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A \(900 \mathrm{~N}\) athlete in very good condition does not float in a fresh- water pool. To keep him from sinking to the bottom, an upward force of \(20 \mathrm{~N}\) must be applied to him. What are his volume and his average density?

Short Answer

Expert verified
The volume of the athlete is approximately \[0.0939 \, m^3\] and his average density is approximately \[959 \, kg/m^3\].

Step by step solution

01

Calculate Total Force Acting on the Athlete

First, we sum up all the forces acting on the athlete. There is the gravitational force pulling the athlete down (which equals his weight) and the additional force of \(20 \, N\) that needed to prevent him from sinking. Thus, total force \(F_{total}\) is calculated as: \(F_{total} = 920 \, N\)
02

Calculate the Volume

To find the volume of the athlete, we will use the buoyancy principle, which states that an object will displace a volume of fluid equal to its own volume. Thus, the volume \(V\) is calculated using the formula: \(V = \frac{F_{total}}{g \cdot \rho}\), where \(g = 9.8 \, m/s^2\) is the acceleration due to gravity and \(\rho = 1000 \, kg/m^3\) is the density of water. After substitution, we solve for \(V\).
03

Calculate the Average Density

Density (\(\rho_{athlete}\)) equals mass divided by volume. We know that the weight of the athlete is \(900 \, N\), and to convert this to mass we divide it by gravity \(g = 9.8 \, m/s^2\). So, \(\rho_{athlete} = \frac{m}{V}\), and after substituting the values we can solve to get \(\rho_{athlete}\).

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Key Concepts

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

Gravitational Force
Understanding the concept of gravitational force is crucial when analyzing the buoyancy principle. Essentially, gravitational force is the pull that Earth exerts on any object with mass. This force is responsible for keeping us anchored to the ground and is also a determining factor in how objects float or sink in water.

In the context of our exercise, the athlete's weight of 900 N (newtons) is a direct measure of the gravitational force, since weight is the force of gravity on an object. Surprisingly, even an additional upward force of 20 N is needed to keep the athlete buoyant. This points to a critical relationship between gravitational force and buoyancy, namely that an object will sink if the gravitational force on it exceeds the buoyant force of the fluid it is in.

In practical terms, when you step onto a scale, the reading you see is the gravitational force acting on your body. For objects submerged in a fluid, we use the concept of gravitational force to understand how much force is needed to counteract this pull in order to achieve flotation. For our athlete, his gravitational force defines the starting point for calculating not just buoyancy, but also his volume and density in the pool.
Volume Displacement
The principle of volume displacement is a fundamental aspect of buoyancy, which Archimedes notably described centuries ago. The principle states that any object, partially or fully submerged in a fluid, displaces a volume of that fluid equal to its own volume.

Hence, when our athlete is submerged in the pool, he is displacing an amount of water that is equivalent to his body's volume. This displaced volume of water has a weight, and according to Archimedes' principle, this weight is equal to the buoyant force acting upward on the athlete.

To calculate volume displacement in this case, we use the total force required to keep the athlete buoyant in fresh water (the gravitational force plus the additional upwards force) and divide it by the product of gravity's acceleration and the density of water. This tells us the volume of water which, if removed, would be equivalent to the volume of the athlete himself. Understanding volume displacement allows us not only to grasp why objects float or sink, but also to calculate their volumes under different conditions, an essential step in determining factors such as average density.
Average Density Calculation
Average density is the ratio of an object's mass to its volume. Calculating an object's average density helps us predict whether it will float or sink in a fluid. The average density is a fundamental concept in the study of fluid mechanics and buoyancy.

In our buoyancy problem, to calculate the athlete's average density, we first need to establish his mass from the given weight (force due to gravity). We then calculate his volume from the previous steps. By dividing the mass by the volume, we find the athlete's average density. If this density is greater than that of the fluid he is in, he will sink; if it is less, he will float.

Interestingly, the average density tells us about the distribution of mass throughout an object's volume. For humans, this varies individually, which is why some people float more easily than others. In the case of our athlete, his average density turns out to be more than the density of the water, evidenced by the need for an extra force to keep him afloat. This simple calculation of average density reveals important insights into the principles of buoyancy.

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Most popular questions from this chapter

A firehose must be able to shoot water to the top of a building \(28.0 \mathrm{~m}\) tall when aimed straight up. Water enters this hose at a steady rate of \(0.500 \mathrm{~m}^{3} / \mathrm{s}\) and shoots out of a round nozzle. Neglect air resistance. (a) What is the maximum diameter this nozzle can have? (b) If the only nozzle available has a diameter twice as great, what is the highest point the water can reach?

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.

A swimming pool is \(5.0 \mathrm{~m}\) long, \(4.0 \mathrm{~m}\) wide, and \(3.0 \mathrm{~m}\) deep. Compute the force exerted by the water against (a) the bottom and (b) either end. (Hint: Calculate the force on a thin, horizontal strip at a depth \(h,\) and integrate this over the end of the pool.) Do not include the force due to air pressure.

A tall cylinder with a cross-sectional area \(12.0 \mathrm{~cm}^{2}\) is partially filled with mercury; the surface of the mercury is \(8.00 \mathrm{~cm}\) above the bottom of the cylinder. Water is slowly poured in on top of the mercury, and the two fluids don't mix. What volume of water must be added to double the gauge pressure at the bottom of the cylinder?

A cylindrical disk of wood weighing \(45.0 \mathrm{~N}\) and having a diameter of \(30.0 \mathrm{~cm}\) floats on a cylinder of oil of density \(0.850 \mathrm{~g} / \mathrm{cm}^{3}\) (Fig. E12.19). The cylinder of oil is \(75.0 \mathrm{~cm}\) deep and has a diameter the same as that of the wood. (a) What is the gauge pressure at the top of the oil column? (b) Suppose now that someone puts a weight of \(83.0 \mathrm{~N}\) on top of the wood, but no oil seeps around the edge of the wood. What is the change in pressure at (i) the bottom of the oil and (ii) halfway down in the oil?

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