective mass is negative.

A Styrofoam slab has a thickness h and a density ρ_{object}.
What is the area of the slab, if it floats with its top surface just awash in
fresh water when a swimmer of mass m is on top?

Solution:

- Concepts:

The buoyant force - Reasoning:

When an object floats, the magnitude of the buoyant force is equal to the magnitude of its weight. - Details of the calculation:

The magnitude of the buoyant force is equal to the magnitude of the weight w_{water}of the displaced water. w_{water }= ρ_{water}Ahg, where A is the area of the slab.

The the magnitude of the weight of the object is w_{object }= ρ_{object}Ahg + mg.

We need w_{water }= w_{object}.

ρ_{water}Ahg = ρ_{object}Ahg + mg.

ρ_{water}Ah - ρ_{object}Ah = m.

A = m/(ρ_{water}h - ρ_{object}h)

**Problem: **

A frog in a hemispherical pod finds that he just floats without sinking into
a sea of blue-green ooze with density 1.35 g/cm^{3}. If the pod
has radius 6 cm and negligible mass, what is the mass of the frog?

Solution:

- Concepts:

The buoyant force - Reasoning:

When an object floats, the magnitude of the buoyant force is equal to the magnitude of its weight. - Details of the calculation:

The magnitude of the buoyant force is equal to the the magnitude of the weight w_{liquid}of the displaced liquid.

w_{liquid }= ρ_{liquid}Vg.

The volume V of the displaced liquid is the volume of one half sphere,

V = 2πr^{3}/3 = 2π(6cm)^{3}/3 = 452cm^{3}.

The magnitude of the weight of the object is w_{object }= m_{frog}g. (We are neglecting the weight of the air-filled pod.)

ρ_{liquid}Vg = m_{frog}g.

m_{frog }= 1.35(g/cm^{3})452cm^{3 }= 610g.

A beaker of mass 1 kg containing 2 kg of water rests on a scale. A 2 kg
block of aluminum (specific gravity of 2.7) is suspended from a spring scale and
is submerged in the water.

(a) What does the upper spring scale read
for the weight of the aluminum block?

(b) What does the lower scale
read for the weight of the whole system?

Solution:

- Concepts:

Buoyancy - Reasoning:

An object partially or wholly immersed in a gas or liquid is acted upon by an upward buoyant force B equal to the weight w of the gas or liquid it displaces. B = w. - Details of the calculation:

(a) The scale reads the weight of the block minus the buoyant force acting on the block.

F_{upper scale}= (m_{AL}- ρ_{W}V_{AL})g = m_{AL}(1 - ρ_{W}/ρ_{AL})g.

ρ_{AL}= 2.7 ρ_{W}.

F_{upper scale}= (2 kg*9.8 m/s^{2})(1 – 1/2.7) = 12.34 N.

(b) Newton’s third law

F_{lower scale}= (m_{W }+ m_{beak}+ ρ_{W}V_{AL})g = (m_{W }+ m_{beak}+ m_{AL}ρ_{W}/ρ_{AL})g

= (3 kg + 2 kg/2.7)* 9.8 N = 36.66 N.

A polar bear partially
supports herself by pulling part of her body out of the water onto a rectangular
slab of ice of volume 10 m^{3}. The ice sinks down so that only
half of what was once exposed now is exposed, and the bear has 60% of her volume
(and weight) out of the water. Assume that the bear and the water have
mass density 1 g/cm^{3} and ice has mass density 0.9 g/cm^{3}.
Estimate the bear's mass.

Solution:

- Concepts:

The buoyant force - Reasoning:

An object partially or wholly immersed in a gas or liquid is acted upon by an upward buoyant force B equal to the weight w of the gas or liquid it displaces. - Details of the calculation:

B = W

Without the bear:

Let f denote the fraction of the ice that is exposed.

B = (1 - f)(10 m^{3})*(1000 kg/m^{3})*g = (1 - f)(10000 kg)*g.

W = (10 m^{3})*(900 kg/m^{3})*g = (9000 kg)*g.

(1 - f)10000 = 9000, f = 1- 9/10 = 0.1.

Without the bear the exposed fraction is 0.1.With the bear:

B = (1 - f/2)(10 m^{3})*(1000 kg/m^{3})*g = (9500 kg)*g.

W = (9000 kg)*g + 0.6*m_{bear}*g.

0.6*m_{bear}= 500 kg, m_{bear}= 833 kg.

A thin vertical uniform wooden rod is pivoted at the top and immersed in water as shown.

The pivot point is slowly lowered. At a certain moment the rod begins to deflect from the vertical. What fraction of the rod is still in the water at that moment if the density of the rod is one-half of the density of water?

Solution:

- Concepts:

The buoyant force, stable versus unstable equilibrium - Reasoning:

If the rod is displaced by a small angle θ gravity and the buoyant force will exert a torque about the pivot point. The direction of the torque determines if we have a restoring force and a stable equilibrium. - Details of the calculation:

Let the rod have length L and cross sectional area A (√A << L), and let aL be the length of the section of the rod in the water. Assume the rod is displaced by a small angle θ. In the small angle approximation, the torque about the pivot point is

τ = mgLθ/2 – B(1 - a/2)Lθ = ½ρ_{water}gAL^{2}θ/2 - ρ_{water}gaAL^{2}(1 - a/2)θ, counterclockwise.

As long as τ is positive we have a restoring force and a stable equilibrium. When τ is negative we have an unstable equilibrium and the rod will deflect from the vertical.

At the moment the rod begins to deflect from the vertical τ = 0.

½ρ_{water}gAL^{2}θ/2 - ρ_{water}gaAL^{2}(1 - a/2)θ = 0, ½ - a(2 - a) = 0, a^{2}- 2a + ½ = 0,

a = (1 - √½).

When the fraction of the rod under water is (1 - √½) ≈ 0.3, the rod begins to deflect from the vertical.