Review. (a) Derive an expression for the buoyant force on a...

Review. (a) Derive an expression for the buoyant force on a spherical balloon, submerged in water, as a function of the depth $h$ below the surface, the volume $V_{i}$ of the balloon at the surface, the pressure $P_{0}$ at the surface, and the density $\rho_{w}$ of the water. Assume the water temperature does not change with depth. (b) Does the buogant force increase or decrease as the balloon is submerged? (c) At what depth is the buoyant force one-half the surface value?

Key Concepts

Archimedes' Principle

Archimedes' Principle is the foundation of buoyancy. It states that any object submerged in a fluid experiences an upward force equal to the weight of the fluid displaced by it. This concept is essential for understanding how buoyant forces are computed, regardless of the object’s shape or changes in its volume.

Hydrostatic Pressure Variation

Hydrostatic pressure describes how the pressure in a fluid increases with depth as a consequence of the weight of the fluid above. This concept is key when analyzing forces on submerged objects because the pressure difference between the bottom and top surfaces of an object produces the net buoyant force.

Isothermal Compression

When a gas-filled object such as a balloon is submerged in a fluid with constant temperature, its internal gas pressure and volume obey Boyle’s Law. This isothermal compression causes the balloon's volume to decrease as the external pressure increases, altering the amount of fluid it displaces and hence the buoyant force.

Buoyant Force and Object Volume Relationship

The magnitude of the buoyant force depends directly on the volume of the displaced fluid. For deformable objects, changes in volume with depth (due to compressibility effects under varying pressure) lead to corresponding changes in the buoyant force, which is crucial when expressing the force as a function of depth.

Transcript

00:01 Okay, so buoyancy force b is given by the density of the water times the volume of the balloon at our depth times g.

00:16 Okay, well, we can look at a couple other equations to solve for v2 because it's the only variable we don't have.

00:23 So first we can look at a modification of the ideal gas law.

00:27 So p -0 -i will be equal to p2 v i or excuse me v2 and we can express p2 as p -not -fi i for v2 and now we can look at the pressure buoyancy equation p2 is equal to p -not plus row w g h and we can plug this in for p2 and solve for v2 that way.

01:01 So p -0 -v -i over v -2 is equal to p -not plus row w -g -h.

01:10 V -2 becomes p -0 -v -i over p -not plus row w -g -h.

01:21 So now we can return to our buoyancy equation.

01:24 So i'll plug in for v2 so we'll get buoyancy equal to row wg p.

01:37 Knot v -i over p -not plus row w g h and that's for a now for b as the depth decreases what does buoyancy do? well you should see that buoyancy and depth h are inversely proportional so when h gets bigger the denominator gets bigger therefore b or buoyancy force gets smaller so actually as h increases buoyancy goes down because it's inversely proportional to b now part c asks us to look at half the buoyancy force to the surface so the buoyancy force at the surface be surface is going to be the density of water r w times the volume at the surface which is v i times g but we want half of that and we need to set that equal to our buoyancy force we found in part a which is row w g p.

02:52 P.

02:52 Knot v not over p .k plus row wg h right off the bat we can cancel a handful of terms.

03:05 Oh, this supposed to be vi, excuse me, but we cancel that out anyway.

03:10 And we are left with one half is equal to p -not over p -0 plus row w -g -h.

03:21 Multiply each side by p -not plus row w -g -h get p -not over 2 plus row w -g -h over 2 equal to p .0...

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