Fluids - Hydrostatics

Physics 6B

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Two Basic Concepts: Density and Pressure

Density measures how “tightly packed” an object is.

The definition is given by a formula:V

m

Prepared by Vince Zaccone

For Campus Learning

Assistance Services at UCSB

Two Basic Concepts: Density and Pressure

Density measures how “tightly packed” an object is.

The definition is given by a formula:V

m

Important note: you can rearrange

this formula to get m=ρV. You will

need to do this a lot.

Prepared by Vince Zaccone

For Campus Learning

Assistance Services at UCSB

Two Basic Concepts: Density and Pressure

Density measures how “tightly packed” an object is.

The definition is given by a formula:

Standard units for density are (you might be used to seeing in chemistry class)

V

m

3m

kg3cm

g

Important note: you can rearrange

this formula to get m=ρV. You will

need to do this a lot.

Prepared by Vince Zaccone

For Campus Learning

Assistance Services at UCSB

Two Basic Concepts: Density and Pressure

Density measures how “tightly packed” an object is.

The definition is given by a formula:

Standard units for density are (you might be used to seeing in chemistry class)

One value you need to know is the density of water:

V

m

3m

kg3cm

g

33watercm

g1

m

kg1000

Important note: you can rearrange

this formula to get m=ρV. You will

need to do this a lot.

Prepared by Vince Zaccone

For Campus Learning

Assistance Services at UCSB

Two Basic Concepts: Density and Pressure

Density measures how “tightly packed” an object is.

The definition is given by a formula:

Standard units for density are (you might be used to seeing in chemistry class)

One value you need to know is the density of water:

Pressure is used for fluids mainly because fluids don’t retain their shape, so it is not as

useful to measure the force on a fluid.

Definition of pressure:

V

m

3m

kg3cm

g

33watercm

g1

m

kg1000

A

FP

Important note: you can rearrange

this formula to get m=ρV. You will

need to do this a lot.

Prepared by Vince Zaccone

For Campus Learning

Assistance Services at UCSB

Two Basic Concepts: Density and Pressure

Density measures how “tightly packed” an object is.

The definition is given by a formula:

Standard units for density are (you might be used to seeing in chemistry class)

One value you need to know is the density of water:

Pressure is used for fluids mainly because fluids don’t retain their shape, so it is not as

useful to measure the force on a fluid.

Definition of pressure:

This takes into account the area as well as the force.

An example will help clarify this definition:

V

m

3m

kg3cm

g

33watercm

g1

m

kg1000

A

FP

Important note: you can rearrange

this formula to get m=ρV. You will

need to do this a lot.

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A thumbtack has a point with a diameter of 1mm. The other end has diameter 1cm. A

force of 10 Newtons is required to push it into the wall. Find the pressure on the wall,

and the pressure on the person’s thumb.

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A thumbtack has a point with a diameter of 1mm. The other end has diameter 1cm. A

force of 10 Newtons is required to push it into the wall. Find the pressure on the wall,

and the pressure on the person’s thumb.

Here are the calculations:

25

2thumb

27

2wall

m

N1027.1

2

m01.

N10P

m

N1027.1

2

m001.

N10P

A

FP

Notice that the pressure on the wall is 100x larger

because the area is 100x smaller.

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Pressure varies with depth in a fluid. The deeper you are under the surface, the larger

the pressure will be. Here is a formula we can use:

h is the depth under the surface of the fluid, and the gauge pressure is measured

relative to the pressure at the surface (usually atmospheric pressure).

ghPgauge

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Try this example:

A cubical box 20.0 cm on a side is completely immersed in a fluid. At the top of the box

the pressure is 105 kPa: at the bottom the pressure is 106.8 kPa. What is the density of

the fluid?

Pressure varies with depth in a fluid. The deeper you are under the surface, the larger

the pressure will be. Here is a formula we can use:

h is the depth under the surface of the fluid, and the gauge pressure is measured

relative to the pressure at the surface (usually atmospheric pressure).

Fluid

Air

20cm

ghPgauge

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Try this example:

A cubical box 20.0 cm on a side is completely immersed in a fluid. At the top of the box

the pressure is 105 kPa: at the bottom the pressure is 106.8 kPa. What is the density of

the fluid?

Pressure varies with depth in a fluid. The deeper you are under the surface, the larger

the pressure will be. Here is a formula we can use:

h is the depth under the surface of the fluid, and the gauge pressure is measured

relative to the pressure at the surface (usually atmospheric pressure).

Fluid

Air

20cm

htop

hbottom

Calculate the pressures using our formula,

then subtract and divide to get our answer:

topbottomtopbottom

bottombottom

toptop

hhgPP

ghP

ghP

ghPgauge

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Try this example:

A cubical box 20.0 cm on a side is completely immersed in a fluid. At the top of the box

the pressure is 105 kPa: at the bottom the pressure is 106.8 kPa. What is the density of

the fluid?

Pressure varies with depth in a fluid. The deeper you are under the surface, the larger

the pressure will be. Here is a formula we can use:

h is the depth under the surface of the fluid, and the gauge pressure is measured

relative to the pressure at the surface (usually atmospheric pressure).

Fluid

Air

20cm

htop

hbottom

Calculate the pressures using our formula,

then subtract and divide to get our answer:

3

2

m

kg

s

m

topbottomtopbottom

bottombottom

toptop

918

m2.08.9kPa105kPa8.106

hhgPP

ghP

ghP

ghPgauge

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Here is the submerged box again. The

pressure is exerted on the box in all directions

by perpendicular forces, as shown. The idea

that pressure in a fluid is applied in all

directions is called Pascal’s Law

Fluid

Air

F┴

F┴

F┴

F┴

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Here is the submerged box again. The

pressure is exerted on the box in all directions

by perpendicular forces, as shown. The idea

that pressure in a fluid is applied in all

directions is called Pascal’s Law

Pascal’s Law helps explain the idea of

Buoyancy (why objects float or sink)

Here’s the basic idea: since the pressure is

larger at the bottom of the box, the upward force

there is larger than the downward force on the

top, creating a net force upward on the box. We

call this the Buoyant Force. Note that the

horizontal forces on the sides cancel out.

Fluid

Air

F┴

F┴

F┴

F┴

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Here is a formula for Buoyant Force:

gVF displacedfluidBuoyant

Very Important Note: the density in this

formula is the density of the FLUID, not the

submerged object.Fluid

Air

F┴

F┴

F┴

F┴

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Here is a formula for Buoyant Force:

Very Important Note: the density in this

formula is the density of the FLUID, not the

submerged object.

Using the definition of density ( ), we

can also see that the buoyant force is the

WEIGHT of the displaced FLUID.

V

m

Fluid

Air

F┴

F┴

F┴

F┴

gVF displacedfluidBuoyant

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Here is a formula for Buoyant Force:

Very Important Note: the density in this

formula is the density of the FLUID, not the

submerged object.

Using the definition of density ( ), we

can also see that the buoyant force is the

WEIGHT of the displaced FLUID.

In the case of an object floating at the

surface of a fluid, the buoyant force is also

equal to the weight of the object.

V

m

Fluid

Air

F┴

F┴

F┴

F┴

gVF displacedfluidBuoyant

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Here is a formula for Buoyant Force:

Very Important Note: the density in this

formula is the density of the FLUID, not the

submerged object.

Using the definition of density ( ), we

can also see that the buoyant force is the

WEIGHT of the displaced FLUID.

In the case of an object floating at the

surface of a fluid, the buoyant force is also

equal to the weight of the object.

One more rule of thumb: if an object is

more dense than the fluid, it sinks; if the

object is less dense, it floats.

V

m

Fluid

Air

F┴

F┴

F┴

F┴

gVF displacedfluidBuoyant

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

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Water

AirHere is a picture of the submerged brick.

Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Prepared by Vince Zaccone

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Assistance Services at UCSB

Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirHere is a picture of the submerged brick.

Draw a free-body diagram of the forces on the brick.

mg

FT

FB

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirHere is a picture of the submerged brick.

Draw a free-body diagram of the forces on the brick.

Now we can write down Newton’s 2nd law:

0mgFFF BTnet

mg

FT

FB

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirHere is a picture of the submerged brick.

Draw a free-body diagram of the forces on the brick.

Now we can write down Newton’s 2nd law:

BT

BTnet

FmgF

0mgFFF

mg

FT

FB

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirHere is a picture of the submerged brick.

Draw a free-body diagram of the forces on the brick.

Now we can write down Newton’s 2nd law:

BT

BTnet

FmgF

0mgFFF

mg

FT

FB

We know the weight, and we can find the

buoyant force a couple of different ways:

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirHere is a picture of the submerged brick.

Draw a free-body diagram of the forces on the brick.

Now we can write down Newton’s 2nd law:

BT

BTnet

FmgF

0mgFFF

mg

FT

FB

We know the weight, and we can find the

buoyant force a couple of different ways:

Option 1: Use the standard formula

gVF dispfluidB

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirHere is a picture of the submerged brick.

Draw a free-body diagram of the forces on the brick.

Now we can write down Newton’s 2nd law:

BT

BTnet

FmgF

0mgFFF

mg

FT

FB

We know the weight, and we can find the

buoyant force a couple of different ways:

Option 1: Use the standard formula

gVF dispfluidB

The volume displaced is just the volume of the

brick (it is fully submerged) and we can find

that from the definition of density:

3

m

kgm00179.0

2800

kg5mV

3

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirHere is a picture of the submerged brick.

Draw a free-body diagram of the forces on the brick.

Now we can write down Newton’s 2nd law:

BT

BTnet

FmgF

0mgFFF

mg

FT

FB

We know the weight, and we can find the

buoyant force a couple of different ways:

Option 1: Use the standard formula

gVF dispfluidB

The volume displaced is just the volume of the

brick (it is fully submerged) and we can find

that from the definition of density:

3

m

kgm00178.0

2800

kg5mV

3

N5.178.9m00179.01000F23 s

m3

m

kgB

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirHere is a picture of the submerged brick.

Draw a free-body diagram of the forces on the brick.

Now we can write down Newton’s 2nd law:

BT

BTnet

FmgF

0mgFFF

mg

FT

FB

We know the weight, and we can find the

buoyant force a couple of different ways:

Option 2: The buoyant force is the weight

of the displaced fluid. Since the brick is

fully submerged, and we know the

densities, the weight of the fluid is just the

weight of the brick times the ratio of the

densities:

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirHere is a picture of the submerged brick.

Draw a free-body diagram of the forces on the brick.

Now we can write down Newton’s 2nd law:

BT

BTnet

FmgF

0mgFFF

mg

FT

FB

We know the weight, and we can find the

buoyant force a couple of different ways:

Option 2: The buoyant force is the weight

of the displaced fluid. Since the brick is

fully submerged, and we know the

densities, the weight of the fluid is just the

weight of the brick times the ratio of the

densities:

N5.178.9kg5ww2

brick

fluid

s

m28001000

brickfluid

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirHere is a picture of the submerged brick.

Draw a free-body diagram of the forces on the brick.

Now we can write down Newton’s 2nd law:

BT

BTnet

FmgF

0mgFFF

mg

FT

FB

Now that we have the buoyant force, we can calculate the tension

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirHere is a picture of the submerged brick.

Draw a free-body diagram of the forces on the brick.

Now we can write down Newton’s 2nd law:

BT

BTnet

FmgF

0mgFFF

mg

FT

FB

Now that we have the buoyant force, we can calculate the tension:

N5.31N5.178.9kg5F2s

mT

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirTo answer the 2nd question, just notice that when the

string is cut, the tension force goes away. This gives us

a new (simpler) free-body diagram.

mg

FB

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirTo answer the 2nd question, just notice that when the

string is cut, the tension force goes away. This gives us

a new (simpler) free-body diagram.

Write down Netwon’s 2nd law again:

mg

FBmamgFF Bnet

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirTo answer the 2nd question, just notice that when the

string is cut, the tension force goes away. This gives us

a new (simpler) free-body diagram.

Write down Netwon’s 2nd law again:

mg

FB

akg58.9kg5N5.17

mamgFF

2s

m

Bnet

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Example: A brick with mass 5kg and density 2800 kg/m3 is tied to a string and lowered into a pool of fresh water.

Find the tension in the string. If the string is cut, how fast will the brick accelerate toward the bottom of the pool?

Water

AirTo answer the 2nd question, just notice that when the

string is cut, the tension force goes away. This gives us

a new (simpler) free-body diagram.

Write down Netwon’s 2nd law again:

mg

FB

2

2

s

m

s

m

Bnet

3.6a

akg58.9kg5N5.17

mamgFF

Note: the acceleration is negative because the brick is sinking

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Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

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Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

Same basic setup as the previous problem, but the

block is floating. We can draw the picture and the

free-body diagram again:Water

Air

Bottom of pool

FB

mgFT

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Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

Same basic setup as the previous problem, but the

block is floating. We can draw the picture and the

free-body diagram again:

Newton’s 2nd law:

Water

Air

Bottom of pool

FB

mgFT

0FmgFF TBnet

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Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

Same basic setup as the previous problem, but the

block is floating. We can draw the picture and the

free-body diagram again:

Newton’s 2nd law:

Water

Air

Bottom of pool

FB

mgFT

0FmgFF TBnet

This time the buoyant force is greater than

the weight of the block because the density

of water is larger than the block’s density.

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Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

Same basic setup as the previous problem, but the

block is floating. We can draw the picture and the

free-body diagram again:

Newton’s 2nd law:

Water

Air

Bottom of pool

FB

mgFT

0FmgFF TBnet

This time the buoyant force is greater than

the weight of the block because the density

of water is larger than the block’s density.

Using option 2 from the previous problem:

N288.9kg2F2s

m700

1000B

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Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

Same basic setup as the previous problem, but the

block is floating. We can draw the picture and the

free-body diagram again:

Newton’s 2nd law:

Water

Air

Bottom of pool

FB

mgFT

N4.88.9kg2N28F

mgFF

0FmgFF

2s

mT

BT

TBnet

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Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

Water

Air

Bottom of pool

FB

mg

To find the acceleration when the string is cut, again

notice that the tension force goes away, then use

Newton’s 2nd law:a

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Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

FB

mg

To find the acceleration when the string is cut, again

notice that the tension force goes away, then use

Newton’s 2nd law:

mamgFF Bnet

Water

Air

Bottom of pool

FB

mg

a

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Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

FB

mg

To find the acceleration when the string is cut, again

notice that the tension force goes away, then use

Newton’s 2nd law:

akg28.9kg2N28

mamgFF

2s

m

Bnet

Water

Air

Bottom of pool

FB

mg

a

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Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

FB

mg

To find the acceleration when the string is cut, again

notice that the tension force goes away, then use

Newton’s 2nd law:

2

2

s

m

s

m

Bnet

2.4a

akg28.9kg2N28

mamgFF

Water

Air

Bottom of pool

FB

mg

a

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Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

For the last part, consider the forces on the block

when it is floating at the surface.

Water

Air

FB

mg

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Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

For the last part, consider the forces on the block

when it is floating at the surface.

Water

Air

FB

mgmgF

0mgFF

B

Bnet

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Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

For the last part, consider the forces on the block

when it is floating at the surface.

Water

Air

FB

mgmgF

0mgFF

B

Bnet

The buoyant force only has to be strong enough to

balance out the weight of the block, so the block

floats above the surface. If you split the block into 2

parts – the part above the surface and the part below,

it is only the part below that is displacing water.

Prepared by Vince Zaccone

For Campus Learning

Assistance Services at UCSB

Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

For the last part, consider the forces on the block

when it is floating at the surface.

Water

Air

FB

mgmgF

0mgFF

B

Bnet

The buoyant force only has to be strong enough to

balance out the weight of the block, so the block

floats above the surface. If you split the block into 2

parts – the part above the surface and the part below,

it is only the part below that is displacing water.

We can find the buoyant force from our standard

formula: gVF dispfluidB

Prepared by Vince Zaccone

For Campus Learning

Assistance Services at UCSB

Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

For the last part, consider the forces on the block

when it is floating at the surface.

Water

Air

FB

mgmgF

0mgFF

B

Bnet

The buoyant force only has to be strong enough to

balance out the weight of the block, so the block

floats above the surface. If you split the block into 2

parts – the part above the surface and the part below,

it is only the part below that is displacing water.

We can find the buoyant force from our standard

formula: gVF dispfluidB

Newton’s law becomes:

mggVdispfluid

Prepared by Vince Zaccone

For Campus Learning

Assistance Services at UCSB

Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

For the last part, consider the forces on the block

when it is floating at the surface.

Water

Air

FB

mgmgF

0mgFF

B

Bnet

The buoyant force only has to be strong enough to

balance out the weight of the block, so the block

floats above the surface. If you split the block into 2

parts – the part above the surface and the part below,

it is only the part below that is displacing water.

We can find the buoyant force from our standard

formula: gVF dispfluidB

Newton’s law becomes:

gVgV

mggV

blockblockdispfluid

dispfluid The mass of the block can be written in terms of the density

of the wood, so that we can get a more general result

Prepared by Vince Zaccone

For Campus Learning

Assistance Services at UCSB

Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

For the last part, consider the forces on the block

when it is floating at the surface.

Water

Air

FB

mgmgF

0mgFF

B

Bnet

The buoyant force only has to be strong enough to

balance out the weight of the block, so the block

floats above the surface. If you split the block into 2

parts – the part above the surface and the part below,

it is only the part below that is displacing water.

We can find the buoyant force from our standard

formula: gVF dispfluidB

Newton’s law becomes:

blockfluid

blockdisp

blockblockdispfluid

dispfluid

VV

gVgV

mggV

This gives us a formula for the portion of the block that is under the surface.

Prepared by Vince Zaccone

For Campus Learning

Assistance Services at UCSB

Now try an example with a floating object: A block of wood with mass 2kg and density 700kg/m3 is tied to a string

and fastened to the bottom of a pool of fresh water. Find the tension in the string. If the string is cut, how fast will

the block accelerate toward the top of the pool? Once it gets to the surface, how much of the block is above the

water?

For the last part, consider the forces on the block

when it is floating at the surface.

Water

Air

FB

mgmgF

0mgFF

B

Bnet

The buoyant force only has to be strong enough to

balance out the weight of the block, so the block

floats above the surface. If you split the block into 2

parts – the part above the surface and the part below,

it is only the part below that is displacing water.

We can find the buoyant force from our standard

formula: gVF dispfluidB

Newton’s law becomes:

blockblockdisp

blockfluid

blockdisp

blockblockdispfluid

dispfluid

V%70V1000

700V

VV

gVgV

mggV

The rest of the block is above the

water, so the answer is 30%Prepared by Vince Zaccone

For Campus Learning

Assistance Services at UCSB