Arehimedes' Principle A spherical buoy has a radius of 1 m and a density one-fourth that of seaw

step 1 Problem 69 question A. So the volume of the sphere is equal to 4 over 3 pi r cube and where r is

Arehimedes' Principle A spherical buoy has a radius of 1 m...

Arehimedes' Principle A spherical buoy has a radius of 1 $\mathrm{m}$ and a density one-fourth that of seawater. By Archimedes' Principle, the weight of the displaced water will equal the weight of the buoy. . Let $x=$ the depth to which the buoy sinks. . Let $d=$ the density of seawater. $\cdot$ Let $r=$ the radius of the circle formed where buoy, air, and water meet. See the figure below. Notice in the figure that $r^{2}=1-(1-x)^{2}=2 x-x^{2},$ and recall from geometry that the volume of submerged spherical cap is $$V=\frac{\pi x}{6} \cdot\left(3 r^{2}+x^{2}\right)$$ (a) Verify that the volume of the buoy is 4$\pi / 3$ . (b) Use your result from (a) to establish the weight of the buoy as $\pi d / 3 .$ (e) Prove the weight of the displaced water is $\pi d \cdot x\left(3 r^{2}+x^{2}\right) / 6 .$ (d) Approximate the depth to which the buoy will sink.

Arehimedes' Principle A spherical buoy has a radius of 1 $\mathrm{m}$ and a density one-fourth that of seawater. By Archimedes' Principle, the weight of the displaced water will equal the weight of the buoy.
. Let $x=$ the depth to which the buoy sinks.
. Let $d=$ the density of seawater.
$\cdot$ Let $r=$ the radius of the circle formed where buoy, air, and water meet. See the figure below. Notice in the figure that $r^{2}=1-(1-x)^{2}=2 x-x^{2},$ and recall from geometry that the volume of submerged spherical cap is $$V=\frac{\pi x}{6} \cdot\left(3 r^{2}+x^{2}\right)$$
(a) Verify that the volume of the buoy is 4$\pi / 3$ .
(b) Use your result from (a) to establish the weight of the buoy as $\pi d / 3 .$
(e) Prove the weight of the displaced water is $\pi d \cdot x\left(3 r^{2}+x^{2}\right) / 6 .$
(d) Approximate the depth to which the buoy will sink.

Key Concepts

Geometric Relationships in Spherical Objects

In problems involving spheres, geometric relationships help relate different dimensions such as the depth of submersion and the corresponding radius of the intersection circle between the sphere and the fluid surface. These relationships, derived from the properties of circles and spheres, are key to establishing volume formulas for partially submerged sections.

Buoyancy Equilibrium

Buoyancy equilibrium occurs when the upward buoyant force exerted by the displaced fluid equals the downward gravitational force acting on the object. By setting the weight of the displaced fluid equal to the weight of the buoy, one can solve for the depth of submersion, which is a common application of Archimedes' Principle in floating body problems.

Archimedes' Principle

Archimedes' Principle states that any body completely or partially submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the body. This principle is fundamental in determining the buoyant force acting on objects in fluids and is used to find the equilibrium conditions for floating and submerged objects.

Volume of a Sphere

The volume of a sphere is calculated using the formula V = (4/3)?r³, where r is the radius of the sphere. This formula is essential when determining the total amount of space enclosed by a spherical object, which is a critical parameter in problems involving mass and buoyancy.

Volume of a Spherical Cap

A spherical cap is a portion of a sphere cut off by a plane. Its volume can be calculated with the formula V = (?h/6)(3r² + h²), where h is the height of the cap and r relates to the radius of the circular intersection. This concept is used to determine the volume of the submerged portion of a sphere, which is necessary for calculating the displaced fluid’s weight.

Transcript

Please give Ace some feedback