- A 10.0-kg block of metal measuring 12.0 cm ×10.0 cm ×10.0 cm is suspended from a scale and imm

- A 10.0-kg block of metal measuring 12.0 cm ×10.0 cm ×10.0...

- A $10.0-\mathrm{kg}$ block of metal measuring $12.0 \mathrm{cm} \times 10.0 \mathrm{cm} \times$ $10.0 \mathrm{cm}$ is suspended from a scale and immersed in water as in Figure P14.25b. The 12.0-cm dimension is vertical, and the top of the block is $5.00 \mathrm{cm}$ below the surface of the water. (a) What are the forces acting on the top and on the bottom of the block? (Take $P_{0}=1.0130 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}$ ) (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.

- A $10.0-\mathrm{kg}$ block of metal measuring $12.0 \mathrm{cm} \times 10.0 \mathrm{cm} \times$
$10.0 \mathrm{cm}$ is suspended from a scale and immersed in water as in Figure P14.25b. The 12.0-cm dimension is vertical, and the top of the block is $5.00 \mathrm{cm}$ below the surface of the water. (a) What are the forces acting on the top and on the bottom of the block? (Take $P_{0}=1.0130 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}$ )
(b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.

Key Concepts

Apparent Weight

Apparent weight is the reading on a scale when an object is immersed in a fluid and is the result of the actual gravitational force on the object being partially offset by the buoyant force. In other words, the apparent weight is the object’s weight in air minus the buoyant force, which explains why objects feel lighter when submerged in a fluid.

Pressure Force Distribution

The net force on a submerged object is obtained by considering the difference in fluid pressure acting on its upper and lower surfaces. Since fluid pressure increases with depth, the bottom surface of an immersed object experiences a higher pressure than the top, resulting in a net upward force. This pressure difference is key to understanding why and how the buoyant force arises from the surrounding fluid.

Hydrostatic Pressure

This concept refers to the variation of pressure within a fluid as a function of depth. In a fluid at rest, the pressure increases linearly with depth due to the weight of the overlying fluid, following the relationship P = P? + ?gh, where P? is the surface pressure, ? is the density of the fluid, g is the acceleration due to gravity, and h is the depth below the surface. This principle is used to calculate the pressure acting on different surfaces of a submerged object.

Archimedes' Principle

Archimedes’ Principle states that a body wholly or partially immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the body. This fundamental idea connects the volume of the displaced fluid, its density, and gravitational acceleration, thereby providing a method to determine the magnitude of the buoyant force acting on a submerged object.

Transcript

00:01 So in this problem, we have a 10 kilogram block measuring 10 centimeters by 10 centimeters by 12 centimeters.

00:14 And we have this block submerged in the water so that the top of the block is 5 centimeters away from the surface of the water.

00:26 We have three parts of the question.

00:29 The first question, now the first part, is asking about the forces that act on the top of the block and the bottom of the block.

00:39 To figure that out, since force is equal to the pressure times the area, the area is pretty simple to figure out the top of the block and the bottom of the block are both 10 centimeters by 10 centimeters.

00:54 So we know that a equals 0 .1 meters squared.

01:02 But what are the pressures? well, first of all, the pressure, the top of the block should be equal to the atmospheric pressure plus whatever the pressure that comes from the water, which is going to be the density of the water times the gravitational acceleration and the depth.

01:22 In this case, i'll write this as height top, where height top equals, well, the top is 5 centimeter away from the surface, so this is going to be 5 centimeters.

01:37 We have something very similar for the bottom of the block.

01:41 We have the atmospheric pressure and the exact same thing, except for that this time, our height is not 5 centimeters, but it's going to be 5 plus 12.

01:55 So that's going to be 17 centimeters.

01:59 Now we have this figured out.

02:03 We can figure out the force, the top, and bottom.

02:06 So we will have force top equals the p top times the area.

02:14 So we're going to have the standard atmosphere plus density of water, g, then h top, times a.

02:29 And the standard atmospheric pressure, all we know is 1 .03 times 10 to the fifth nons per meter squared.

02:45 Then we have plus 1 ,000 kilograms per meters cube.

02:52 That's a density of water, 9 .8 meters per second squared times h top, which is 0 .4.

03:00 0 .05 05 centimeters.

03:07 Sorry, i made a mistake here.

03:10 This should be 0.

03:16 0 .1 meters, sorry.

03:21 0 .0 .05 meters for each top times or area 0 .1 meters squared.

03:31 And hopefully you can check that the units all cancel units all.

03:36 All work out to the correct unit, which is newton's.

03:40 And once you plug all this into the calculator, you should get 1 ,017 .9 newtons.

03:48 And we can do the exact same thing, except replacing the height for the force on the bottom.

04:03 So let me write all of that down...

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