Review problem. (a) Derive an expression for the buoyant...

Review problem. (a) Derive an expression for the buoyant force on a spherical balloon, submerged in water, as a function of the depth below the surface, the volume of the balloon at the surface, the pressure at the surface, and the density of the water. (Assume the water temperature does not change with depth (b) Does the buoyant force increase or decrease as the balloon is submerged? (c) At what depth is the buoyant force one-half the surface value?

Key Concepts

Hydrostatic Pressure Variation

In a fluid at rest, pressure increases with depth according to the equation p = p? + ?gd, where p? is the surface pressure, ? is the fluid density, g is the acceleration due to gravity, and d is the depth. This concept is crucial for determining the pressure exerted on objects as they are submerged deeper into the fluid.

Isothermal Gas Compression

When a gas undergoes an isothermal process, its pressure and volume are inversely related, such that pV remains constant. This relationship is critical for analyzing how the volume of a gas-filled object changes with variations in external pressure, as when the object is submerged in a fluid.

Archimedes' Principle

This principle states that the buoyant force on an object submerged in a fluid equals the weight of the fluid displaced by the object. It is fundamental to understanding how and why objects experience an upward force when immersed in a fluid.

Transcript

00:01 So we need the buoyant force fb as a function of the depth at which the bubble is underwater, which i'll call h.

00:13 Its initial volume v -0, the initial pressure, which is atmospheric pressure, p.

00:18 Not, and the density of the water.

00:22 Well, first, we just use the definition of the buoyant force, which is the density of the water, in this case, times volume.

00:34 Now we know this volume is going to change, and the volume of the air inside the balloon is going to change, so we can use the ideal gas law, pv equals nrt, which we know that nrt is going to remain constant.

00:49 So we can say that pv equals p0 v0, and so we can solve for v to be p0v0 over the pressure that the balloon's at.

01:07 Now we know the pressure that the balloon is at can be given by the equation p0 plus the density times gravitational acceleration times h.

01:21 So the volume becomes p0 v0 over p0 plus row gh.

01:29 So the buoyant force plugging this in just becomes p0 v0 times row the density.

01:39 Divided by p0 plus row g h so that's part a for part b it asks what happens as the depth increases and you can see the h is on the denominator here and it's not showing up anywhere else so that means that the buoyant force increases sorry it decreases as h increases finally, for part c, it asks, what is the depth when the point force is one -half of the point force at the surface? so at the surface, which is where h equals 0, we know that fb, which i'll call fb -0, is equal to p0, b -0, d -0, divided by p0, which just equals v0 row, which makes sense because the row, the density is not going to change, and we just have that the volume is the initial volume.

03:04 So now we set fb equal to one -half v -0 row.

03:16 F -b equals p -0 v -0, divided by p0 plus row gh and we're just solving for h here so we multiply the denominator to the other side so one -half v -0 row times p -0 plus row g h is equal to p -0 v0 times row x x0 times row expanding out the left -hand side we have one -half p -0 v -0 row plus 1 1 1� v0, row squared, gh equals p0 v0 row...

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