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Chapter 13: Problem 35

A hollow plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of \(0.650 \mathrm{~m}^{3}\) and the tension in the cord is \(900 \mathrm{~N}\). (a) Calculate the buoyant force exerted by the water on the sphere. (b) What is the mass of the sphere? (c) The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

Short Answer

Expert verified
(a) 6376.5 N, (b) 558.27 kg, (c) 0.858.

Step by step solution

01

Apply Archimedes' Principle

The buoyant force exerted on an object submerged in a fluid is equal to the weight of the fluid displaced by that object. For a sphere with a volume \( V = 0.650 \, m^3 \), and water having a density \( \rho_{water} = 1000 \, kg/m^3 \), the buoyant force \( F_b \) is calculated using the formula\[F_b = \rho_{water} \times V \times g\]where \( g = 9.81 \, m/s^2 \) is the acceleration due to gravity.
02

Calculate the Buoyant Force

Substituting the known values into the formula:\[F_b = 1000 \, kg/m^3 \times 0.650 \, m^3 \times 9.81 \, m/s^2\]\[F_b = 6376.5 \, N\]Thus, the buoyant force is 6376.5 N.
03

Analyze Forces on the Sphere

The sphere is held in place by the tension in the cord and the buoyant force. The gravitational force on the sphere, \( W_{sphere} \), equals the buoyant force minus the tension:\[W_{sphere} = F_b - T\]Substitute the values:\[W_{sphere} = 6376.5 \, N - 900 \, N = 5476.5 \, N\]
04

Calculate the Mass of the Sphere

The mass \( m \) of the sphere can be found using the formula for weight:\[W_{sphere} = m \times g\]Substitute the expression for \( W_{sphere} \):\[m = \frac{5476.5 \, N}{9.81 \, m/s^2} = 558.27 \, kg\]
05

Determine the Fraction of Volume Submerged

When the cord breaks, the sphere reaches the surface and some fraction of its volume is submerged due to buoyancy. At equilibrium, the weight of the sphere is equal to the weight of the displaced water:\[W_{sphere} = \rho_{water} \times V_{submerged} \times g\]From Step 4:\[558.27 \, kg \times 9.81 \, m/s^2 = 1000 \, kg/m^3 \times V_{submerged} \times 9.81 \, m/s^2\]Simplifying:\[V_{submerged} = 0.55827 \, m^3\]The fraction of the volume submerged is then:\[\frac{V_{submerged}}{V} = \frac{0.55827}{0.650} \approx 0.858\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Buoyant Force
Understanding the concept of buoyant force is fundamental in physics, especially when dealing with objects submerged in a fluid. Archimedes' Principle states that the buoyant force acting on an object fully or partially submerged in a fluid is equal to the weight of the fluid that the object displaces. This is vital as it helps in determining whether objects will float or sink.
For our scenario, when the hollow plastic sphere is submerged in the lake, it displaces a certain volume of water. The buoyant force for this sphere is calculated by multiplying the density of water, the volume of the sphere, and the acceleration due to gravity (9.81 m/s虏). This gives us a force of 6376.5 N upward, opposing gravity and tending to make the object float.
Density of Water
Density of water, denoted as \(\rho_{water}\), is an intrinsic property of water that plays a crucial role in calculating the buoyant force. It is typically 1000 kg/m鲁 at standard conditions, which means each cubic meter of water weighs 1000 kilograms.

The density of water is essential when calculating how much buoyant force will act on an object. This is because the weight of the displaced water directly depends on this density. Hence, any calculation involving objects in water, like determining whether they'll sink or float, relies heavily on the density value.
Submerged Volume Fraction
The submerged volume fraction is the ratio of the volume of the object that remains underwater to its total volume when it achieves equilibrium. This fraction tells us how much of the object will be below the water surface.

In the sphere's case, once it's free from the cord's restraint, it will float, but not entirely above water. Instead, it adjusts to a position where the weight of the sphere matches the weight of the displaced water. Simplified further, the fraction is calculated as 0.858, revealing that about 85.8% of the sphere remains under water. This formula is derived by equating the gravitational weight of the sphere to the buoyant force acting on the submerged portion.
Weight of Displaced Fluid
The concept of the weight of displaced fluid helps us understand the buoyant force more intuitively. When an object is submerged in a fluid, it pushes aside a volume of that fluid equivalent to its own volume, in line with the principles laid out by Archimedes.

The weight of these displaced fluids can be calculated by multiplying the fluid's density by the displaced volume and gravity. For the plastic sphere, knowing the displaced volume and the constant density of water, we compute the weight of the water displaced. This calculation provides a direct correlation between the sphere's buoyant force and the necessary upward force needed to keep the sphere floating or submerged at equilibrium.

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Most popular questions from this chapter

Blood pressure. Systemic blood pressure is expressed as the ratio of the systolic pressure (when the heart first ejects blood into the arteries) to the diastolic pressure (when the heart is relaxed). Normal systemic blood pressure is \(\frac{120}{80}\). (a) What are the maximum and minimum forces (in newtons) that the blood exerts against each square centimeter of the heart for a person with normal blood pressure? (b) Blood pressure is normally measured on the upper arm at the same height as the heart. Due to therapy for an injury, a patient's upper arm is extended \(30.0 \mathrm{~cm}\) above his heart. In that position, what should be his systemic blood-pressure reading, expressed in the standard way, if he has normal blood pressure? The density of blood is \(1060 \mathrm{~kg} / \mathrm{m}^{3}\).

Elephants under pressure. An elephant can swim or walk with its chest several meters underwater while the animal breathes through its trunk, which remains above the water surface and acts like a snorkel. The elephant's tissues are at an increased pressure due to the surrounding water, but the lungs are at atmospheric pressure because they are connected to the air through the trunk. Figure 13.47 shows the gauge pressures in an elephant's lungs and abdomen when the elephant's chest is submerged to a particular depth in a lake. In this situation, the elephant's diaphragm, which separates the lungs from the abdomen, must maintain the difference in pressure between the lungs and the abdomen. The diaphragm of an elephant is typically \(3.0 \mathrm{~cm}\) thick and \(120 \mathrm{~cm}\) in diameter. For the situation shown, the tissues in the elephant's abdomen are at a gauge pressure of \(150 \mathrm{~mm} \mathrm{Hg}\). This pressure corresponds to what distance below the surface of a lake? A. \(1.5 \mathrm{~m}\) B. \(2.0 \mathrm{~m}\) C. \(3.0 \mathrm{~m}\) D. \(15 \mathrm{~m}\)

An ore sample weighs \(17.50 \mathrm{~N}\) in air. When the sample is suspended by a light cord and totally immersed in water, the tension in the cord is \(11.20 \mathrm{~N}\). Find the total volume and the density of the sample.

Elephants under pressure. An elephant can swim or walk with its chest several meters underwater while the animal breathes through its trunk, which remains above the water surface and acts like a snorkel. The elephant's tissues are at an increased pressure due to the surrounding water, but the lungs are at atmospheric pressure because they are connected to the air through the trunk. Figure 13.47 shows the gauge pressures in an elephant's lungs and abdomen when the elephant's chest is submerged to a particular depth in a lake. In this situation, the elephant's diaphragm, which separates the lungs from the abdomen, must maintain the difference in pressure between the lungs and the abdomen. The diaphragm of an elephant is typically \(3.0 \mathrm{~cm}\) thick and \(120 \mathrm{~cm}\) in diameter. The maximum force the muscles of the diaphragm can exert is \(24,000 \mathrm{~N}\). What maximum pressure difference can the diaphragm withstand? A. \(160 \mathrm{~mm} \mathrm{Hg}\) B. \(760 \mathrm{~mm} \mathrm{Hg}\) C. \(920 \mathrm{~mm} \mathrm{Hg}\) D. \(5000 \mathrm{~mm} \mathrm{Hg}\)

You are designing a machine for a space exploration vehicle. It contains an enclosed column of oil that is \(1.50 \mathrm{~m}\) tall, and you need the pressure difference between the top and the bottom of this column to be 0.125 atm. (a) What must be the density of the oil? (b) If the vehicle is taken to Mars, where the acceleration due to gravity is \(0.379 g,\) what will be the pressure difference (in earth atmospheres) between the top and bottom of the oil column?
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