Problem 44 - Fundamentals of Physics

Question

A wood block (mass \(3.67 \mathrm{~kg}\), density \(\left.600 \mathrm{~kg} / \mathrm{m}^{\text {' }}\right)\) is fitted with lead (density \(1.14 \times 10^{4} \mathrm{~kg} / \mathrm{m}^{3}\) ) so that it floats in water with 0.900 of its volume submerged. Find the lead mass if the lead is fitted to the block's (a) top and (b) bottom.

Step-by-Step Solution

Answer

Lead mass needed is approximately 1.835 kg, regardless of placement on top or bottom.

Step 1: Understand the Problem

The problem involves a wood block with a certain mass and density, and lead being added to it so that it floats with 90% of its volume submerged in water. We need to determine the mass of the lead when it is placed on the top and bottom of the block.

Step 2: Understand Buoyancy

According to Archimedes' principle, the buoyant force is equal to the weight of the water displaced. For the block with lead to float, the total weight of the block plus lead must equal the weight of the water displaced by 90% of the block's volume.

Step 3: Calculate the Volume of the Wood Block

The density of the block is given as \(600 \text{ kg/m}^3\) and its mass is \(3.67 \text{ kg}\). The volume \(V_b\) of the block can be calculated as: \[ V_b = \frac{3.67 \text{ kg}}{600 \text{ kg/m}^3} \approx 0.0061167 \text{ m}^3 \]

Step 4: Calculate the Buoyant Force

The volume submerged is 0.900 times the volume of the block: \[ V_{ ext{submerged}} = 0.900 \times 0.0061167 \approx 0.00550503 \text{ m}^3 \] The buoyant force is the weight of the displaced water, which is:\[ F_{ ext{buoyant}} = \rho_{ ext{water}} \cdot V_{ ext{submerged}} \cdot g \] Assuming \(\rho_{\text{water}} = 1000 \text{ kg/m}^3\) and \(g = 9.81 \text{ m/s}^2\),\[ F_{ ext{buoyant}} = 1000 \cdot 0.00550503 \cdot 9.81 \approx 54.00184 \text{ N} \]

Step 5: Equate Buoyant Force to Total Weight

The total weight of the block plus the lead must equal the buoyant force. The weight of the block is:\[ W_b = 3.67 \cdot 9.81 = 35.9977 \text{ N} \] Thus, the lead must supply the difference:\[ W_{ ext{lead}} = 54.00184 - 35.9977 \approx 18.00414 \text{ N} \]

Step 6: Calculate Mass of Lead

Convert the weight of lead needed to mass. Using \(F = mg\), rearrange to solve for mass:\[ m_{\text{lead}} = \frac{W_{ ext{lead}}}{g} \approx \frac{18.00414}{9.81} \approx 1.835 \text{ kg} \]This mass applies whether the lead is on top or on the bottom due to the same amount of buoyancy needed.

Key Concepts

Archimedes' principle

Archimedes' principle is a fundamental concept in fluid mechanics. It states that a body immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the body. This principle is crucial when understanding why objects like ships and logs float on water. When the buoyant force matches the weight of the object, the object remains afloat. If the object's weight exceeds the buoyant force, it sinks. Archimedes' principle helps us calculate important factors such as the submerged volume of floating objects, ensuring stability and balance in designs like boats and ships. Understanding this concept allows us to solve problems related to floating objects and ensure things work as expected.

Density

Density is a measure of how much mass is contained in a given volume of a substance. It is defined mathematically as mass per unit volume, represented by the formula: \( \text{density} = \frac{\text{mass}}{\text{volume}} \). The density of an object determines its relative buoyancy when placed in a fluid, like water. Objects with a density lesser than the fluid will float, whereas objects with a higher density will sink. In our exercise, we have a wood block with a known density, and we're trying to balance it using lead to achieve a specific floating condition. By understanding density, we can predict and manipulate whether and how an object will float.

Floating Objects

Floating objects are governed by their density relative to the fluid in which they are placed. If an object floats, it means the fluid displaced equals the weight of the object per Archimedes' principle. In our exercise, 90% of the wood block's volume is submerged, indicating that the block and lead's combined density is less than or slightly equal to water.

• A floating object in equilibrium has a buoyant force equal to its weight.
• The configuration of mass (placement of lead) affects how a block will float but not its ability to do so.

Whether the lead is on top or bottom, the block will float if this condition is met. This emphasizes that the submerged volume is a key determinant of floating status.

Mass Calculation

Mass calculation is an essential part of determining how objects will behave in fluids. In this exercise, we calculated the lead mass needed so that the wood block would float with a specific volume submerged. To do this, we first calculated the buoyant force from the displaced water; then, we equated this to the total weight of the block plus lead, per Archimedes' principle.The mass of the lead needed was calculated by equating the total force exerted by the water and the weight with the equation:\[ \text{Weight of lead} = \text{Buoyant force} - \text{Weight of block} \]And then rearranging to solve for mass:\[ \text{Mass of lead} = \frac{\text{Weight of lead}}{g} \]This mass calculation ensures that the conditions for floating are correctly met regardless of where the lead is positioned -- on top or at the bottom of the block. Proper mass calculation is vital for designing objects that will float correctly and maintain stability in fluids.