# Buoyancy

## Practice

### practice problem 1

- Which is
*more massive*on the surface of the*Earth*? - Which is
*more massive*on the surface of the*Moon*? - Which will have the greater "weight" when placed
*on a spring scale*on the surface of the*Earth*? - Which will have the greater "weight" when placed
*on a spring scale*on the surface of the*Moon*?

- The phrase "more massive" should be read literally as "has more mass" not "fills more space".
- All "weights" measured on the surface of the Moon are not made inside the protective environment of a human habitation.

#### solution

This is a question about how the presence of an atmosphere affects the precise measurement of weight.

- Don't think too hard about this one. Objects with the same mass just have the same mass. A one kilogram block of aluminum has the same mass as a one kilogram block of lead. Mass is not affected by what material something is made from. Mass is an invariant property.
- Objects with the same mass always have the same mass even when they are somewhere where gravity is different. A one kilogram block of aluminum has the same mass as a one kilogram block of lead even if both are on the moon. Gravity does not affect mass. Nothing does. Mass is an invariant property.
- Now we have a genuine problem that requires thought. A one kilogram block of aluminum is larger than a one kilogram block of lead because aluminum is less dense than lead. Consequently, it displaces more of the Earth's sweet atmosphere. Thus there is a greater, upward buoyant force on the aluminum than there is on the lead. When placed on a sensitve enough scale, a one kilogram block of aluminum would weigh less on a scale on the Earth than a one kilogram block of lead even though the effect of gravity on both would be identical.
- The moon barely has an atmosphere — just a few stray particles carried in the solar wind, background cosmic radiation, and junk baked out of the regolith (the lunar equivalent of soil, a.k.a. moondust). With no atmosphere worth talking about, there is no upward buoyant force. All objects with the same mass have the same weight on a scale. Precision metrologists rejoice. A one kilogram block of aluminum would weigh the same on a scale on the moon as a one kilogram block of lead.

The past version of this question is about how people sometimes confuse the meanings of words.

### practice problem 2

*above*water?

#### solution

An object floats on the surface of a liquid when the downward force of gravity of the object is balanced by the upward force of buoyancy.

*W* = *B*

The weight of an object is its mass times gravity, and mass is density times volume.

*W* = *m _{object}g* = ρ

_{object}gV_{object}Buoyancy is the weight of the fluid being displaced. Weight is still mass times gravity and mass is still density times volume.

*B* = *m _{fluid}g* = ρ

_{fluid}gV_{fluid}Set the two expressions equal to one another.

ρ* _{object}gV_{object}* = ρ

_{fluid}gV_{fluid}Rearrange into a fraction like this.

V_{fluid} | = | ρ_{object} |

V_{object} |
ρ_{fluid} |

This is the fraction submerged — the fraction *in* the water. This book comes with a table of densities. Use numbers for liquid seawater and freshwater ice.

V_{fluid} |
= | 916 kg/m^{3} |
= 89.4% |

V_{object} |
1025 kg/m^{3} |

This does not answer the question asked, however. We need the fraction *not in* the water. Just subtract the fraction above from one.

10.6% ≈ ^{1}_{9}

Only a fraction of an iceberg is visible above the surface of the water. This is the origin of the expression "tip of the iceberg" in reference to situations that are much more complex or extensive than they appear at first glance.

### practice problem 3

#### solution

This book has a table with the densities of helium (ρ_{He} = 0.164 kg/m^{3}) and air (ρ_{air} = 1.161 kg/m^{3}) at a temperature of 300 K (27 °C, 80 °F). Some may think 300 K a bit high as a "normal" temperature, but it's more important that the temperatures match than that they are some metaphysical ideal value. This essentially is an order of magnitude calculation anyway, so anything within an order of magnitude of "normal" is close enough. (For this problem, that would be anything from 100 K to 1000 K.)

Weight (*W*) is mass times gravity (*mg*). Mass is density times volume (ρ*V*). Thus, one cubic meter of helium weighs…

W = mgW = ρ_{He}gVW = (0.164 kg/m^{3})(9.8 m/s^{2})(1 m^{3})W = 1.607 N |

Buoyancy (*B*) is the weight of the fluid displaced — in this case, one cubic meter of air. One cubic meter of air weighs…

B = mgB = ρ_{air}gVB = (1.161 kg/m^{3})(9.8 m/s^{2})(1 m^{3})B = 11.378 N |

The difference in these two forces (buoyancy up minus weight down) is the lifting force described in this problem. Let's use the symbol *F* for that, because why not.

F = B − WF = 11.378 N − 1.607 NF = 9.771 N |

Is this the weight of one kilogram? If weight is mass times gravity then mass is weight divided by gravity.

m = W/gm = (9.771 N)/(9.8 m/s^{2})m = 0.997 kg |

Looks like one kilogram to me.

Want to try a fancier proof? The lifting force provided by a lighter than air gas is the upward buoyant force provided by the air minus the downward weight of the lifting gas.

*F* = *B* − *W*

The overall force at the start is the weight of some load carried by the balloon, the buoyant force is the weight of the air displaced, and the weight at the end refers to the weight of the helium.

*mg* = ρ* _{air}gV* − ρ

_{He}gVNote how gravity cancels out of each term and we're left with mass — the thing were were trying to find. Factor out volume to make it look nicer.

*m* = (ρ* _{air}* − ρ

*)*

_{He}*V*

Numbers in…

*m* = (1.161 kg/m^{3} − 0.164 kg/m^{3})(1 m^{3})

Answer out…

*m* = 0.997 kg

Looks like one kilogram again.

### practice problem 4

#### solution

Answer it.