Problem 43
Question
A block of cherry wood that is \(20.0 \mathrm{~cm}\) long, \(10.0 \mathrm{~cm}\) wide, and \(2.00 \mathrm{~cm}\) thick has a density of \(800 . \mathrm{kg} / \mathrm{m}^{3}\). What is the volume of a piece of iron that, if glued to the bottom of the block makes the block float in water with its top just at the surface of the water? The density of iron is \(7860 \mathrm{~kg} / \mathrm{m}^{3},\) and the density of water is \(1000 . \mathrm{kg} / \mathrm{m}^{3}\).
Step-by-Step Solution
Verified Answer
Answer: The volume of the piece of iron needed is approximately \(1.01 * 10^{-5} \mathrm{~m}^3\).
1Step 1: Calculate the volume and mass of the cherry wood block
To calculate the volume (V) of the cherry wood block, use the formula: V = l × w × h:
V_wood = 20 cm * 10 cm * 2 cm = 400 cm^3
Convert the volume to m^3:
V_wood = 400 cm^3 * (1 m^3 / 1000000 cm^3) = 4.00 * 10^{-4} m^3
Now, calculate the mass of the cherry wood block using its density (ρ_wood) and the volume (V_wood).
mass_wood = ρ_wood * V_wood = 800 kg/m^3 * 4.00 * 10^{-4} m^3 = 0.320 kg
2Step 2: Determine the buoyant force acting on the block
According to Archimedes' principle, the buoyant force (F) is equal to the weight of the water displaced by the block. The weight of the water is equal to the mass of water (m_water) multiplied by gravity (g). And the mass of water is equal to its density (ρ_water) and the volume of the cherry wood block (V_wood).
F = m_water * g = ρ_water * V_wood * g
F = 1000 kg/m^3 * 4.00 * 10^{-4} m^3 * 9.81 m/s^2 ≈ 3.924 N
3Step 3: Calculate the mass of the iron required
The block will float when the buoyant force F is equal to the sum of the weight of the cherry wood block and the iron piece. Hence,
mass_iron * g + mass_wood * g = F
Solving for mass_iron:
mass_iron = (F / g) - mass_wood
mass_iron = (3.924 N / 9.81 m/s^2) - 0.320 kg ≈ 0.0796 kg
4Step 4: Find the volume of iron
To find the volume of iron (V_iron), use the formula:
V_iron = mass_iron / ρ_iron
V_iron = 0.0796 kg / 7860 kg/m^3 ≈ 1.01 * 10^{-5} m^3
So, the volume of the piece of iron needed to make the block float with its top just at the surface of the water is approximately \(1.01 * 10^{-5} \mathrm{~m}^3\).
Key Concepts
Buoyant ForceDensityVolume CalculationMass Calculation
Buoyant Force
When you put an object in a fluid, it experiences an upward force called the buoyant force. This force is what helps objects to float. It is an essential part of understanding why some objects float while others sink.
According to Archimedes' Principle, the buoyant force is equal to the weight of the fluid that is displaced by the object. This means if an object displaces a lot of water (or any fluid), it will experience a strong buoyant force. For an object to float, the buoyant force needs to balance the object's weight. If the buoyant force is greater than the object's weight, the object will float. If it's less, the object will sink.
In our solution, we calculate the buoyant force by using the formula:
According to Archimedes' Principle, the buoyant force is equal to the weight of the fluid that is displaced by the object. This means if an object displaces a lot of water (or any fluid), it will experience a strong buoyant force. For an object to float, the buoyant force needs to balance the object's weight. If the buoyant force is greater than the object's weight, the object will float. If it's less, the object will sink.
In our solution, we calculate the buoyant force by using the formula:
- F = \( \rho_{\text{water}} \times V_{\text{water\ displaced}} \times g \)
- \( F \) is the buoyant force,
- \( \rho_{\text{water}} \) is the density of water,
- \( V_{\text{water\ displaced}} \) is the volume of water displaced, which is the same as the volume of the cherry wood block in our case, and
- \( g \) is the acceleration due to gravity.
Density
Density is an intrinsic property of materials, and it tells us how much mass is within a given volume. The formula for density \( \rho \) is simply:
In the exercise, the cherry wood's density is given as \( 800 \, \text{kg/m}^3 \), meaning each cubic meter of cherry wood weighs 800 kilograms. Density helps us understand how the wood and iron compare to water. Water has a density of \( 1000 \, \text{kg/m}^3 \), so anything with a density lower than this will float on water if it displaces a sufficient amount of it.
When making the block float exactly at the water surface, the density of the combined block must equal the density of water, allowing precise balancing with the water's buoyancy force.
- \( \rho = \frac{\text{mass}}{\text{volume}} \)
In the exercise, the cherry wood's density is given as \( 800 \, \text{kg/m}^3 \), meaning each cubic meter of cherry wood weighs 800 kilograms. Density helps us understand how the wood and iron compare to water. Water has a density of \( 1000 \, \text{kg/m}^3 \), so anything with a density lower than this will float on water if it displaces a sufficient amount of it.
When making the block float exactly at the water surface, the density of the combined block must equal the density of water, allowing precise balancing with the water's buoyancy force.
Volume Calculation
The volume of an object is the amount of space it occupies. For a rectangular block like the cherry wood, the volume is calculated using the straightforward formula \( V = l \times w \times h \). This formula factors in the length, width, and height of the block.
In our exercise, the block's volume is calculated as:
In our exercise, the block's volume is calculated as:
- \( V_{\text{wood}} = 20 \, \text{cm} \times 10 \, \text{cm} \times 2 \, \text{cm} = 400 \, \text{cm}^3 \)
- \( V_{\text{wood}} = 400 \, \text{cm}^3 \times \frac{1 \, \text{m}^3}{1,000,000 \, \text{cm}^3} = 4.00 \times 10^{-4} \, \text{m}^3 \)
Mass Calculation
Calculating mass involves knowing an object's density and its volume. Mass tells us how much material is present in an object.
Using the density formula rearranged as:
Using the density formula rearranged as:
- \( \text{mass} = \rho \times \text{volume} \)
- \( \text{mass}_{\text{wood}} = 800 \, \text{kg/m}^3 \times 4.00 \times 10^{-4} \, \text{m}^3 = 0.320 \, \text{kg} \)
- \( \text{mass}_{\text{iron}} = 0.0796 \, \text{kg} \)
- \( V_{\text{iron}} = \frac{\text{mass}_{\text{iron}}}{\rho_{\text{iron}}} = \frac{0.0796 \, \text{kg}}{7860 \, \text{kg/m}^3} \approx 1.01 \times 10^{-5} \, \text{m}^3 \)
Other exercises in this chapter
Problem 40
A supertanker filled with oil has a total mass of \(10.2 \cdot 10^{8} \mathrm{~kg}\). If the dimensions of the ship are those of a rectangular box \(250 . \math
View solution Problem 41
A box with a volume \(V=0.0500 \mathrm{~m}^{3}\) lies at the bottom of a lake whose water has a density of \(1.00 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\).
View solution Problem 44
The average density of the human body is \(985 \mathrm{~kg} / \mathrm{m}^{3}\) and the typical density of seawater is about \(1020 \mathrm{~kg} / \mathrm{m}^{3}
View solution Problem 45
A tourist of mass \(60.0 \mathrm{~kg}\) notices a chest with a short chain attached to it at the bottom of the ocean. Imagining the riches it could contain, he
View solution