Mastering Physics HW 1 Ch 15 - Fluids and Elasticity
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Transcript of Mastering Physics HW 1 Ch 15 - Fluids and Elasticity
HW 1 Ch 15 Fluids and ElasticityDue: 11:59pm on Tuesday, September 8, 2015
To understand how points are awarded, read the Grading Policy for this assignment.
Force on a Goldfish Vector Drawing
A fishbowl contains a single goldfish and is filled with water to the level indicated.
Part A
At each of the designated points, rotate the given vector to indicate the direction of the force exerted by the wateron either the inside of the fishbowl (for points A and B) or the outside of the goldfish (for points C, D, and E).
The orientation of your vectors will be graded.
Hint 1. Direction of the force due to the water
The water in the fishbowl will exert a force perpendicular to every surface it contacts. The pressure of thewater, like the pressure in any fluid, pushes the water outward, equally in all directions, exerting forces onwhatever it comes into contact with.
ANSWER:
Correct
Pressure on a Dam
As the reservoir behind a dam is filled with water, the pressure that the water exerts on the dam increases. Eventually,the force on the dam becomes substantial, and it could cause the dam to collapse. There are two significant issues tobe considered: First, the base of the dam should be able to withstand the pressure , where is the density of thewater behind the dam, is its depth, and is the magnitude of the acceleration due to gravity. This means that thematerial of which the dam is made needs to be strong enough so that it doesn't crack (compressive strength).
The second issue has to do with the strength of the foundation of the dam. The water pressure exerts a clockwisetorque on the dam, as shown in the figure. The foundation of the dam should be strong enough so that the dam doesnot topple. The material has to be strong enough that the dam does not snap (shear strength).
To study this phenomenon, consider the simple model of a dam depicted in the diagram. A reservoir of water (density ) behind the dam is filled to a height . Assume that the
width of the dam (the dimension pointing into the screen) is .
Part A
Consider a horizontal layer of the dam wall of thickness located a distance above the reservoir floor. What isthe magnitude of the force on this layer due to the water in the reservoir?
Express your answer in terms of , , the magnitude of the acceleration due to gravity , and anyquantities from the problem introduction.
Hint 1. Find the pressure as a function of depth
What is the pressure of the water at a vertical height above the floor of the reservoir?
Express your answer in terms of and any quantities from the problem introduction.
ANSWER:
ANSWER:
ρgh ρh g
ρ h
L
dx xdF
x dx g
p(x) x
x
= p(x)
Correct
Part B
The force of the water produces a torque on the dam. In a simple model, if the torque due to the water were enoughto cause the dam to break free from its foundation, the dam would pivot about its base (point P). What is themagnitude of the torque about the point P due to the water in the reservoir?
Express your answer in terms of quantities given in the problem introduction and the magnitude of theacceleration due to gravity .
Hint 1. How to approach the problem
Using the answer to Part A, find the torque on a horizontal layer of the dam. Then integrate over the entireheight of the dam to find the total torque.
Hint 2. Definition of torque
The torque due to a force about a point is the product of the magnitude of the force and the moment arm,which is the perpendicular distance from the line of action of the force to the pivot.
Hint 3. Find the torque on a horizontal layer of the dam
What is the torque due to the water acting on a horizontal layer of the dam of thickness at position above the bottom of the reservoir?
Express your answer in terms of , , the gravitational accleration , and any quantities neededfrom the problem introduction.
Hint 1. Definition of torque
The torque due to a force about a point is the product of the magnitude of the force and the momentarm, which is the perpendicular distance from the line of action of the force to the pivot.
ANSWER:
Hint 4. Some helpful integrals
and
.
= dF ρg(h − x)Ldx
τ
g
dτ dx x
x dx g
= dτ
x dx =∫ a
0a2
2
dx =∫ a
0 x2 a3
3
ANSWER:
Correct
Thus the average moment arm is . Therefore, since the water exerts a greater force at the
bottom of the dam than at the top, the net effect is as if the force were acting at a height of from thebottom.
Fluid Pressure in a UTube
A Utube is filled with water, and the two arms are capped.The tube is cylindrical, and the right arm has twice the radiusof the left arm. The caps have negligible mass, arewatertight, and can freely slide up and down the tube.
Part A
A oneinch depth of sand is poured onto the cap on each arm. After the caps have moved (if necessary) toreestablish equilibrium, is the right cap higher, lower, or the same height as the left cap?
= τ ρgL16
h3
=torquetotal force
h3
h3
Hint 1. Pressure at the surface
The pressure at the base of each arm depends on the pressure at the surface of each arm, not the weight atthe surface of each arm.
Hint 2. Evaluate the weight of the sand
Compare the weight of the sand on the two caps. Which of the following is true?
ANSWER:
Correct
Since the left cap has half the radius of the right cap, and both caps have a oneinchdeep layer ofsand on them, there is much less sand on the left cap.
Hint 3. Evaluate the pressure applied by the sand
Now compare the pressure exerted by the sand on the two caps. Which of the following is true?
ANSWER:
The weight of the sand on the right cap is
greater than
less than
equal to the weight of the sand on the left cap.
The pressure exerted by the sand on the right cap is
greater than
less than
equal to
the pressure exerted by the sand on the left cap.
Correct
Since the right cap has twice the radius of the left cap, there is four times as much sand on the rightcap. However, the area of the right cap is four times as great as that of the left cap (which is whythere’s four times as much sand!), so the pressures on the left and right caps are equal.
ANSWER:
Correct
Although one inch of sand on the right cap is much heavier than one inch of sand on the left cap, thepressures exerted by the sand are the same on both caps. Since the pressures exerted by the sand are equal,the pressures at the base of each arm due to the water must be equal. This requires equal heights of water inthe two arms.
Part B
The sand is removed and a 1.0 mass block is placed on each cap. After the caps have moved (if necessary) toreestablish equilibrium, is the right cap higher, lower, orthe same height as the left cap?
Hint 1. Evaluate the pressure applied by the 1.0 blocks
Compare the pressure exerted on the water from the two 1 blocks. Which of the following is true?
ANSWER:
higher
lower
the same height
kg
kg
kg
Correct
The right cap has four times the area of the left cap. Since the masses are identical, this results in apressure onefourth as large on the right cap.
ANSWER:
Correct
Although the masses of the blocks are equal, the pressures exerted by them on the caps are not equal. Thereis a greater pressure on the left cap, which results in a greater pressure at the base of the left arm. Tocompensate for this increased pressure, the height of the water column in the right arm will have to be greaterthan in the left arm.
Part C
If a 1.0 mass block is on the left cap, how much total mass must be placed on the right cap so that the capsequilibrate at equal height?
Express your answer in kilograms.
Hint 1. Meaning of equal water levels
Equal water levels in the two arms require equal pressures on the two caps. Remember that pressure isforce divded by area and that area is proportional to the square of the radius.
ANSWER:
Correct
Part D
The locations of the two caps at equilibrium are now as given in this figure. The dashed line represents the level of
The pressure exerted on the right cap is
greater than
less than
equal to the pressure exerted on the left cap.
higher
lower
the same height
kg
4.0 kg
the water in the left arm.What is the mass of the water located in the right armbetween the dashed line and the right cap?
Express your answer in kilograms.
Hint 1. Pressure at the dashed line
The pressure at the location of the dashed line must be equal in the two arms. Therefore, the total massabove the line in the left arm, divided by the area of the left arm, must equal the total mass above the line inthe right arm, divided by the area of the right arm.
ANSWER:
Correct
Pressure in the Ocean
The pressure at 10 below the surface of the ocean is about 2.00×105 .
Part A
Which of the following statements is true?
Hint 1. How to approach the problem
Since pressure is defined as force per unit area, the pressure at a given depth below the surface of theocean is the normal force on a small area at that depth divided by that area. Thus, given the pressure at 10 you can determine the normal force on a unit area at that depth. Also recall that, in general, the pressure
in a fluid varies with height, but it also depends on the external pressure applied to the fluid. In the case ofthe ocean, the external pressure applied at its surface is the atmospheric pressure.
ANSWER:
3.0 kg
m Pa
m
Correct
The pressure at a given level in a fluid is caused by the weight of the overlying fluid plus any externalpressure applied to the fluid. In this case, the external pressure is the atmospheric pressure. Tropical stormsare formed by lowpressure systems in the atmosphere. When such a storm forms over the ocean, both theatmospheric pressure and the fluid pressure in the ocean decrease.
Part B
Now consider the pressure 20 below the surface of the ocean. Which of the following statements is true?
Hint 1. Variation of pressure with height
If the pressure at a point in a fluid is , then, at a distance below this point, the pressure in the fluid is
,where is the density of the fluid and is the acceleration due to gravity. This means that the pressure ata depth below a reference height of pressure is greater than by an amount . Note that thequantity is related to the weight of a column of fluid of height and 1 in cross section.
ANSWER:
Correct
Tactics Box 15.2 Finding Whether an Object Floats or Sinks
Learning Goal:
To practice Tactics Box 15.2 Finding whether an object floats or sinks.
If you hold an object underwater and then release it, it can float to the surface, sink, or remain “hanging” in the water,depending on whether the fluid density is larger than, smaller than, or equal to the object’s average density .
The weight of a column of seawater 1 in cross section and 10 high is about 2.00×105 .
The weight of a column of seawater 1 in cross section and 10 high plus the weight of a column ofair with the same cross section extending up to the top of the atmosphere is about 2.00×105 .
The weight of 1 of seawater at 10 below the surface of the ocean is about 2.00×105 .
The density of seawater is about 2.00×105 times the density of air at sea level.
m2 m N
m2 mN
m3 m N
m
p0 h p
p = + ρghp0ρ g
h p0 p0 ρgh
ρgh h m2
The pressure is twice that at a depth of 10 .
The pressure is the same as that at a depth of 10 .
The pressure is equal to that at a depth of 10 plus the weight per 1 cross sectional area of acolumn of seawater 10 high.
The pressure is equal to the weight per 1 cross sectional area of a column of seawater 20 high.
mm
m m2
m
m2 m
ρf ρavg
These conditions are summarized in this Tactics Box.
TACTICS BOX 15.2 Finding whether an object floats or sinksObject sinks Object floats Object has neutral buoyancy
An object sinks if it weighsmore than the fluid itdisplaces, that is, if itsaverage density is greaterthan the density of the fluid:
.
An object floats on the surfaceif it weighs less than the fluidit displaces, that is, if itsaverage density is less thanthe density of the fluid:
.
An object hangs motionless in thefluid if it weighs exactly the same asthe fluid it displaces. It has neutralbuoyancy if its average densityequals the density of the fluid:
.
Part A
Ice at 0.0 has a density of 917 . A 3.00 ice cube is gently released inside a small container filledwith oil and is observed to be neutrally buoyant. What is the density of the oil, ?
Express your answer in kilograms per meter cubed to three significant figures.
ANSWER:
Correct
Part B
Once the ice cube melts, what happens to the liquid water that it produces?
Hint 1. Density of water
Water has a density of 1000 .
ANSWER:
>ρavg ρf <ρavg ρf =ρavg ρf
C∘ kg/m3 cm3
ρoil
= 917 ρoil kg/m3
kg/m3
Correct
Part C
What happens if some ethyl alcohol of density 790 is poured into the container after the ice cube hasmelted?
ANSWER:
Correct
Score Summary:Your score on this assignment is 97.1%.You received 4.86 out of a possible total of 5 points.
The liquid water sinks to the bottom of the container.
The liquid water rises to the surface and floats on top of the oil.
The liquid water is in static equilibrium at the location where the ice cube was originally placed.
kg/m3
A layer of ethyl alcohol forms between the oil and the water.
The layer of ethyl alcohol forms at the bottom of the container.
The layer of ethyl alcohol forms on the surface.