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Chapter 12: Problem 76

A plastic ball has radius 12.0 \(\mathrm{cm}\) and floats in water with 24.0\(\%\) of its volume submerged. (a) What force must you apply to the ball to hold it at rest totally below the surface of the water? (b) If you let go of the ball, what is its acceleration the instant you release it?

Short Answer

Expert verified
(a) 54.014 N applied upwards, (b) acceleration tends towards g.

Step by step solution

01

Calculate the Volume of the Ball

The formula for the volume of a sphere is given by \( V = \frac{4}{3}\pi r^3 \). Using a radius \( r = 12.0\,\text{cm} = 0.12\,\text{m} \), calculate the volume: \[ V = \frac{4}{3}\pi (0.12)^3 = 7.238\,\text{m}^3 \]
02

Determine the Submerged Volume

Since the ball is 24.0\(\%\) submerged, the volume of water displaced is \( 0.24 \times V \). Calculate it:\[ V_{\text{submerged}} = 0.24 \times 7.238 = 1.737\,\text{m}^3 \]
03

Calculate the Buoyant Force

The buoyant force is given by \( F_{\text{b}} = \rho_{\text{water}} \times g \times V_{\text{submerged}} \), where \( \rho_{\text{water}} = 1000\,\text{kg/m}^3 \) and \( g = 9.81\,\text{m/s}^2 \). \[ F_{\text{b}} = 1000 \times 9.81 \times 1.737 = 17.039\,\text{N} \]
04

Calculate the Total Volume Buoyed Force Must Overcome

To hold it fully submerged, calculate the force needed to balance the total buoyage, \( F = \rho_{\text{water}} \times g \times V \): \[ F = 1000 \times 9.81 \times 7.238 = 71.053\,\text{N} \]
05

Compute Additional Force Needed for Full Submersion

The applied force must overcome the buoyant force for the entire volume.\[ F_{\text{applied}} = 71.053 - 17.039 = 54.014\, \text{N} \]
06

Calculate the Net Force When Released

Net force when the ball is released is the total buoyant force minus the gravitational force on the ball:\[ F_{\text{net}} = F_{\text{b}} - m \cdot g \]Where \( m = \rho_{\text{ball}} \times V \approx 0 \) as assumption for floating object. Here, add mass equivalence based on buoyant assumptions.
07

Determine the Acceleration

Since initial density of ball and volume give limited gravitational effect, try using simplified reasons:\[ a = \frac{F_{\text{net}}}{m} \approx \frac{\Delta F}{\rho_{\text{ball}} \times V} \approx g \]Inference here is direct from cautious assessments.

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Key Concepts

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

Archimedes' principle
Archimedes' principle is a fundamental concept in fluid dynamics. It states that any object fully or partially submerged in a fluid experiences a buoyant force equal to the weight of the fluid displaced by the object. This principle helps to explain why objects float or sink. When a object is placed in a fluid, it pushes some of the fluid out of the way. The fluid tries to return to its original place, creating an upward force on the object. This force is what we call buoyancy. Understanding this principle is crucial when dealing with floating or submerged objects, as it allows us to calculate how much of an object will be submerged or what force is needed to keep it submerged. When the upward buoyant force matches the downward weight of the object, the object floats. If more upward force exists, the object rises, while more weight causes it to sink.
Net force calculation
Calculating the net force involves determining the total force acting on an object by considering both the forces that push it up and the ones that pull it down. In the context of floating objects, like the plastic ball in our example, the net force is the result of the difference between the buoyant force and the weight of the object.

Assuming the ball is fully submerged, we calculate the force that needs to be applied to keep it there. The force acting upwards is the buoyant force, whereas gravity pulls the object downwards. The net force when released, can be calculated as the difference between these two. If the buoyant force is greater than the gravitational force, the ball will accelerate upwards when released. If they are equal, there is no net force, and hence no acceleration.
Sphere volume calculation
The volume of a sphere is a basic concept often used in physics problems including those involving buoyancy.

To find the volume of a sphere, use the formula \[ V = \frac{4}{3}\pi r^3 \]where \( V \) is the volume and \( r \) is the radius of the sphere.

This formula is derived from the geometric properties of a sphere and allows you to calculate how much space the sphere occupies. It's particularly important for problems involving buoyancy because the amount of fluid displaced by the sphere directly depends on its volume.

In our example, calculating the spherical volume of the ball determines how much water it displaces, which is critical for finding the buoyant force acting on it. This volume also plays a part when considering how much force must be applied to fully submerge the object, since the total volume displaced maximizes the buoyant effect.

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Most popular questions from this chapter

It has been proposed that we could explore Mars using inflated balloons to hover just above the surface. The buoyancy of the atmosphere would keep the balloon aloft. The density of the Martian atmosphere is 0.0154 \(\mathrm{kg} / \mathrm{m}^{3}\) (although this varies with temperature). Suppose we construct these balloons of a thin but tough plastic having a density such that each square meter has a mass of 5.00 g. We inflate them with a very light gas whose mass we can neglect. (a) What should be the radius and mass of these balloons so they just hover above the surface of Mars? (b) If we released one of the balloons from part (a) on earth, where the atmospheric density is \(1.20 \mathrm{kg} / \mathrm{m}^{3},\) what would be its initial acceleration assuming it was the same size as on Mars? Would it go up or down? (c) If on Mars these balloons have five times the radius found in part (a), how heavy an instrument package could they carry?

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?

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 block of balsa wood placed in one scale pan of an equalarm balance is exactly balanced by a 0.115-kg brass mass in the other scale pan. Find the true mass of the balsa wood if its density is 150 \(\mathrm{kg} / \mathrm{m}^{3} .\) Explain why it is accurate to ignore the buoyancy in air of the brass but not the buoyancy in air of the balsa wood.

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. (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?
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