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Fluids - Hydrostatics. Physics 6B. 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:. Prepared by Vince Zaccone - PowerPoint PPT Presentation
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Fluids - Hydrostatics
Physics 6B
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: Vm
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: Vm
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)
Vm
3mkg
3cmg
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:
Vm
3mkg
3cmg
33watercm
g1mkg1000
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:
Vm
3mkg
3cmg
33watercm
g1mkg1000
AFP
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:
Vm
3mkg
3cmg
33watercm
g1mkg1000
AFP
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
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.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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
mN1027.1
2m01.
N10P
m
N1027.1
2m001.
N10P
AFP
Notice that the pressure on the wall is 100x larger because the area is 100x smaller.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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
hhgPPghP
ghP
ghPgauge
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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
mkg
sm
topbottomtopbottom
bottombottom
toptop
918
m2.08.9kPa105kPa8.106
hhgPPghP
ghP
ghPgauge
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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┴
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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┴
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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┴
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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.
Vm
Fluid
Air
F┴
F┴
F┴
F┴
gVF displacedfluidBuoyant
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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.
Vm
Fluid
Air
F┴
F┴
F┴
F┴
gVF displacedfluidBuoyant
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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.
Vm
Fluid
Air
F┴
F┴
F┴
F┴
gVF displacedfluidBuoyant
Prepared by Vince Zaccone
For Campus Learning 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?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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
For Campus Learning 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
Prepared by Vince Zaccone
For Campus Learning 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.
Now we can write down Newton’s 2nd law:
0mgFFF BTnet
mg
FT
FB
Prepared by Vince Zaccone
For Campus Learning 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.
Now we can write down Newton’s 2nd law:
BT
BTnet
FmgF0mgFFF
mg
FT
FB
Prepared by Vince Zaccone
For Campus Learning 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.
Now we can write down Newton’s 2nd law:
BT
BTnet
FmgF0mgFFF
mg
FT
FB
We know the weight, and we can find the buoyant force a couple of different ways:
Prepared by Vince Zaccone
For Campus Learning 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.
Now we can write down Newton’s 2nd law:
BT
BTnet
FmgF0mgFFF
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
Prepared by Vince Zaccone
For Campus Learning 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.
Now we can write down Newton’s 2nd law:
BT
BTnet
FmgF0mgFFF
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
mkg m00179.0
2800kg5mV
3
Prepared by Vince Zaccone
For Campus Learning 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.
Now we can write down Newton’s 2nd law:
BT
BTnet
FmgF0mgFFF
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
mkg m00178.0
2800kg5mV
3
N5.178.9m00179.01000F 23 sm3
mkg
B
Prepared by Vince Zaccone
For Campus Learning 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.
Now we can write down Newton’s 2nd law:
BT
BTnet
FmgF0mgFFF
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:
Prepared by Vince Zaccone
For Campus Learning 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.
Now we can write down Newton’s 2nd law:
BT
BTnet
FmgF0mgFFF
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:
N5.178.9kg5ww 2brick
fluid
sm
28001000
brickfluid
Prepared by Vince Zaccone
For Campus Learning 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.
Now we can write down Newton’s 2nd law:
BT
BTnet
FmgF0mgFFF
mg
FT
FB
Now that we have the buoyant force, we can calculate the tension
Prepared by Vince Zaccone
For Campus Learning 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.
Now we can write down Newton’s 2nd law:
BT
BTnet
FmgF0mgFFF
mg
FT
FB
Now that we have the buoyant force, we can calculate the tension:
N5.31N5.178.9kg5F 2sm
T
Prepared by Vince Zaccone
For Campus Learning 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
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
Prepared by Vince Zaccone
For Campus Learning 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
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
Prepared by Vince Zaccone
For Campus Learning 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
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
2sm
Bnet
Prepared by Vince Zaccone
For Campus Learning 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
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
sm
sm
Bnet
3.6a
akg58.9kg5N5.17
mamgFF
Note: the acceleration is negative because the brick is sinking
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?
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?
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
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?
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
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?
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.
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?
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:
N288.9kg2F 2sm
7001000
B
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?
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
mgFF0FmgFF
2sm
T
BT
TBnet
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?
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
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?
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
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?
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
2sm
Bnet
Water
Air
Bottom of pool
FB
mg
a
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?
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
sm
sm
Bnet
2.4a
akg28.9kg2N28
mamgFF
Water
Air
Bottom of pool
FB
mg
a
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.
WaterAir
FB
mg
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.
WaterAir
FB
mgmgF0mgFF
B
Bnet
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.
WaterAir
FB
mgmgF0mgFF
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.
WaterAir
FB
mgmgF0mgFF
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.
WaterAir
FB
mgmgF0mgFF
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.
WaterAir
FB
mgmgF0mgFF
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.
WaterAir
FB
mgmgF0mgFF
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.
WaterAir
FB
mgmgF0mgFF
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%70V1000700V
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
