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A rubber ball filled with air has a diameter of \(25.0 \mathrm{~cm}\) and a mass of \(0.540 \mathrm{~kg}\). What force is required to hold the ball in equilibrium immediately below the surface of water in a swimming pool?

Short Answer

Expert verified
The force required to hold the rubber ball in equilibrium immediately below the surface of water in a swimming pool is approximately 7.0306 N.

Step by step solution

01

Calculate the weight of the rubber ball

The weight of the ball can be calculated by the formula \(F = m*g\), where m is the mass of the ball (0.540 kg) and g is acceleration due to gravity (9.81 m/s^2). So, \(F = 0.540 \mathrm{~kg} * 9.81 \mathrm{~m/s^{2}} = 5.2974 \mathrm{~N}\).
02

Calculate the buoyant force

The buoyant force is calculated by the formula \(F_b = 蟻_f*V_f*g\), where 蟻_f is the density of the fluid (water), which is approximately 1000 kg/m^3, V_f is the volume of the fluid displaced, which is equal to the volume of the ball that can be calculated using the formula \(\frac{4}{3} *蟺*(\frac{d}{2})^3\) (where d is the diameter of the ball, 0.25 m), and g is acceleration due to gravity (9.81 m/s^2). So, the buoyant force \(F_b = 1000 \mathrm{~kg/m^{3}}* \frac{4}{3} *蟺* (\frac{0.25 \mathrm{~m}}{2})^3*9.81 \mathrm{~m/s^{2}} = 12.328 \mathrm{~N}\).
03

Calculate the force required to hold the ball in equilibrium

To hold the ball in equilibrium below the surface of water, the weight of the ball should be equal to the sum of the buoyant force and the force required, hence the force required can be calculated by subtracting the weight of the ball from the buoyant force. So, the force required = \(12.328 \mathrm{~N} - 5.2974 \mathrm{~N} = 7.0306 \mathrm{~N}\).

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Key Concepts

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

Fluid Mechanics
At its core, fluid mechanics is the branch of physics concerned with the behavior of fluids鈥攍iquids and gases鈥攁nd the forces on them. It has wide applications, extending from everyday household tasks to sophisticated engineering designs. In our exercise, fluid mechanics concepts allow us to predict how water will exert force on a submerged object, like the rubber ball.

When an object is immersed in a fluid, it feels an upward force because the pressure at the bottom of the object is greater than the pressure at the top. This disparity is due to the depth-dependent nature of fluid pressure; deeper in the fluid, the pressure increases. This fundamental understanding provides us the tools to analyze the rubber ball鈥檚 behavior in water and is essential in calculating the force needed to keep it submerged in equilibrium.
Buoyancy
Buoyancy is the uplifting force that a fluid exerts on any object submerged in it. The principle of buoyancy is crucial in understanding why some objects float while others sink. This concept is tightly connected to the density of objects in comparison to the fluid they are in. A solid understanding of buoyancy is not just important for solving problems in physics but also in real-life scenarios such as designing ships or understanding the science behind swimming.

In our exercise, the rubber ball has a lower density compared to the water, due to which it experiences an upward buoyant force. To calculate this force, which we often call the 'buoyant force', we must know the volume of water displaced by the ball. This concept is directly linked to Archimedes' principle and is calculated based on the volume of the submerged part of the object.
Archimedes' Principle
Archimedes' principle states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid that the object displaces. This principle is a gateway to understanding why ships float and balloons rise in the air. It's one of the fundamental principles used in hydrostatics鈥攖he study of fluids at rest鈥攁nd is derived from the observations made by the ancient Greek mathematician and inventor, Archimedes.

Based on this principle, in the step-by-step solution, the buoyant force on the rubber ball is precisely the weight of the volume of water it displaces. In mathematical terms, for the ball in equilibrium, the buoyant force must counterbalance the weight of the ball, which leads us to the calculation seen in the problem. For the ball to be in equilibrium, the upward buoyant force must exactly balance the downward force of gravity on the ball鈥攁 beautiful illustration of how physical concepts interplay to provide a detailed understanding of phenomena encountered in everyday life.

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Most popular questions from this chapter

A \(10.0-\mathrm{kg}\) block of metal is suspended from a scale and immersed in water, as in Figure P9.38. The dimensions of the block are \(12.0 \mathrm{~cm} \times 10.0 \mathrm{~cm} \times\) \(10.0 \mathrm{~cm}\). The \(12.0-\mathrm{cm}\) dimen- sion is vertical, and the top of the block is \(5.00 \mathrm{~cm}\) below the surface of the water. (a) What are the forces exerted by the water on the top and bottom of the block? (Take \(P_{0}=1.0130 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\), (b) What is the reading of the spring scale? (c) Show that the buoyant force cquals the difference between the forces at the top and bottom of the block.

The block of ice (temperature \(0^{\circ} \mathrm{C}\) ) shown in Figure \(\mathrm{P} 9.63\) is drawn over a level surface lubricated by a layer of water \(0.10 \mathrm{~mm}\) thick. Determine the magnitude of the force \(\vec{F}\) needed to pull the block with a constant speed of \(0.50 \mathrm{~m} / \mathrm{s}\). At \(0^{\circ} \mathrm{C}\), the viscosity of water has the value \(\eta=1.79 \times 10^{-3} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\)

A \(62.0-\mathrm{kg}\) survivor of a cruise line disaster rests atop a block of Styrofoam insulation, using it as a raft. The Styrofoam has dimensions \(2.00 \mathrm{~m} \times 2.00 \mathrm{~m} \times 0.0900 \mathrm{~m}\). The bottom \(0.024 \mathrm{~m}\) of the raft is submerged. (a) Draw a free-body diagram of the system consisting of the survivor and raft. (b) Write Newton's second law for the system in one dimension, using \(B\) for buoyancy, \(w\) for the weight of the survivor, and \(w_{r}\) for the weight of the raft. (Set \(a=0 .\) ) (c) Calculate the numeric value for the buoyancy, \(B\). (Seawater has density \(1025 \mathrm{~kg} / \mathrm{m}^{3}\).) (d) Using the value of \(B\) and the weight w of the survivor, calculate the weight \(w_{r}\) of the Styrofoam. (c) What is the density of the Styrofoam? (f) What is the maximum buoyant force, corresponding to the raft being submerged up to its top surface? (g) What total mass of survivors can the raft support?

The viscous force on an oil drop is measured to be equal to \(3.0 \times 10^{-15} \mathrm{~N}\) when the drop is falling through air with a speed of \(4.5 \times 10^{-4} \mathrm{~m} / \mathrm{s}\). If the radius of the drop is \(2.5 \times 10^{-6} \mathrm{~m}\), what is the viscosity of air?

The aorta in humans has a diameter of about \(2.0 \mathrm{~cm}\), and at certain times the blood speed through it is about \(55 \mathrm{~cm} / \mathrm{s}\), Is the blood flow turbulent? The density of whole blood is \(1050 \mathrm{~kg} / \mathrm{m}^{3}\), and its coefficient of viscosity is \(2.7 \times 10^{-4} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\)

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