skiladæmi 10
Due: 11:59pm on Wednesday, November 11, 2015
You will receive no credit for items you complete after the assignment is due. Grading Policy
Alternative Exercise 11.125
⃗ A bar with cross­sectional area A is subjected to equal and opposite tensile forces F at its ends. Consider a plane through
the bar making an angle θ with a plane at right angles to the bar (the figure ).
Part A
What is the tensile (normal) stress at this plane in terms of F , A, and θ?
ANSWER:
F
A
2
cos (θ)
Correct
Part B
What is the shear (tangential) stress at the plane in terms of F , A, and θ?
ANSWER:
F
A
cos(θ)sin(θ)
Correct
Part C
For what value of θ is the tensile stress a maximum?
ANSWER:
= 0 ∘ θ
Correct
Part D
For what value of θ is the shear stress a maximum?
ANSWER:
= 45 ∘ θ
Correct
Alternative Exercise 11.116
A moonshiner produces pure ethanol (ethyl alcohol) late at night and stores it in a stainless steel tank in the form of a
cylinder 0.310 m in diameter with a tight­fitting piston at the top. The total volume of the tank is 260 L . In an attempt to
squeeze a little more into the tank, the moonshiner piles lead breaks of total mass 1430 kg on top of the piston.
Part A
What additional volume of ethanol can the moonshiner squeeze into the tank? (Assume that the wall of the tank is
perfectly rigid.)
ANSWER:
V
= 5.31×10−2 L Correct
± A Wire under Stress
A steel wire of length 2.01 m with circular cross section must stretch no more than 0.200 cm when a tensile (stretching)
force of 380 N is applied to each end of the wire.
Part A
What minimum diameter dmin is required for the wire?
Express your answer in millimeters. Take Young's modulus for steel to be Y = 2.00×1011 Pa .
Hint 1. How to approach the problem
Recall that Young's modulus is defined as
Y =
tensile stress
.
tensile strain
Compute the strain in terms of quantities given in the problem introduction, and write the stress in terms of given
quantities and the unknown wire diameter. Use these along with the given value of Young's modulus for steel to
solve for the diameter of the wire.
Hint 2. Calculate the tensile strain
Calculate the tensile strain on the wire.
Hint 1. Definition of tensile strain
Tensile strain is defined as the ratio of the elongation ΔL to the original length L0 of a material that is
under stress:
tensile strain =
ΔL
L0
.
ANSWER:
tensile strain = 9.95×10−4
Correct
Hint 3. Definition of tensile stress
Tensile stress is defined as the force perpendicular to the surface of a material divided by the cross­sectional
area of the surface:
tensile stress =
F⊥
A
.
Hint 4. Relation between the area and the diameter
The relation between the area A and the diameter d of a circle is
A=π
d
2
4
.
ANSWER:
dmin
= 1.56 mm Correct
Note that you were asked for the minimum diameter. Where does this figure?
The extension is directly proportional to the stress, i.e., the force per unit area. One way to decrease the stress is
to increase the surface area over which the stretching force is applied. So any diameter (and so area) greater than
the one you calculated would serve to keep the extension within the tolerance specified (i.e., the maximum
allowable extension).
± Young's Modulus
Learning Goal:
To understand the meaning of Young's modulus, to perform some real­life calculations related to stretching steel, a common
construction material, and to introduce the concept of breaking stress.
Hooke's law states that for springs and other "elastic" objects
F = kΔx,
where F is the magnitude of the stretching force, Δx is the corresponding elongation of the spring from equilibrium, and k
is a constant that depends on the geometry and the material of the spring. If the deformations are small enough, most
materials, in fact, behave like springs: Their deformation is directly proportional to the external force. Therefore, it may be
useful to operate with an expression that is similar to Hooke's law but describes the properties of various materials, as
L
opposed to objects such as springs. Such an expression does exist. Consider, for instance, a bar of initial length L and
cross­sectional area A stressed by a force of magnitude F . As a result, the bar stretches by ΔL .
Let us define two new terms:
Tensile stress is the ratio of the stretching force to the
cross­sectional area:
stress =
F
A
.
Tensile strain is the ratio of the elongation of the rod to the
initial length of the bar:
ΔL
strain =
L
.
It turns out that the ratio of the tensile stress to the tensile strain
is a constant as long as the tensile stress is not too large. That
constant, which is an inherent property of a material, is called
Young's modulus and is given by
Y =
F /A
ΔL/L
.
Part A
What is the SI unit of Young's modulus?
Hint 1. Look at the dimensions
If you look at the dimensions of Young's modulus, you will see that they are equivalent to the dimension of
pressure. Use the SI unit of pressure.
ANSWER:
Pa
Correct
Part B
Consider a metal bar of initial length L and cross­sectional area A. The Young's modulus of the material of the bar is Y . Find the "spring constant" k of such a bar for low values of tensile strain.
Express your answer in terms of Y , L, and A.
Hint 1. Use the definition of Young's modulus
Consider the equation defining Y . Then isolate F and compare the result with Hooke's law: F
ANSWER:
= k
Y
A
L
Correct
Part C
L
k
= kΔx
.
Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end, so
that the resultant wire has length 10L. What is the "spring constant" of the resulting wire?
Hint 1. The spring constant
Use the expression for the spring constant determined in Part B. From the expression derived in the Part B, you
can determine what happens to the spring constant when the length of the spring increases.
ANSWER:
0.1k k 10k 100k
Correct
Part D
Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are slightly twisted together,
so that the resultant wire has length L and its cross­sectional area is ten times that of the individual wire. What is the
"spring constant" of the resulting wire?
Hint 1. The spring constant
Use the expression for the spring constant determined in Part B. From the expression derived in Part B, you can
determine what happens to the spring constant when the area of the spring increases.
ANSWER:
0.1k k 10k 100k
Correct
Part E
Ten identical steel wires have equal lengths L and equal "spring constants" k. The Young's modulus of each wire is Y .
The wires are connected end to end, so that the resultant wire has length 10L. What is the Young's modulus of the
resulting wire?
ANSWER:
0.1Y Y 10Y 100Y
Correct
Part F
Ten identical steel wires have equal lengths L and equal "spring constants" k. The Young's modulus of each wire is Y .
The wires are slightly twisted together, so that the resultant wire has length L and is ten times as thick as the
individual wire. What is the Young's modulus of the resulting wire?
ANSWER:
0.1Y Y 10Y 100Y
Correct
By rearranging the wires, we create a new object with new mechanical properties. However, Young's modulus
depends on the material, which remains unchanged. To change the Young's modulus, one would have to change
the properties of the material itself, for instance by heating or cooling it.
Part G
Consider a steel guitar string of initial length L = 1.00 meter and cross­sectional area A = 0.500 square millimeters.
The Young's modulus of the steel is Y = 2.0 × 10 11 pascals. How far ( ΔL ) would such a string stretch under a
tension of 1500 newtons?
Use two significant figures in your answer. Express your answer in millimeters.
ANSWER:
ΔL
= 15 mm Correct
Steel is a very strong material. For these numeric values, you may assume that Hooke's law holds. However, for
greater values of tensile strain, the material no longer behaves elastically. If the strain and stress are large
enough, the material deteriorates. The final part of this problem illustrates this point and gives you a sense of the
"stretching limit" of steel.
Part H
Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earth
has proven much more difficult. The deepest mines ever drilled are only about 10 miles deep. To illustrate the
difficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubic
meter, and its breaking stress, defined as the maximum stress the material can bear without deteriorating, is about 9
2.0 × 10 pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that the
magnitude of the acceleration due to gravity remains constant at 9.8 meters per second per second.
Use two significant figures in your answer, expressed in kilometers.
Hint 1. Why does the cable break?
The cable breaks because of the stress exerted on it by its own weight. At the moment that the breaking stress
is reached, the stress at the top of the cable reaches its maximum, and the material begins to deteriorate.
Introduce an arbitrary cross­sectional area of the cable (which will cancel out of the final answer). The mass of
the cable below the top point can be found as the product of its volume and its density. Use this to find the force
at the top that will lead to the breaking stress.
Hint 2. Find the stress in the cable
Assume that the cable has cross­sectional area A and length L. The density is ρ. The maximum stress in the
cable is at the very top, where it has to support its own weight. What is this maximum stress?
Express your answer in terms of ρ, L, and g, the magnitude of the acceleration due to gravity. Recall that
the stress is the force per unit area, so the area will not appear in your expression.
ANSWER:
maximum stress = ρLg
ANSWER:
26 km Correct
This is only about 16 miles, and we have assumed that no extra load is attached. By the way, this length is small
enough to justify the assumption of virtually constant acceleration due to gravity. When making such assumptions,
one should always check their validity after obtaining a result.
Problem 11.76
Two identical, uniform beams weighing 260 N each are connected at one end by a frictionless hinge. A light horizontal
crossbar attached at the midpoints of the beams maintains an angle of 53.0 ∘ between the beams. The beams are
suspended from the ceiling by vertical wires such that they form a "V", as shown in the figure .
Part A
What force does the crossbar exert on each beam?
ANSWER:
F
= 130 N Correct
Part B
Is the crossbar under tension or compression?
ANSWER:
tension
compression
Correct
Part C
What is the magnitude of the force that the hinge at point A exerts on each beam?
ANSWER:
F
= 130 N Correct
Part D
What is the direction of the force that the hinge at point A exerts on the right­hand beam?
ANSWER:
= 180 ∘ with the direction to the right ϕ
Correct
Part E
What is the direction of the force that the hinge at point A exerts on the left­hand beam?
ANSWER:
= 0 ∘ with the direction to the right ϕ
Correct
Understanding Bernoulli's Equation
Bernoulli's equation is a simple relation that can give useful insight into the balance among fluid pressure, flow speed, and
elevation. It applies exclusively to ideal fluids with steady flow, that is, fluids with a constant density and no internal friction
forces, whose flow patterns do not change with time. Despite its limitations, however, Bernoulli's equation is an essential
tool in understanding the behavior of fluids in many practical applications, from plumbing systems to the flight of airplanes.
For a fluid element of density ρ that flows along a streamline, Bernoulli's equation states that
p1 + ρgh1 +
1
2
2
ρv
1
= p2 + ρgh2 +
1
2
2
ρv
2
,
where p is the pressure, v is the flow speed, h is the height, g is the acceleration due to gravity, and subscripts 1 and 2
refer to any two points along the streamline. The physical interpretation of Bernoulli's equation becomes clearer if we
rearrange the terms of the equation as follows:
p1 − p2 = ρg(h2 − h1 ) +
1
2
2
ρ(v
2
− v ).
2
1
The term p1 − p2 on the left­hand side represents the total work done on a unit volume of fluid by the pressure forces of
the surrounding fluid to move that volume of fluid from point 1 to point 2. The two terms on the right­hand side represent,
2
respectively, the change in potential energy, ρg(h2 − h1 ), and the change in kinetic energy, ρ(v2
− v ), of the unit
2
1
1
2
volume during its flow from point 1 to point 2. In other words, Bernoulli's equation states that the work done on a unit volume
of fluid by the surrounding fluid is equal to the sum of the change in potential and kinetic energy per unit volume that occurs
during the flow. This is nothing more than the statement of conservation of mechanical energy for an ideal fluid flowing along
a streamline.
Part A
Consider the portion of a flow tube shown in the figure. Point
1 and point 2 are at the same height. An ideal fluid enters the
flow tube at point 1 and moves steadily toward point 2. If the
cross section of the flow tube at point 1 is greater than that at
point 2, what can you say about the pressure at point 2?
Hint 1. How to approach the problem
Apply Bernoulli's equation to point 1 and to point 2. Since the points are both at the same height, their elevations
cancel out in the equation and you are left with a relation between pressure and flow speeds. Even though the
problem does not give direct information on the flow speed along the flow tube, it does tell you that the cross
section of the flow tube decreases as the fluid flows toward point 2. Apply the continuity equation to points 1 and
2 and determine whether the flow speed at point 2 is greater than or smaller than the flow speed at point 1. With
that information and Bernoulli's equation, you will be able to determine the pressure at point 2 with respect to the
pressure at point 1.
Hint 2. Apply Bernoulli's equation
Apply Bernoulli's equation to point 1 and to point 2 to complete the expression below. Here p and v are the
pressure and flow speed, respectively, and subscripts 1 and 2 refer to point 1 and point 2. Also, use h for
elevation with the appropriate subscript, and use ρ for the density of the fluid.
Express your answer in terms of some or all of the variables p1 , v 1 , h1 , p2 , v 2 , h2 , and ρ.
Hint 1. Flow along a horizontal streamline
Along a horizontal streamline, the change in potential energy of the flowing fluid is zero. In other words,
when applying Bernoulli's equation to any two points of the streamline, h1 = h2 and they cancel out.
ANSWER:
p1 +
1
2
2
ρv
1
= p2 +
ρv2
2
2
Hint 3. Determine v2 with respect to v1
By applying the continuity equation, determine which of the following is true.
Hint 1. The continuity equation
The continuity equation expresses conservation of mass for incompressible fluids flowing in a tube. It
says that the amount of fluid ΔV flowing through a cross section A of the tube in a time interval Δt
must be the same for all cross sections, or
ΔV
Δt
= A1 v1 = A2 v2 .
Therefore, the flow speed must increase when the cross section of the flow tube decreases, and vice
versa.
ANSWER:
v2
> v1
v2
= v1
v2
< v1
ANSWER:
lower than the pressure at point 1.
The pressure at point 2 is equal to the pressure at point 1.
higher than the pressure at point 1.
Correct
Thus, by combining the continuity equation and Bernoulli's equation, one can characterize the flow of an ideal
fluid.When the cross section of the flow tube decreases, the flow speed increases, and therefore the pressure
decreases. In other words, if A2 < A1 , then v 2 > v 1 and p2 < p1 .
Part B
As you found out in the previous part, Bernoulli's equation tells us that a fluid element that flows through a flow tube
with decreasing cross section moves toward a region of lower pressure. Physically, the pressure drop experienced by
the fluid element between points 1 and 2 acts on the fluid element as a net force that causes the fluid to __________.
Hint 1. Effects from conservation of mass
Recall that, if the cross section A of the flow tube varies, the flow speed v must change to conserve mass.
This means that there is a nonzero net force acting on the fluid that causes the fluid to increase or decrease
speed depending on whether the fluid is flowing through a portion of the tube with a smaller or larger cross
section.
ANSWER:
decrease in speed
increase in speed
remain in equilibrium
Correct
Part C
Now assume that point 2 is at height h with respect to point 1, as shown in the figure. The ends of the flow tube have
the same areas as the ends of the horizontal flow tube shown
in Part A. Since the cross section of the flow tube is
decreasing, Bernoulli's equation tells us that a fluid element
flowing toward point 2 from point 1 moves toward a region of
lower pressure. In this case, what is the pressure drop
experienced by the fluid element?
Hint 1. How to approach the problem
Apply Bernoulli's equation to point 1 and to point 2, as you did in Part A. Note that this time you must take into
account the difference in elevation between points 1 and 2. Do you need to add this additional term to the other
term representing the pressure drop between the two ends of the flow tube or do you subtract it?
ANSWER:
smaller than the pressure drop occurring in a purely horizontal flow.
The pressure drop is equal to the pressure drop occurring in a purely horizontal flow.
larger than the pressure drop occurring in a purely horizontal flow.
Correct
Part D
From a physical point of view, how do you explain the fact that the pressure drop at the ends of the elevated flow tube
from Part C is larger than the pressure drop occurring in the similar but purely horizontal flow from Part A?
Hint 1. Physical meaning of the pressure drop in a tube
As explained in the introduction, the difference in pressure p1 − p2 between the ends of a flow tube represents
the total work done on a unit volume of fluid by the pressure forces of the surrounding fluid to move that volume
of fluid from one end to the other end of the flow tube.
ANSWER:
increase in potential energy from the elevation change.
decrease in potential energy from the elevation change.
A greater amount of work is needed to balance the larger increase in kinetic energy.
larger decrease in kinetic energy.
Correct
In the case of purely horizontal flow, the difference in pressure between the two ends of the flow tube had to
balance only the increase in kinetic energy resulting from the acceleration of the fluid. In an elevated flow tube, the
difference in pressure must also balance the increase in potential energy of the fluid; therefore a higher pressure is
needed for the flow to occur.
Water Flowing from a Tank
Water flows steadily from an open tank as shown in the figure.
The elevation of point 1 is 10.0 meters, and the elevation of points
2 and 3 is 2.00 meters. The cross­sectional area at point 2 is
0.0480 square meters; at point 3, where the water is discharged, it
is 0.0160 square meters. The cross­sectional area of the tank is
very large compared with the cross­sectional area of the pipe.
Part A
Assuming that Bernoulli's equation applies, compute the discharge rate dV
dt
.
Express your answer in cubic meters per second.
Hint 1. How to approach the problem
The discharge rate is the rate at which a given volume of water flows across the exit of the pipe per unit time. It
is also defined as volume flow rate, and it depends on both the cross­sectional area of the pipe at the exit and
the fluid speed at that point.
Hint 2. The volume flow rate
Consider a steadily moving incompressible fluid, and let A denote the cross­sectional area of a flow tube. The
volume ΔV of fluid flowing across the cross section of area A at speed v during a small interval of time Δt is
given by AvΔt. Therefore, the rate at which fluid volumes cross a portion of the flow tube is
dV
dt
Hint 3. Find the fluid speed at the end of the pipe
= Av .
Assuming that Bernoulli's equation applies, find the speed v 3 of the water at point 3. Recall that the area of the
tank is very large compared to the cross­sectional area of the pipe, and consequently, the velocity of water at a
point on the surface of the water in the tank may be considered to be zero.
Express your answer in meters per second to three significant figures.
Hint 1. Apply Bernoulli's principle
Let pa be the atmospheric pressure and ρ the density of water. Consider the entire volume of water as a
single flow tube and apply Bernoulli's principle to point 3 and to point 1. Complete the expression below,
where v 3 is the fluid speed at point 3.
Express your answer in terms of pa , ρ, and g the free­fall acceleration due to gravity.
Hint 1. Bernoulli's principle
For the steady flow of an incompressible fluid with no internal friction, the pressure p and the flow
speed v at depth H below the surface are linked by an important relationship, known as
Bernoulli's principle. In particular, at any point at depth H along a flow tube, the following relation
holds:
p + ρgH +
1
2
2
ρv
= constant,
where ρ is the density of the fluid and g is the accereleration due to gravity.
Since Bernoulli's principle is valid at any point along a flow tube, it takes the form
p
1
+ ρgH 1 +
1
2
2
ρv
1
= p
2
+ ρgH 2 +
1
2
2
ρv
2
when applied to two distinct points along a flow tube. The subscripts 1 and 2 refer to such points.
ANSWER:
pa + 2ρg +
1
2
2
ρv
3
= p
a
+ 10ρg
ANSWER:
v3
= 12.5 m/s ANSWER:
dV
dt
3
= 0.200 m /s Correct
Part B
What is the gauge pressure at point 2?
Express your answer in pascals.
Hint 1. Definition of gauge pressure
Gauge pressure is defined as the excess pressure above atmospheric pressure. Let pa be the atmospheric
pressure and p the total pressure of a fluid. Then the gauge pressure is p − pa .
Hint 2. How to approach the problem
You can relate the fluid pressure at point 2 with the atmospheric pressure by applying Bernoulli's principle to
point 2 and point 1, or alternatively to point 2 and point 3. To determine the fluid speed at point 2 you can use
the continuity equation, using the fluid speed at the exit of the pipe found in Part A.
Hint 3. Apply Bernoulli's principle
Consider the entire volume of water as a single flow tube. Let p2 and v 2 be respectively the pressure and the
fluid speed at point 2. Let the atmospheric pressure be pa and the density of water ρ. Apply Bernoulli's principle
to point 1 and point 2 and complete the expression below.
Express your answer in terms of v 2 , ρ, and g, the free­fall acceleration.
Hint 1. Bernoulli's principle
For the steady flow of an incompressible fluid with no internal friction, the pressure p and the flow speed v
at depth H below the surface are linked by an important relationship, known as Bernoulli's principle. In
particular, at any point at depth H along a flow tube, the following relation holds:
p + ρgH +
1
2
2
ρv
= constant,
where ρ is the density of the fluid and g is the accereleration due to gravity.
Since Bernoulli's principle is valid at any point along a flow tube, it takes the form
p1 + ρgH 1 +
1
2
2
ρv
1
= p2 + ρgH 2 +
1
2
2
ρv
2
when applied to two distinct points along a flow tube. The subscripts 1 and 2 refer to such points.
ANSWER:
p
2
−p
a
= −0.5ρv 2
2
+ 8gρ
Hint 4. Find the fluid speed at point 2
Find v 2 , the speed of the water at point 2.
Express your answer in meters per second to three significant figures.
Hint 1. The continuity equation
In a steadily moving incompressible fluid, the mass of fluid flowing along a flow tube is constant. In
particular, consider a flow tube between two stationary cross sections with areas A1 and A2 . Let v 1 and v 2 be the fluid speeds at these sections, respectively. Then conservation of mass takes the form
A1 v 1 = A2 v 2
which is known as the continuity equation.
ANSWER:
v2
= 4.17 m/s Hint 5. Density of Water
Recall that the density of water is 1000
kg/
3
,
Recall that the density of water is 1000
kg/m
ANSWER:
6.98×104 Pa Correct
Score Summary:
Your score on this assignment is 101%.
You received 7.07 out of a possible total of 7 points.
3