Experiment suggestion: Measure oneself outside and inside a pool using a waterproof scale to calculate personal volume.
Volume calculation: Estimating surface increase to measure displaced water volume.
Volume and Surface Area
Conversion to familiar units:
Volume of the object = approximately 0.02 square feet or 34 square inches.
Suggests a size of approximately 3-inch cube.
Example Problem: Balsa Wood
Density of balsa wood: 130 kg/m³
Problem: Determine percentage of balsa wood cube submerged in water.
Solution Steps
Force Equilibrium:
Buoyant force = weight of balsa wood
Formula: Volume of wood x Density of wood x Gravity = Volume of submerged wood x Density of water x Gravity
Simplification: Cross out gravity from both sides.
Algebraic Manipulation:
(\frac{\text{Volume submerged}}{\text{Volume of balsa wood}} = \frac{\text{Density of balsa wood}}{\text{Density of water}})
Calculation: (\frac{130}{1000} = 0.13)
Percentage Submerged:
Result: 13% of balsa wood block will be submerged in water.
Demonstrates applicability to different shapes (e.g., shaped like a horse).
Conclusion
Understanding buoyant forces and Archimedes' principle helps in determining the volume of submerged objects and their interaction with water. Further problems and examples can deepen comprehension.
Final Note: These principles apply to objects of any shape.