## Home Work Problems and Solutions: 1-6

**1.1 At the instant of Fig. 11-42, a 2.0 kg particle P has a position vector of magnitude **
3.0 m and angle θ1 = 45° and a velocity vector of magnitude 4.0 m/s and angle θ2

30°. Force of magnitude 2.0 N and angle θ3 =30°, acts on P. All three vectors lie in the xy plane. About the origin, what are the (a) magnitude and (b) direction of the angular momentum of P and the (c) magnitude and (d) direction of the torque acting on P? (HR 11-28)

*Sol: We note that the component of v perpendicular to r*
*has magnitude v sin * where = 30°. A similar
observation applies to

*F .*
(a) Eq. 11-20

*r* *p*

leads to

###

3.0 m 2.0 kg 4.0 m/s sin 30###

12 kg m s.^{2}

*rmv*

_{}

(b) Using the right-hand rule for vector products, we find
*r* *p*

points out of the
*page, or along the +z axis, perpendicular to the plane of the figure.*

(c) Eq. 10-38 leads to *rF*sin2

###

3.0 m 2.0 N sin 30###

3.0N m. (d) Using the right-hand rule for vector products, we find *r F* is also out of the
*page, or along the +z axis, perpendicular to the plane of the figure.*

**1.2 A track is mounted on a large wheel that is free to turn with negligible friction **
about a vertical axis (Fig. 11-49). A toy train of mass m is placed on the track and,
with the system initially at rest, the train’s electrical power is turned on.The train
reaches speed 0.15 m/s with respect to the track. What is the angular speed of the
wheel if its mass is 1.1m and its radius is 0.43 m? (Treat the wheel as a hoop, and
neglect the mass of the spokes and hub.) (HR 11-49)

*Sol: No external torques act on the system consisting*
of the train and wheel, so the total angular

momentum of the system (which is initially zero)
*remains zero. Let I = MR*^{2} be the rotational

inertia of the wheel. Its final angular momentum is

Lf^{} * ^{I}*k

^{}

^{ }

^{M R}^{2} k,

^{}

where k

* is up in Fig. 11-47 and that last step (with the minus sign) is done in *
recognition that the wheel’s clockwise rotation implies a negative value for .

The linear speed of a point on the track is R and the speed of the train (going
counterclockwise in Fig. 11-47 with speed *v relative to an outside observer) is *
therefore *v* *v* *R*

* where v is its speed relative to the tracks. Consequently, *
the angular momentum of the train is *m v*

## c

*R R*

## h

^{k.}Conservation of angular momentum yields

0 *MR*^{2} k*m v*

## c h

*R R*k.

When this equation is solved for the angular speed, the result is

###

^{2}

###

^{(0.15 m/s)}

| | 0.17 rad/s.

/ 1 (1.1+1)(0.43 m)

*mvR* *v*

*M m R* *M m* *R*

**2-3 In Fig. 12-42, a 55 kg rock climber is in a lie-back climb along a fissure, with **
hands pulling on one side of the fissure and feet pressed against the opposite side.

The fissure has width w0.20 m, and the center of mass of the climber is a horizontal distance d =0.40 m from the fissure. The coefficient of static friction between hands and rock is μ1 =0.40, and between boots and rock it is μ2 =1.2. (a) What is the least horizontal pull by the hands and push by the feet that will keep the climber stable? (b) For the horizontal pull of (a), what must be the vertical distance h between hands and feet? If the climber encounters wet rock, so that μ1

and μ2 are reduced, what happens to (c) the answer to (a) and (d) the answer to (b)? (HR 12-26)

*Sol: (a) The problem asks for the person’s pull (his force exerted*
on the rock) but since we are examining forces and torques
*on the person, we solve for the reaction force F*^{N}^{1}

(exerted leftward on the hands by the rock). At that point, there is

*also an upward force of static friction on his hands f*1 which we will take to be at
its maximum value ^{1}*F*^{N}^{1}

. We note that equilibrium of horizontal forces requires

1 2

*N* *N*

*F* *F*

(the force exerted leftward on his feet); on this feet there is also an
upward static friction force of magnitude 2*F**N2*. Equilibrium of vertical forces
gives

2

1 2 1

1 2

+ = 0 = = 3.4 10 N.

*N* +

*f* *f* *mg* *F* *mg*

(b) Computing torques about the point where his feet come in contact with the rock, we find

###

1 11 1

1

+ * _{N}* = 0 = +

*= 0.88 m.*

^{N}*N*

*mg d w* *F w*

*mg d w* *f w F h* *h*

*F*

(c) Both intuitively and mathematically (since both coefficients are in the
denominator) we see from part (a) that *F*^{N}^{1}

would increase in such a case.

(d) As for part (b), it helps to plug part (a) into part (b) and simplify:

*h*=

## a f

*d w*+

^{}

_{2}+

*d*

^{}

_{1}

*from which it becomes apparent that h should decrease if the coefficients *
decrease.

**2-4 In Fig. 12-50, a uniform plank, with a length L of 6.10 m and a weight of 445 N, **
rests on the ground and against a frictionless roller at the top of a wall of height
h = 3.05 m. The plank remains in equilibrium for any value of θ≧ 70° but slips if
θ< 70°. Find the coefficient of static friction between the plank and the ground.

(HR 12-37)

*Sol: The free-body diagram on the right*
shows the forces acting on the plank.

Since the roller is frictionless the force it exerts is normal to the plank and makes the angle with the vertical. Its

*magnitude is designated F. W is the force of gravity; this force acts at the center *
*of the plank, a distance L/2 from the point where the plank touches the floor. F*^{N}*is the normal force of the floor and f is the force of friction. The distance from the*
*foot of the plank to the wall is denoted by d. This quantity is not given directly *
*but it can be computed using d = h/tan. *

The equations of equilibrium are:

horizontal force components vertical force components torques

###

2###

sin 0

cos 0

cos 0.

*N*
*N* *L*

*F* *f*

*F* *W F*

*F d* *fh W d*

The point of contact between the plank and the roller was used as the origin for writing the torque equation.

When = 70º the plank just begins to slip and f = sF*N*, where *s* is the
coefficient of static friction. We want to use the equations of equilibrium to
*compute F**N** and f for = 70º, then use **s** = f/F**N* to compute the coefficient of
friction.

*The second equation gives F = (W – F**N*)/cos and this is substituted into the first
to obtain

*f = (W – F**N*) sin /cos = (W – F*N*) tan .

*This is substituted into the third equation and the result is solved for F**N*:

###

^{2}

2

/2 cos + tan (1 tan ) ( / 2)sin

= ,

+ tan (1 tan )

*N*

*d* *L* *h* *h* *L*

*F* *W* *W*

*d h* *h*

*where we have use d = h/tan and multiplied both numerator and denominator by*
tan . We use the trigonometric identity 1+ tan^{2} = 1/cos^{2} and multiply both
numerator and denominator by cos^{2} to obtain

= 1 cos sin2 .

*N* 2
*F* *W* *L*

*h*

*Now we use this expression for F**N** in f = (W – F**N*) tan to find the friction:

*f* *WL*

= *h*

2 sin cos^{2} .

*We substitute these expressions for f and F**N* into *s** = f/F**N* and obtain

*s*

*L*

= *h L*

2 sin cos^{2} _{2} .
sin cos

Evaluating this expression for = 70º, we obtain

###

###

2

2

6.1m sin 70 cos70

= = 0.34.

2 3.05m 6.1m sin70 cos 70

*s*

**3-1 In Fig. 15-31, two identical springs of spring constant 7580 N/m are attached to **
ablock of mass 0.245 kg. What is the frequency of oscillation on the frictionless
floor? (HR 15-13)

*Sol: *When displaced from equilibrium, the net force
*exerted by the springs is –2kx acting in a direction*
so as to return the block to its equilibrium

*position (x = 0). Since the acceleration a d x dt* ^{2} / ^{2}, Newton’s second law yields

= 2 .

2

*md x*2

*dt* *kx*

*Substituting x = x**m* cos(t + ) and simplifying, we find

=2

2 ^{k}

*m*

where is in radians per unit time. Since there are 2 radians in a cycle, and
*frequency f measures cycles per second, we obtain*

1 2 1 2(7580 N/m)

= = 39.6 Hz.

2 2 2 0.245 kg

*f* *k*

*m*

**3-2 In Fig. 15-36, two springs are joined and connected to a block of mass 0.245 kg**
that is set oscillating over a frictionless floor. The springs each have spring
constant k = 6430 N/m. What is the frequency of the oscillations? (HR 15-26)

*Sol: *We wish to find the effective spring constant for
the combination of springs shown in the figure.

*We do this by finding the magnitude F of the*

force exerted on the mass when the total elongation of the springs is x. Then keff

*= F/x. Suppose the left-hand spring is elongated by x* and the right-hand
spring is elongated by x*r**. The left-hand spring exerts a force of magnitude k x* _{}
on the right-hand spring and the right-hand spring exerts a force of magnitude

*kx**r* on the left-hand spring. By Newton’s third law these must be equal, so

*x*_{} *x** _{r}*. The two elongations must be the same and the total elongation is twice
the elongation of either spring:

*x* 2

*x*

_{}. The left-hand spring exerts a force on

*the block and its magnitude is F k x*

_{}

*. Thus k*

_{eff}

*k x*

_{}/2

*x*

**

_{r}*k*/2 . The block behaves as if it were subject to the force of a single spring, with spring

*constant k/2. To find the frequency of its motion replace k*eff in

*f* 1 2

## a f

/ *k*eff /

*m with k/2 to obtain*= 1

2 2

*f* *k*

*m*.

*With m =* *0.245 kg and k = 6430 N/m, the *

*frequency is f =* 18.2 Hz.

**3-3 For Eq. 15-45, suppose the amplitude x**m is given by

where Fm is the (constant) amplitude of the external oscillating force exerted on the spring by a rigid support in Fig. 15-15. At resonance, what are the (a) amplitude and (b) velocity amplitude of the oscillating object? (HR 15-61)

*Sol: *(a) We set = *d* and find that the given expression reduces
*to x**m** = F**m**/b at resonance.*

*(b) since the velocity amplitude v**m* = x*m*, at resonance, we
*have v**m* = F*m**/b = F**m**/b.*

**3-4 In Fig. 15-60, a solid cylinder attached to a horizontal spring (k = 3.00 N/m) rolls**
without slipping along a horizontal surface. If the system is released from rest
when the spring is stretched by 0.250 m, find (a) the translational kinetic energy
and (b) the rotational kinetic energy of the cylinder as it passes through the
equilibrium position. (c) Show that under these conditions the cylinder’s center
of mass executes simple harmonic motion with period T = 2π(3M/2k)^{1/2}

where M is the cylinder mass. (Hint: Find the time derivative of the total mechanical energy.) (HR 15-106)

*Sol: *(a) The potential energy at the turning point is
equal (in the absence of friction) to the total
kinetic energy (translational plus rotational) as
it passes through the equilibrium position:

2

2 2 2 2 2 2 cm

cm cm cm

2 2 2

cm cm cm

1 1 1 1 1 1

2 2 2 2 2 2

1 1 3

2 4 4

*m*

*kx* *Mv* *I* *Mv* *MR* *v*

*R*

*Mv* *Mv* *Mv*

*which leads to Mv*_{cm}^{2} 2*kx*_{m}^{2} */ = 0.125 J. The translational kinetic energy is*3
therefore ^{1}2*Mv*2_{cm} *kx** _{m}*2 /3 0 0625 . J.

(b) And the rotational kinetic energy is ^{1}4*Mv*cm^{2} *kx*_{m}^{2} / 6 0.03125J 3.13 10 J ^{}^{2} .
*(c) In this part, we use v*cm to denote the speed at any instant (and not just the

maximum speed as we had done in the previous parts). Since the energy is constant, then

2 2

3 1 3

4 2 2 0

*dE* *d* *d*

*Mv* *kx* *Mv a* *kxv*

*dt* *dt* *dt*

** ^{cm}**

****

^{cm cm}****

^{cm}which leads to

= 2

3 .

*a* *k*

*M* *x*

cm

## F

## H G IKJ

Comparing with Eq. 15-8, we see that 2 3*k*/ *M*

for this system. Since

= 2/T, we obtain the desired result: *T* 2 3*M*/2*k*
.

**4-1 A uniform rope of mass m and length L hangs from a ceiling. (a) Show that the **
speed of a transverse wave on the rope is a function of y, the distance from the
lower end, and is given by v = (gy)^{1/2}. (b) Show that the time a transverse wave
takes to travel the length of the rope is given by t = 2(L /g)^{1/2}. (HR 16-25)
*Sol: (a) The wave speed at any point on the rope is given by v =*

, where is the tension at that point and is the linear
mass density. Because the rope is hanging the tension varies
*from point to point. Consider a point on the rope a distance y*
from the bottom end. The forces acting on it are the weight of the
rope below it, pulling down, and the tension, pulling up. Since

the rope is in equilibrium, these forces balance. The weight of the rope below is
given by gy, so the tension is = gy. The wave speed is ^{v}^{} ^{gy}^{/} ^{} ^{gy}^{.}
*(b) The time dt for the wave to move past a length dy, a distance y from the *
bottom end, is ^{d}^{t}^{}^{d}^{y v}^{}^{d}^{y}* ^{gy}* and the total time for the wave to move the
entire length of the rope is

0 0

d 2 2 .

*L*

*L* *y* *y* *L*

*t*

###

*gy*

*g*

*g*

**4-2 The type of rubber band used inside some baseballs and golf balls obeys Hooke’s **
law over a wide range of elongation of the band. A segment of this material has
an unstretched length L and a mass m. When a force F is applied, the band
stretches an additional length ΔL . (a) What is the speed (in terms of m, ΔL , and
the spring constant k) of transverse waves on this stretched rubber band? (b)
Using your answer to (a), show that the time required for a transverse pulse to
travel the length of the rubber band is proportional to 1/(ΔL)^{ 1/2} ifΔL << L and is
constant ifΔL>>L. (HR 16-89)

*Sol: (a) The wave speed is*

( )

/( ) .

*F* *k* *k*

*v* *m* *m*

(b) The time required is

1 .

( ) /

*t* *m*

*v* *k* *m* *k*

Thus if ^{}^{/}^{}^{ } ^{1}, then *t* / 1/ and if ; ^{}^{/}^{}^{ } ^{1}, then
/ const.

*t* *m k*

**4-3 Underwater illusion. One clue used by your brain to determine the direction of a **
source of sound is the time delay Δt between the arrival of the sound at the ear
closer to the source and the arrival at the farther ear. Assume that the source is
distant so that a wavefront from it is approximately planar when it reaches you,

and let D represent the separation between your ears. (a) If the source is located at angleθin front of you (Fig. 17-31), what is Δt in terms of D and the speed of sound v in air? (b) If you are submerged in water and the sound source is directly to your right, what is Δt in terms of D and the speed of sound vw in water? (c) Based on the time-delay clue, your brain interprets the submerged sound to arrive at an angleθfrom the forward direction. Evaluateθfor fresh water at 20°C. (HR 17-12)

*Sol: The key idea here is that the time delay *^{}* ^{t}* is due to the

*distance d that each wavefront must travel to reach your*

*left ear (L) after it reaches your right ear (R).*

(a) From the figure, we find *d* *D*sin

*t* *v* *v*

.

(b) Since the speed of sound in water is now *v*^{w}

, with ^{90}

, we have sin 90

*w*

*w* *w*

*D* *D*

*t* *v* *v*

.

(c) The apparent angle can be found by substituting *D v*/ _{w}

for ^{}* ^{t}*:

sin

*w*

*D* *D*

*t* *v* *v*

.

Solving for with *v** _{w}*1482 m/s

(see Table 17-1), we obtain

1 1 343 m/s 1

sin sin sin (0.231) 13

1482 m/s

*w*

*v*

^{} *v* ^{} ^{}

**4-4 In Fig. 17-42, a French submarine and a U.S. submarine move toward each other **
during maneuvers in motionless water in the North Atlantic. The French sub
moves at speed vF = 50.00 km/h, and the U.S. sub at vUS = 70.00 km/h. The
French sub sends out a sonar signal (sound wave in water) at 1.000 × 10^{3} Hz.

Sonar waves travel at 5470 km/h. (a) What is the signal’s frequency as detected by the U.S. sub? (b) What frequency is detected by the French sub in the signal reflected back to it by the U.S. sub? (HR 17-61)

*Sol: We denote the speed of*
the French submarine by
*u*1 and that of the U.S. sub
*by u*2.

(a) The frequency as detected by the U.S. sub is

3 3

1 1 2

1

5470 km/h 70.00 km/h

(1.000 10 Hz) 1.022 10 Hz.

5470 km/h 50.00 km/h
*f* *f* *v u*

*v u*

(b) If the French sub were stationary, the frequency of the reflected wave would
*be f**r** = f*1*(v+u*2*)/(v – u*2*). { f*detected* = f*source*(v+u*2*)/v, f**reflected** = f*detected*v/(v – u*2) }
*Since the French sub is moving towards the reflected signal with speed u*1, then

3

1 1 2

1

2 3

( )( ) (1.000 10 Hz)(5470 50.00)(5470 70.00)

( ) (5470)(5470 70.00)

1.045 10 Hz.

*r* *r*

*v u* *v u v u*

*f* *f* *f*

*v* *v v u*

**5-1 The orbit of Earth around the Sun is almost circular: The closest and farthest **
distances are 1.47× 10^{8} km and 1.52 × 10^{8} km respectively. Determine the
corresponding variations in (a) total energy, (b) gravitational potential energy, (c)
kinetic energy, and (d) orbital speed. (Hint: Use conservation of energy and
conservation of angular momentum.) (HR 13-87)

*Sol: (a) The total energy is conserved, so there is no difference between its values at*
aphelion and perihelion.

(b) Since the change is small, we use differentials:

###

###

11 30 24

9

2 11 2

6.67 10 1.99 10 5.98 10

1.5 10 5 10

*E* *S*

*dU* *GM M* *dr*

*r*

which yields U 1.8 10^{32} J. A more direct subtraction of the values
of the potential energies leads to the same result.

(c) From the previous two parts, we see that the variation in the kinetic energy

*K must also equal 1.8 10*^{32} J.

(d) With *K dK = mv dv, *where* v 2R/T, *we have

###

^{11}

###

32 24

7

2π 1.5 10

1.8 10 5.98 10

3.156 10 *v*

which yields a difference of v 0.99 km/s in Earth’s speed (relative to the Sun) between aphelion and perihelion.

**5-2 The fastest possible rate of rotation of a planet is that for which the gravitational **

force on material at the equator just barely provides the centripetal force needed for the rotation.(Why?) (a) Show that the corresponding shortest period of

rotation is T = (3π/Gρ)^{1/2 } where is the uniform density (mass per unit volume) of
the spherical planet. (b) Calculate the rotation period assuming a density of 3.0
g/cm3, typical of many planets, satellites, and asteroids. No astronomical object
has ever been found to be spinning with a period shorter than that determined by
this analysis. (HR 13-90)

*Sol: If the angular velocity were any greater, loose objects on the surface would not*
go around with the planet but would travel out into space.

(a) The magnitude of the gravitational force exerted by the planet on an object of
*mass m at its surface is given by** F = GmM / R*^{2}*, where M is the mass of the planet and*
*R is its radius. According to Newton’*s second law this must equal^{ mv}^{2}^{ / R, }*where v*
is the speed of the object. Thus,

2

2 = .

*GM* *v*

*R* *R*

*Replacing M with* (4/3) R^{3} (where ^{ }is the density of the planet) and^{ v}

with 2R/T *(where T is the period of revolution), we find*

2

2

4 4

= .

3
*G R* *R*

*T*

*We solve for T and obtain*

*T* 3
*G*

_{.}

(b) The density is^{ 3.0 10}^{3}^{ kg/m}^{3}^{. }*We evaluate the equation for T:*

###

^{11}

^{3}

^{2}

^{3}

###

^{3}

^{3}

###

6.86 10 s 1.9 h.^{3}6.67 10 m / s kg 3.0 10 kg/m

*T*

**5-3 Several planets (Jupiter, Saturn, Uranus) are encircled by rings, perhaps composed **
of material that failed to form a satellite. In addition, many galaxies contain ring-
like structures. Consider a homogeneous thin ring of mass M and outer radius R
(Fig. 13-55). (a) What gravitational attraction does it exert on a particle of mass m
located on the ring’s central axis a distance x from the ring center? (b) Suppose
the particle falls from rest as a result of the attraction of the ring of matter. What is
the speed with which it passes through the center of the ring? (HR 13-99)

*Sol: (a) All points on the ring are the same distance (r = )*

*from the particle, so the gravitational potential energy is simply U = –GMm/. The*
*corresponding force (by symmetry) is expected to be along the x axis, so we take*
*a (negative) derivative of U (with respect to x) to obtain it. The result for the*
*magnitude of the force is GMmx(x*^{2}* + R*^{2})^{3/2}.

*(b) Using our expression for U, then the magnitude of the loss in potential energy*
*as the particle falls to the center is GMm(1/R 1/). This must “turn into” kinetic*
*energy ( mv*^{2 }), so we solve for the speed and obtain

*v = [2GM(R*^{1}* – (R*^{2}* + x*^{2})^{1/2})]^{1/2} .

**5-4 A certain triple-star system consists of two stars, each**
of mass m, revolving in the same circular orbit of
radius r around a central star of mass M (Fig. 13-
54).The two orbiting stars are always at opposite ends
of a diameter of the orbit. Derive an expression for the
period of revolution of the stars. (HR 13-93)

*Sol:* The magnitude of the net gravitational force on one of the smaller stars (of mass
*m) is*

###

^{2}

2 2 .

2 4

*GMm* *Gmm* *Gm* *m*

*r* *r* *r* *M*

This supplies the centripetal force needed for the motion of the star:

2 2

where 2 . 4

*Gm* *m* *v* *r*

*M* *m* *v*

*r* *r* *T*

*Plugging in for speed v, we arrive at an equation for period T:*

2 3 2

( / 4).
*T* *r*

*G M* *m*

6-1 In Fig. 14-37, water stands at depth D = 35.0 m behind the vertical upstream face of a dam of width W = 314 m. Find (a) the net horizontal force on the dam from the gauge pressure of the water and (b) the net torque due to that force about a line through O parallel to the width of the dam. (c) Find the moment arm of this torque. (HRW14-24)

*Sol: (a) At depth y the gauge pressure of the water is p = *

*gy, where is the density of the water. We consider*
*a horizontal strip of width W at depth y, with (vertical)*
*thickness dy, across the dam. Its area is dA = W dy and*
*the force it exerts on the dam is dF = p dA = gyW dy.*

The total force of the water on the dam is

### ^{} ^{ } ^{}

^{2}

2 3 3 2

0

9

1 1

1.00 10 kg m 9.80 m s 314 m 35.0 m

2 2

1.88 10 N.

*F* *D* *gyW dy* *gWD*

###

*(b) Again we consider the strip of water at depth y. Its moment arm for the torque*
*it exerts about O is D – y so the torque it exerts is *

*d = dF(D – y) = gyW (D – y)dy*
and the total torque of the water is

###

### ^{} ^{ } ^{}

3 3 3

0

3 3 2 3 10

1 1 1

2 3 6

1 1.00 10 kg m 9.80 m s 314 m 35.0 m 2.20 10 N m.

6

*D* *gyW D y dy* *gW* *D* *D* *gWD*

###

(c) We write = rF, where r is the effective moment arm. Then,

1 3 6 1 2 2

35.0 m

11.7 m.

3 3

*gWD* *D*

*r* *F* *gWD*

14-2 Figure 14-53 shows a stream of water flowing through a hole at depth h = 10 cm in a tank holding water to height H = 40 cm. (a) At what distance x does the stream strike the floor? (b) At what depth should a second hole be made to give the same value of x? (c) At what depth should a hole be

made to maximize x? (HRW14-65)

*Sol: (a) Since Sample Problem 14-8 deals with a similar*
situation, we use the final equation from it:

2 0 for the projectile motion.

*v* *gh* *v v*

The stream of water emerges horizontally (0 = 0° in the notation of Chapter 4),
*and setting y – y*0* = –(H – h) in Eq. 4-22, we obtain the “time-of-flight” *

2( ) 2

( ).

*t* *H h* *H h*

*g* *g*

*Using this in Eq. 4-21, where x*0 = 0 by choice of coordinate origin, we find

0

2( )

2 *H h* 2 ( ) 2 (10 cm)(40 cm 10 cm) 35 cm.

*x v t* *gh* *h H h*

*g*

*(b) The result of part (a) (which, when squared, reads x*^{2}* = 4h(H – h)) is a *
*quadratic equation for h once x and H are specified. Two solutions for h are *
therefore mathematically possible, but are they both physically possible? For
*instance, are both solutions positive and less than H? We employ the quadratic *

formula:

2 2 2

2 0

4 2

*x* *H* *H* *x*

*h* *Hh* *h*

*which permits us to see that both roots are physically possible, so long as x H. *

*Labeling the larger root h*1 (where the plus sign is chosen) and the smaller root as
*h*2 (where the minus sign is chosen), then we note that their sum is simply

2 2 2 2

1 2 .

2 2

*H* *H* *x* *H* *H* *x*

*h* *h* *H*

*Thus, one root is related to the other (generically labeled h' and h) by h' = H – h. *

Its numerical value is *h*' 40cm 10 cm 30 cm.

*(c) We wish to maximize the function f = x*^{2}* = 4h(H – h). We differentiate with *
*respect to h and set equal to zero to obtain *

4 8 0

2

*df* *H*

*H* *h* *h*

*dh*

*or h = (40 cm)/2 = 20 cm, as the depth from which an emerging stream of water *
will travel the maximum horizontal distance.

14-3 A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the pipe (Fig. 14-55); the cross-sectional area A of the entrance and exit of the meter matches the pipe’s crosssectional area. Between the entrance and exit, the fluid flows from the pipe with speed V and then through a narrow“throat” of cross-sectional area a with speed v. A manometer connects the wider portion of the meter to the narrower portion. The change in the fluid’s speed is accompanied by a change Δp in the fluid’s pressure, which causes a height difference h of the liquid in the two arms of the

manometer. (Here Δp means pressure in the throat minus pressure in the pipe.) (a) By applying Bernoulli’s equation and the equation of continuity to points 1 and 2 in Fig. 14-55, show that ,

where is the density of the fluid. (b) Suppose that the fluid is fresh water, that the
cross-sectional areas are 64 cm^{2} in the pipe and 32 cm^{2} in the throat, and that the
pressure is 55 kPa in the pipe and 41 kPa in the throat.What is the rate of water
flow in cubic meters per second? (HRW14-67)

*Sol: (a) The continuity equation yields Av = aV, and*
Bernoulli’s equation yields

2 2

1 1

2 2

*p* *v* *V*

,
where p = p1* – p*2*. The first equation gives V =*

*(A/a)v. We use this to substitute for V in the second equation, and obtain*

###

^{2}

2 2

1 1

2 2

*p* *v* *A a v*

*. We solve for v. The result is*

###

2

2 2 2

2 2

( / ) 1 .

*p* *a p*

*v* *A a* *A* *a*

(b) We substitute values to obtain

###

4 2 2 3 3

3 4 2 2 4 2 2

2(32 10 m ) (55 10 Pa 41 10 Pa)

3.06 m/s.

(1000 kg / m ) (64 10 m ) (32 10 m )

*v* ^{} _{} _{}

Consequently, the flow rate is

4 2 2 3

(64 10 m ) (3.06 m/s) 2.0 10 m / s.

*Av* ^{} ^{}

14-4 A pitot tube (Fig. 14-56) is used to determine the airspeed of an airplane. It consists of an outer tube with a number of small holes B (four are shown) that allow air into the tube; that tube is connected to one arm of a U-tube. The other arm of the U-tube is connected to hole A at the front end of the device, which points in the direction the plane is headed. At A the air becomes stagnant so that vA = 0. At B, however, the speed of the air presumably equals the airspeed v of the plane. (a) Use Bernoulli’s equation to show that,

where vis the density of the liquid in the U-tube and h is the difference in the liquid levels in that tube. (b) Suppose that the tube contains alcohol and the level difference h is 26.0 cm. What is the plane’s speed relative to the air? The density of the air is 1.03 kg/m3 and that of alcohol is 810 kg/m3. (HRW14-70)

*Sol: (a) Bernoulli’s equation gives*

1 2 air

2 But

*A* *B* *A* *B*

*p* *p* *v* *p* *p* *p* *gh*
in order to balance the pressure in the two
arms of the U-tube. Thus

1 2 air

*gh* 2 *v*

, or

air

2 *gh*.

*v*

(b) The plane’s speed relative to the air is

###

^{3}

###

^{2}

3 air

2 810 kg/m (9.8m/s ) (0.260 m)

2 63.3m/s.

1.03kg/m
*v* *gh*