Students will learn about the Buoyant force, how to calculate problems involving buoyancy and Archimedes principle.

### Key Equations

\begin{align*}
F_{\text{buoy}}= \rho_{\text{water}} g V_{\text{displaced}} && \text{Archimedes' principle}\end{align*}

*Archimedes’ Principle*states that the upward buoyant force on an object in the water is equal to the weight of the displaced volume of water. The reason for this upward force is that the bottom of the object is at lower depth, and therefore higher pressure, than the top. If an object has a higher density than the density of water, the weight of the displaced volume will be less than the object’s weight, and the object will sink. Otherwise, the object will float. The ratio of an object or substance's density to the density of water is called it's specific gravity.

#### Example 1

A simple boat (really a metal box with an open top) is floating at rest in a pond. The boat is 3 m long, 2 m wide, and has walls 1.5 m high and has an empty mass of 500 kg. If you begin filling the boat with gravel of density 1922 kg/m^{3} at a constant rate of .1 m^{3}/s, how long will it be before your boat sinks?

##### Solution

To solve this problem, we're going to start out by determining the maximum mass that the boat can support. The weight of the maximum mass is equal to the buoyant force when the boat is submerged all the way up the the edge of the side wall.

\begin{align*}
mg&=F_{buoy}\\
mg&=\rho_{water}gV\\
m&=\rho_{water}V\\
m&=1000\;\text{kg/m}^3 * 3\;\text{m} * 2\;\text{m} * 1.5\;\text{m}\\
m&=9000\;\text{kg}\\
\end{align*}

Now we need to convert the rate at which sand is being added from m^{3}/s to kg/s.

\begin{align*}
.1\;\frac{\text{m}^3}{\text{s}} * 1922\;\frac{\text{kg}}{\text{m}^3} = 192.2\;\text{kg/s}
\end{align*}

Now, we can set up something that resembles a linear equation where the 192.2 kg/s is m, t is x, the empty weight of the boat is b, and the maximum weight of the boat is y.

\begin{align*}
9000\;\text{kg}&=192.2\;\text{kg/s} * t + 500\;\text{kg}\\
t&=\frac{9000\;\text{kg} - 500\;\text{kg}}{192.2\;\text{kg/s}}\\
t&=44.2\;\text{s}\\
\end{align*}

### Watch this Explanation

### Simulation

### Time for Practice

- A block of wood with a density of \begin{align*}920 \;\mathrm{kg/m}^3\end{align*}
920kg/m3 is floating in a fluid of density \begin{align*}1100 \;\mathrm{kg/m}^3\end{align*}1100kg/m3 . What fraction of the block is submerged, and what fraction is above the surface? - The density of ice is \begin{align*}90\end{align*}
90 % that of water.- Why does this fact make icebergs so dangerous?
- A form of the liquid naphthalene has a specific gravity of \begin{align*}1.58\end{align*}
1.58 . What fraction of an ice cube would be submerged in a bath of naphthalene?

- A cube of aluminum with a specific gravity of \begin{align*}2.70\end{align*}
2.70 and side length \begin{align*}4.00 \;\mathrm{cm}\end{align*}4.00cm is put into a beaker of methanol, which has a specific gravity of \begin{align*}0.791\end{align*}0.791 .- Draw a free body diagram for the cube.
- Calculate the buoyant force acting on the cube.
- Calculate the acceleration of the cube toward the bottom when it is released.

- A cube of aluminum (specific gravity of \begin{align*}2.70\end{align*}
2.70 ) and side length \begin{align*}4.00 \;\mathrm{cm}\end{align*}4.00cm is put in a beaker of liquid naphthalene (specific gravity of \begin{align*}1.58\end{align*}1.58 ). When the cube is released, what is its acceleration? - Your class is building boats out of aluminum foil. One group fashions a boat with a square \begin{align*}10 \;\mathrm{cm}\end{align*}
10cm by \begin{align*}10\;\mathrm{cm}\end{align*}10cm bottom and sides \begin{align*}1 \;\mathrm{cm}\end{align*}1cm high. They begin to put \begin{align*}2.5 \;\mathrm{g}\end{align*}2.5g coins in the boat, adding them until it sinks. Assume they put the coins in evenly so the boat doesn’t tip. How many coins can they put in? (You may ignore the mass of the aluminum boat … assume it is zero.) - You are riding a hot air balloon. The balloon is a sphere of radius \begin{align*}3.0 \;\mathrm{m}\end{align*}
3.0m and it is filled with hot air. The density of hot air depends on its temperature: assume that the density of the hot air is \begin{align*}0.925 \;\mathrm{kg/m}^3\end{align*}0.925kg/m3 , compared to the usual \begin{align*}1.29 \;\mathrm{kg/m}^3\end{align*}1.29kg/m3 for air at room temperature. The balloon and its payload (including you) have a combined mass of \begin{align*}100 \;\mathrm{kg}\end{align*}100kg .- Draw a free body diagram for the cube.
- Is the balloon accelerating upward or downward?
- What is the magnitude of the acceleration?
- Why do hot air ballooners prefer to lift off in the morning?
- What would limit the maximum height attainable by a hot air balloon?

- You are doing an experiment in which you are slowly lowering a tall, empty cup into a beaker of water. The cup is held by a string attached to a spring scale that measures tension. You collect data on tension as a function of depth. The mass of the cup is \begin{align*}520 \;\mathrm{g}\end{align*}
520g , and it is long enough that it never fills with water during the experiment. The data**Table**(below was collected; use it to complete the following problems:- Complete the chart (
**Table**below) by calculating the buoyant force acting on the cup at each depth. - Make a graph of buoyant force vs. depth, find a best-fit line for the data points, and calculate its slope.
- What does this slope physically represent? (That is, what would a
*greater*slope mean?) - With this slope, and the value for the density of water, calculate the area of the circular cup’s bottom and its radius.
- Design an experiment using this apparatus to measure the density of an unknown fluid.

String tension \begin{align*}(N)\end{align*} (N) Depth \begin{align*}(cm)\end{align*} (cm) Buoyant force \begin{align*}(N)\end{align*} (N) \begin{align*}5.2\end{align*} 5.2 \begin{align*}0\end{align*} 0 \begin{align*}4.9\end{align*} 4.9 \begin{align*}1\end{align*} 1 \begin{align*}4.2\end{align*} \begin{align*}3\end{align*} \begin{align*}3.7\end{align*} \begin{align*}5\end{align*} \begin{align*}2.9\end{align*} \begin{align*}8\end{align*} \begin{align*}2.3\end{align*} \begin{align*}10\end{align*} \begin{align*}1.7\end{align*} \begin{align*}12\end{align*} \begin{align*}0.7\end{align*} \begin{align*}15\end{align*} \begin{align*}0.3\end{align*} \begin{align*}16\end{align*} \begin{align*}0\end{align*} \begin{align*}17\end{align*} - Complete the chart (
- A submarine is moving directly upwards in the water at constant speed. The weight of the submarine is \begin{align*}500,000 \;\mathrm{N}\end{align*}. The submarine’s motors are off.
- Draw a sketch of the situation and a free body diagram for the submarine.
- What is the magnitude of the buoyant force acting on the submarine?

- A glass of water with weight \begin{align*}10 \;\mathrm{N}\end{align*} is sitting on a scale, which reads \begin{align*}10 \;\mathrm{N}\end{align*}. An antique coin with weight \begin{align*}1 \;\mathrm{N}\end{align*} is placed in the water. At first, the coin accelerates as it falls with an acceleration of \begin{align*}g/2\end{align*}. About half-way down the glass, the coin reaches terminal velocity and continues at constant speed. Eventually, the coin rests on the bottom of the glass. What was the scale reading when:
- The coin had not yet been released into the water?
- The coin was first accelerating?
- The coin reached terminal velocity?
- The coin came to rest on the bottom?

#### Answers to Selected Problems

- \begin{align*}0.84\end{align*}
- a. \begin{align*}90\end{align*}% of the berg is underwater b. \begin{align*}57\end{align*}%
- b. \begin{align*}5.06 \times 10^{-4} \;\mathrm{N}\end{align*} c. \begin{align*}7.05 \;\mathrm{m/s}^2\end{align*}
- \begin{align*}4.14 \;\mathrm{m/s}\end{align*}
- 6. \begin{align*}40\end{align*} coins
- b. upward c. \begin{align*}4.5 \;\mathrm{m/s}^2\end{align*} d. Cooler air outside, so more initial buoyant force e. Thin air at high altitudes weighs almost nothing, so little weight displaced.
- a. At a depth of \begin{align*}10 \;\mathrm{cm}\end{align*}, the buoyant force is \begin{align*}2.9 \;\mathrm{N}\end{align*} d. The bottom of the cup is \begin{align*}3 \;\mathrm{cm}\end{align*} in radius
- b. \begin{align*}500,000 \;\mathrm{N}\end{align*}
- a. \begin{align*}10 \;\mathrm{N}\end{align*} b. \begin{align*}10.5 \;\mathrm{N}\end{align*} c. \begin{align*}11 \;\mathrm{N}\end{align*} d. \begin{align*}11 \;\mathrm{N}\end{align*}