Problem 80 - University Physics with Modern Physics

Question

A cubical block of wood 0.100 \(\mathrm{m}\) on a side and with a density of 550 \(\mathrm{kg} / \mathrm{m}^{3}\) floats in a jar of water. Oil with a density of 750 \(\mathrm{kg} / \mathrm{m}^{3}\) is poured on the water until the top of the oil layer is 0.035 \(\mathrm{m}\) below the top of the block. (a) How deep is the oil layer? (b) What is the gauge pressure at the block's lower face?

Step-by-Step Solution

Answer

(a) 0.045 m, (b) 869.75 Pa

Step 1: Determine the block's volume and mass

The block of wood is a cube with each side measuring 0.100 meters. To find the volume, use the formula for the volume of a cube, which is \( V = a^3 \). With \( a = 0.100 \) meters, the volume \( V = (0.100)^3 = 0.001 \, \text{m}^3 \).Next, to find the mass, use the formula \( m = \rho \times V \), where \( \rho \) is the density. The block's density is 550 \( \text{kg/m}^3 \). Thus, \( m = 550 \times 0.001 = 0.55 \, \text{kg} \).

Step 2: Calculate buoyant force from water displaced

Since the block is floating, the buoyant force equals the block's weight. The weight \( W \) of the block is the mass \( m \) times the gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \). Thus, \( W = 0.55 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 5.39 \, \text{N} \). The buoyant force is the weight of the water displaced, which equals the weight of the block: \( F_b = 5.39 \, \text{N} \).

Step 3: Determine the submerged height of the block in the water

Since the buoyant force equals the weight of the water displaced, we use \( F_b = \rho_{\text{water}} \times g \times V_{\text{submerged}} \). Here, the density of water \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \). Thus, \[ 5.39 = 1000 \times 9.8 \times V_{\text{submerged}} \]Solving for \( V_{\text{submerged}} \),\[ V_{\text{submerged}} = \frac{5.39}{9800} = 0.00055 \, \text{m}^3 \].Since \( V_{\text{submerged}} = A \times h_{\text{water}} \) where \( A = 0.01 \, \text{m}^2 \) (the area of the block's base), we find:\[ h_{\text{water}} = \frac{0.00055}{0.01} = 0.055 \, \text{m} \].

Step 4: Find the oil layer's depth

The total height of the block is 0.100 m. With 0.035 m of the block above the oil-water interface, the height submerged in oil is \( h_{\text{oil}} = 0.100 - 0.055 = 0.045 \, \text{m} \).Therefore, the oil layer's depth is the remaining part of this 0.045 m submerged in oil which is 0.045 m since oil sits above the water.

Step 5: Calculate the gauge pressure at block's lower face

The gauge pressure at the lower face of the block is due to both the water and oil. First, calculate the pressure due water:\( P_{\text{water}} = \rho_{\text{water}} \times g \times h_{\text{water}} = 1000 \times 9.8 \times 0.055 = 539 \text{Pa} \).Next, calculate the pressure due to the oil layer:\( P_{\text{oil}} = \rho_{\text{oil}} \times g \times h_{\text{oil}} = 750 \times 9.8 \times 0.045 = 330.75 \text{Pa} \).The total gauge pressure is:\[ P_{\text{gauge}} = P_{\text{water}} + P_{\text{oil}} = 539 + 330.75 = 869.75 \, \text{Pa} \].

Key Concepts

Buoyant Force

The concept of buoyant force is a fundamental principle in fluid mechanics. It refers to the upward force exerted by a fluid that opposes the weight of an object immersed in it. This force is what enables objects to float or appear less heavy when submerged in a fluid. To calculate the buoyant force, we use the formula: \[ F_b = \rho \times g \times V_{\text{submerged}} \] where

• \(F_b\) is the buoyant force
• \(\rho\) is the fluid density
• \(g\) is the acceleration due to gravity (9.8 \(\text{m/s}^2\))
• \(V_{\text{submerged}}\) is the volume of fluid displaced by the object

For a floatation scenario, like a cubical block partially submerged in water, the buoyant force equals the weight of the fluid displaced, which is equivalent to the gravitational force acting on the submerged block. This principle is commonly referred to as Archimedes' principle.

Gauge Pressure

In fluid mechanics, gauge pressure is the additional pressure in a fluid system above the atmospheric pressure. It is crucial in determining the total pressure acting at a specific depth from the surface in a fluid. Calculate gauge pressure using:\[ P_{\text{gauge}} = (\rho_{\text{water}} \times g \times h_{\text{water}}) + (\rho_{\text{oil}} \times g \times h_{\text{oil}}) \] Here:

• \(P_{\text{gauge}}\) is the gauge pressure
• \(\rho_{\text{water}}\) and \(\rho_{\text{oil}}\) are the densities of the water and oil, respectively
• \(g\) is the gravitational acceleration
• \(h_{\text{water}}\) and \(h_{\text{oil}}\) are the respective heights of the water and oil layers over the point in question

Gauge pressure is particularly important when working with submerged objects, as it affects both the buoyant force calculation and the structural integrity of objects in varying fluid depths.

Density

Density is a core concept in fluid mechanics, describing the mass of a substance per unit of volume. It is usually denoted with the Greek letter \(\rho\) and calculated as: \[ \rho = \frac{m}{V} \] where

• \(\rho\) is the density
• \(m\) is the mass
• \(V\) is the volume

Density determines how substances will interact when combined. In the case of the wooden block, its density (550 \(\text{kg/m}^3\)) allows it to float on water (1000 \(\text{kg/m}^3\)). A lower density than the fluid enables the block to stay on the surface due to buoyant force acting greater than the gravitational pull.Knowing the density of various materials is essential for predicting their behavior in a fluid, understanding buoyancy, and solving practical problems involving fluid displacements.

Floating Objects

Floating objects in fluid mechanics are governed by the principle of buoyancy, which determines whether an object will sink, float, or remain suspended in a fluid. An object floats when its density is less than the density of the fluid it is in. For instance, the wooden block in this exercise floats because its density is 550 \(\text{kg/m}^3\), considerably less than 750 \(\text{kg/m}^3\) of the oil and 1000 \(\text{kg/m}^3\) of the water.Floating conditions can be understood better through:

• Archimedes' Principle: States that the buoyant force on an object is equal to the weight of the liquid displaced by the object.
• Equilibrium: A floating object achieves equilibrium when the gravitational force equals the buoyant force, leading to a balance of forces and steady positioning on the fluid's surface.

These principles explain how and why objects of varying densities interact differently with fluids, helping engineers design ships, submarines, and flotation devices effectively.