Chapter 13.3, Problem 31PP

To determine

To Calculate: The approximate minimum volume of foam needed for flotation of a $\text{480-N}$ canoe.

Answer to Problem 31PP

  $\text{4}{\text{.9×10}}^{\text{-2}}{\text{m}}^{\text{3}}$

Explanation of Solution

Given data:

Weight of canoe, $F_{buoyant}\text{=}\text{480}\text{N}$

Gravitational acceleration, $g=\text{9}\text{.8}{\text{m/s}}^{\text{2}}$

Density of water, ${\rho }_{water}\text{=1}{\text{.00×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}$

Formula used:

The increase of pressure with depth has an important consequence. It allows an individual to swim. That is, it creates an upward force, described as a buoyant force, on all objects. Every object in a liquid, even those that sink, has an apparent weight. This weight is less than when the object is in air.

One force is the upward force of the liquid on the object. There is also the downward force of the object’s weight. The difference between the buoyant force (or upward force) and the object’s weight determines whether an object floats or sinks.

The upward force or buoyant force can be calculated by using formula, $F_{buoyant}\text{ =}\rho Vg$ .

Where, V is the volume, $\rho$ is the density of fluid and g is the gravitational acceleration.

Calculation:

Now, by using formula, $F_{buoyant}\text{ =}\rho Vg$

  $V\text{=}\frac{F_{buoyant}}{{\rho }_{water}g}$

Substitute all the values in above formula,

  $\begin{array}{c}V\text{=}\frac{\text{480}\text{N}}{\left(\text{1}{\text{.00×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}\right)\left(\text{9}\text{.80}{\text{m/s}}^{\text{2}}\right)}\\ \text{= 4}{\text{.9×10}}^{\text{-2}}{\text{m}}^{\text{3}}\end{array}$

Conclusion:

The approximate minimum volume is $\text{4}{\text{.9×10}}^{\text{-2}}{\text{m}}^{\text{3}}$ .