Download Document

Chap 12
11. The forces on the ladder are shown in the diagram below. F1 is the force of the
window, horizontal because the window is frictionless. F2 and F3 are components of
the force of the ground on the ladder. M is the mass of the window cleaner and m is
the mass of the ladder.
The force of gravity on the man acts at a point 3.0 m up the ladder and the force of
gravity on the ladder acts at the center of the ladder. Let  be the angle between the
ladder and the ground. We use cos  d / L or sin  L2  d 2 /L to find  = 60º.
Here L is the length of the ladder (5.0 m) and d is the distance from the wall to the
foot of the ladder (2.5 m).
(a) Since the ladder is in equilibrium the sum of the torques about its foot (or any
other point) vanishes. Let be the distance from the foot of the ladder to the position
of the window cleaner. Then, Mg cos  mg  L / 2 cos  F1L sin  0 , and
F1 
( M  mL / 2) g cos [(75kg) (3.0 m)+(10 kg) (2.5m)](9.8m/s 2 ) cos 60

L sin 
(5.0 m)sin 60
 2.8 102 N.
This force is outward, away from the wall. The force of the ladder on the window has
the same magnitude but is in the opposite direction: it is approximately 280 N,
inward.
(b) The sum of the horizontal forces and the sum of the vertical forces also vanish:
F1  F3  0
F2  Mg  mg  0
The first of these equations gives F3  F1  2.8  102 N and the second gives
F2  (M  m) g  (75kg  10 kg) (9.8m/s 2 )  8.3 102 N
The magnitude of the force of the ground on the ladder is given by the square root of
the sum of the squares of its components:
F  F2  F3  ( 2.8  102 N)2  (8.3  102 N)2  8.8  102 N.
2
2
(c) The angle  between the force and the horizontal is given by
tan  = F3/F2 = 830/280 = 2.94,
so  = 71º. The force points to the left and upward, 71º above the horizontal. We note
that this force is not directed along the ladder.
21. The beam is in equilibrium: the sum of the forces and the sum of the torques
acting on it each vanish. As we see in the figure, the beam makes an angle of 60º with
the vertical and the wire makes an angle of 30º with the vertical.
(a) We calculate the torques around the hinge. Their sum is
TL sin 30º – W(L/2) sin 60º = 0.
Here W is the force of gravity acting at the center of the beam, and T is the tension
force of the wire. We solve for the tension:
T=
W sin60  222N  sin 60
=
= 192 N.
2 sin30
2 sin 30
(b) Let Fh be the horizontal component of the force exerted by the hinge and take it to
be positive if the force is outward from the wall. Then, the vanishing of the horizontal
component of the net force on the beam yields Fh – T sin 30º = 0 or
Fh = T sin30 = 192.3N  sin 30 = 96.1N.
(c) Let Fv be the vertical component of the force exerted by the hinge and take it to be
positive if it is upward. Then, the vanishing of the vertical component of the net force
on the beam yields Fv + T cos 30º – W = 0 or
Fv = W  T cos30 = 222 N  192.3N  cos30 = 55.5 N.
31. The diagram below 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.
FN 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
F sin   f  0
F cos   W  FN  0
FN d  fh  W  d  L2 cos    0.
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 = sFN, where s is the coefficient of
static friction. We want to use the equations of equilibrium to compute FN and f for 
= 70º, then use s = f/FN to compute the coefficient of friction.
The second equation gives F = (W – FN)/cos  and this is substituted into the first to
obtain
f = (W – FN) sin /cos  = (W – FN) tan .
This is substituted into the third equation and the result is solved for FN:
FN =
d   L/2  cos + h tan
d + h tan
W
h(1  tan 2  )  ( L / 2)sin 
W,
h(1  tan 2  )
where we have use d = h/tan and multiplied both numerator and denominator by tan
. We use the trigonometric identity 1+ tan2 = 1/cos2 and multiply both numerator
and denominator by cos2 to obtain
L


FN = W 1  cos 2 sin  .
 2h

Now we use this expression for FN in f = (W – FN) tan  to find the friction:
f =
WL 2
sin  cos .
2h
We substitute these expressions for f and FN into s = f/FN and obtain
s =
L sin 2 cos
.
2h  L sin cos2
Evaluating this expression for  = 70º, we obtain
s =
 6.1m  sin2 70cos70
= 0.34.
2  3.05m    6.1m  sin70cos2 70
35. The diagrams below show the forces on the two sides of the ladder, separated. FA
and FE are the forces of the floor on the two feet, T is the tension force of the tie rod,
W is the force of the man (equal to his weight), Fh is the horizontal component of the
force exerted by one side of the ladder on the other, and Fv is the vertical component
of that force. Note that the forces exerted by the floor are normal to the floor since the
floor is frictionless. Also note that the force of the left side on the right and the force
of the right side on the left are equal in magnitude and opposite in direction.
Since the ladder is in equilibrium, the vertical components of the forces on the left
side of the ladder must sum to zero: Fv + FA – W = 0. The horizontal components must
sum to zero: T – Fh = 0. The torques must also sum to zero. We take the origin to be at
the hinge and let L be the length of a ladder side. Then
FAL cos  – W(L/4) cos  – T(L/2) sin  = 0.
Here we recognize that the man is one–fourth the length of the ladder side from the
top and the tie rod is at the midpoint of the side.
The analogous equations for the right side are FE – Fv = 0, Fh – T = 0, and FEL cos  –
T(L/2) sin  = 0.
There are 5 different equations:
Fv  FA  W  0,
T  Fh  0
FA L cos   W ( L / 4) cos   T ( L / 2)sin   0
FE  Fv  0
FE L cos   T ( L / 2)sin   0.
The unknown quantities are FA, FE, Fv, Fh, and T.
(a) First we solve for T by systematically eliminating the other unknowns. The first
equation gives FA = W – Fv and the fourth gives Fv = FE. We use these to substitute
into the remaining three equations to obtain
T  Fh  0
WL cos   FE L cos   W ( L / 4) cos   T ( L / 2)sin   0
FE L cos   T ( L / 2)sin   0.
The last of these gives FE = Tsin/2cos = (T/2) tan. We substitute this expression
into the second equation and solve for T. The result is
T
3W
.
4 tan 
To find tan, we consider the right triangle formed by the upper half of one side of the
ladder, half the tie rod, and the vertical line from the hinge to the tie rod. The lower
side of the triangle has a length of 0.381 m, the hypotenuse has a length of 1.22 m,
and the vertical side has a length of
. mg b
0.381 mg 116
. m . This means
b122
2
2
tan  = (1.16m)/(0.381m) = 3.04.
Thus,
T
3(854 N)
 211N.
4(3.04)
(b) We now solve for FA. Since Fv = FE and FE = T sin2cos, Fv = 3W/8. We
substitute this into Fv + FA – W = 0 and solve for FA. We find
FA  W  Fv  W  3W / 8  5W / 8  5(884 N)/8=534 N.
(c) We have already obtained an expression for FE: FE = 3W/8. Evaluating it, we get
FE = 320 N.
36. (a) The Young’s modulus is given by
stress
150 106 N/m2
E
 slope of the stress-strain curve 
 7.5 1010 N/m 2 .
strain
0.002
(b) Since the linear range of the curve extends to about 2.9 × 108 N/m2, this is
approximately the yield strength for the material.
39. (a) Let FA and FB be the forces exerted by the wires on the log and let m be the
mass of the log. Since the log is in equilibrium FA + FB – mg = 0. Information given
about the stretching of the wires allows us to find a relationship between FA and FB. If
wire A originally had a length LA and stretches by L A , then LA  FA LA / AE ,
where A is the cross–sectional area of the wire and E is Young’s modulus for steel
(200 × 109 N/m2). Similarly, LB  FB LB / AE . If  is the amount by which B was
originally longer than A then, since they have the same length after the log is attached,
LA  LB  . This means
FA LA FB LB

 .
AE
AE
We solve for FB:
FB 
FA LA AE

.
LB
LB
We substitute into FA + FB – mg = 0 and obtain
FA 
mgLB  AE
.
LA  LB
c
h
2
The cross–sectional area of a wire is A  r 2   120
.  103 m  4.52  106 m2 .
Both LA and LB may be taken to be 2.50 m without loss of significance. Thus
FA 
(103kg) (9.8 m/s2 ) (2.50 m)+(4.52 106 m2 ) (200 109 N/m2 ) (2.0 103 m)
2.50 m+2.50 m
 866 N.
(b) From the condition FA + FB – mg = 0, we obtain
FB  mg  FA  (103kg) (9.8m/s2 )  866 N=143 N.
(c) The net torque must also vanish. We place the origin on the surface of the log at a
point directly above the center of mass. The force of gravity does not exert a torque
about this point. Then, the torque equation becomes FAdA – FBdB = 0, which leads to
dA FB 143 N


 0.165.
dB FA 866 N
40. (a) Since the brick is now horizontal and the cylinders were initially the same
length  , then both have been compressed an equal amount  . Thus,

FA


AA E A
and

F
 B

AB EB
which leads to
FA AA E A ( 2 AB )( 2 EB )


 4.
FB AB EB
AB EB
When we combine this ratio with the equation FA + FB = W, we find FA/W = 4/5 =
0.80 .
(b) This also leads to the result FB/W = 1/5 = 0.20.
(c) Computing torques about the center of mass, we find FAdA = FBdB which leads to
d A FB 1

  0.25.
d B FA 4
Chap 13
6. The gravitational forces on m5 from the two 5.00g masses m1 and m4 cancel each
other. Contributions to the net force on m5 come from the remaining two masses:
Fnet
 6.67 10
=
11
N  m 2 /kg 2  2.50 103 kg  3.00 10 3 kg  1.00 10 3 kg 

2 101 m

2
= 1.67 1014 N.
The force is directed along the diagonal between m2 and m3, towards m2. In
unit-vector notation, we have
Fnet  Fnet (cos 45ˆi  sin 45ˆj)  (1.18 1014 N) ˆi  (1.18 1014 N) ˆj
15. The acceleration due to gravity is given by ag = GM/r2, where M is the mass of
Earth and r is the distance from Earth’s center. We substitute r = R + h, where R is
the radius of Earth and h is the altitude, to obtain ag = GM /(R + h)2. We solve for h
and obtain h  GM / ag  R . According to Appendix C, R = 6.37  106 m and M =
5.98  1024 kg, so
h
 6.67 10
11
m3 / s 2  kg  5.98 1024 kg 
 4.9m / s 
2
 6.37 106 m  2.6 106 m.
21. Using the fact that the volume of a sphere is 4 R3/3, we find the density of the
sphere:

M total 1.0 104 kg

 2.4 103 kg/m3 .
3
3
4
4
3R
3  1.0 m 
When the particle of mass m (upon which the sphere, or parts of it, are exerting a
gravitational force) is at radius r (measured from the center of the sphere), then
whatever mass M is at a radius less than r must contribute to the magnitude of that
force (GMm/r2).
(a) At r = 1.5 m, all of Mtotal is at a smaller radius and thus all contributes to the force:
Fon m 
GmM total
 m  3.0  107 N/kg  .
r2
(b) At r = 0.50 m, the portion of the sphere at radius smaller than that is
4

M =    r 3  = 1.3 103 kg.
3

Thus, the force on m has magnitude GMm/r2 = m (3.3  107 N/kg).
(c) Pursuing the calculation of part (b) algebraically, we find
Fon m 
Gm   43  r 3 
r
2

N 
 mr  6.7 107
.
kg  m 

31. (a) The work done by you in moving the sphere of mass mB equals the change in
the potential energy of the three-sphere system. The initial potential energy is
Ui  
GmA mB GmA mC GmB mC


d
L
Ld
and the final potential energy is
Uf  
GmA mB GmAmC GmB mC


.
Ld
L
d
The work done is
 1
1 
1 
 1
W  U f  U i  GmB  mA  
 
  mC 
 L  d d 
 d Ld 



1
1
 (6.67  1011 m3 / s 2  kg) (0.010 kg) (0.080 kg) 


 0.040 m 0.080 m 



1
1
(0.020 kg) 


 0.080 m 0.040 m  
  5.0  1013 J.
(b) The work done by the force of gravity is (Uf  Ui) = 5.0  1013 J.
32. Energy conservation for this situation may be expressed as follows:
K1  U1  K 2  U 2
K1 
GmM
GmM
 K2 
r1
r2
where M = 5.0  1023 kg, r1 = R = 3.0  106 m and m = 10 kg.
(a) If K1 = 5.0  107 J and r2 = 4.0  106 m, then the above equation leads to
1 1
K 2  K1  GmM     2.2  107 J.
 r2 r1 
(b) In this case, we require K2 = 0 and r2 = 8.0  106 m, and solve for K1:
1 1
K1  K2  GmM     6.9  107 J.
 r1 r2 
33. (a) We use the principle of conservation of energy. Initially the particle is at the
surface of the asteroid and has potential energy Ui = GMm/R, where M is the mass of
the asteroid, R is its radius, and m is the mass of the particle being fired upward. The
initial kinetic energy is 1 2 mv 2 . The particle just escapes if its kinetic energy is zero
when it is infinitely far from the asteroid. The final potential and kinetic energies are
both zero. Conservation of energy yields GMm/R + ½mv2 = 0. We replace GM/R
with agR, where ag is the acceleration due to gravity at the surface. Then, the energy
equation becomes agR + ½v2 = 0. We solve for v:
v  2ag R  2(3.0 m/s2 ) (500  103 m)  1.7  103 m/s.
(b) Initially the particle is at the surface; the potential energy is Ui = GMm/R and the
kinetic energy is Ki = ½mv2. Suppose the particle is a distance h above the surface
when it momentarily comes to rest. The final potential energy is Uf = GMm/(R + h)
and the final kinetic energy is Kf = 0. Conservation of energy yields

GMm 1 2
GMm
 mv  
.
R
2
Rh
We replace GM with agR2 and cancel m in the energy equation to obtain
ag R 2
1 2
 ag R  v  
.
2
( R  h)
The solution for h is
h
2ag R 2
2ag R  v 2
R 
2(3.0 m/s 2 ) (500  103 m) 2
 (500  103 m)
2
3
2
2(3.0 m/s ) (500  10 m)  (1000 m/s)
 2.5  105 m.
(c) Initially the particle is a distance h above the surface and is at rest. Its potential
energy is Ui = GMm/(R + h) and its initial kinetic energy is Ki = 0. Just before it hits
the asteroid its potential energy is Uf = GMm/R. Write
1
2
mv2f for the final kinetic
energy. Conservation of energy yields

GMm
GMm 1 2

 mv .
Rh
R
2
We substitute agR2 for GM and cancel m, obtaining

ag R 2
Rh
  ag R 
1 2
v .
2
The solution for v is
v  2a g R 
2a g R 2
Rh
 2(3.0 m/s 2 ) (500  103 m) 
2(3.0 m/s 2 )(500  103 m) 2
(500  103 m) + (1000  103 m)
 1.4  103 m/s.
45. (a) The greatest distance between the satellite and Earth’s center (the apogee
distance) is Ra = (6.37  106 m + 360  103 m) = 6.73  106 m. The least distance
(perigee distance) is Rp = (6.37  106 m + 180  103 m) = 6.55  106 m. Here 6.37 
106 m is the radius of Earth. From Fig. 13-14, we see that the semi-major axis is
a
Ra  Rp
2
6.73  106 m + 6.55  106 m

 6.64  106 m.
2
(b) The apogee and perigee distances are related to the eccentricity e by Ra = a(1 + e)
and Rp = a(1  e). Add to obtain Ra + Rp = 2a and a = (Ra + Rp)/2. Subtract to obtain
Ra  Rp = 2ae. Thus,
e
Ra  R p
2a

Ra  R p
Ra  R p

6.73  106 m  6.55  106 m
 0.0136.
6.73  106 m  6.55  106 m
51. In our system, we have m1 = m2 = M (the mass of our Sun, 1.99  1030 kg). With r
= 2r1 in this system (so r1 is one-half the Earth-to-Sun distance r), and v = r/T for the
speed, we have
 r T   T 
Gm1m2
 m1
2
r
r 2
2
2 2 r 3
.
GM
With r = 1.5  1011 m, we obtain T = 2.2  107 s. We can express this in terms of
Earth-years, by setting up a ratio:
 2.2  107 s 
T 
T    (1y) = 
 1 y   0.71 y.
7
 1y 
 3.156  10 s 
56. Although altitudes are given, it is the orbital radii which enter the equations. Thus,
rA = (6370 + 6370) km = 12740 km, and rB = (19110 + 6370) km = 25480 km
(a) The ratio of potential energies is
GmM
UB
r
1
rB

 A  .
U A  GmM
rB 2
rA

(b) Using Eq. 13-38, the ratio of kinetic energies is
GmM
KB
r
1
2rB

 A  .
K A GmM
rB 2
2rA
(c) From Eq. 13-40, it is clear that the satellite with the largest value of r has the
smallest value of |E| (since r is in the denominator). And since the values of E are
negative, then the smallest value of |E| corresponds to the largest energy E. Thus,
satellite B has the largest energy.
(d) The difference is
E  EB  EA  
GmM  1
1
  .
2  rB rA 
Being careful to convert the r values to meters, we obtain E = 1.1  108 J. The mass
M of Earth is found in Appendix C.
60. (a) The pellets will have the same speed v but opposite direction of motion, so the
relative speed between the pellets and satellite is 2v. Replacing v with 2v in Eq. 13-38
is equivalent to multiplying it by a factor of 4. Thus,
11
3
2
24
 GM E m  2(6.67  10 m / kg  s )  5.98  10 kg   0.0040 kg 
K rel  4 

 4.6  105 J.

3
(6370  500)  10 m
 2r 
(b) We set up the ratio of kinetic energies:
K rel

K bullet
Chap 14
4.6  105 J
1
2
 0.0040 kg  950 m/s 
2
 2.6  102.
4. The magnitude F of the force required to pull the lid off is F = (po – pi)A, where po
is the pressure outside the box, pi is the pressure inside, and A is the area of the lid.
Recalling that 1N/m2 = 1 Pa, we obtain
pi  po 
F
480 N
 1.0 105 Pa 
 3.8 104 Pa.
A
77 104 m 2
18. (a) The force on face A of area AA due to the water pressure alone is



FA  p A AA   w ghA AA   w g (2d )d 2  2 1.0 103 kg m3 9.8 m s 2  5.0 m 
3
 2.5 106 N.
Adding the contribution from the atmospheric pressure,
F0= (1.0  105 Pa)(5.0 m)2 = 2.5  106 N,
we have
FA '  F0  FA  2.5 106 N  2.5 106 N  5.0 106 N.
(b) The force on face B due to water pressure alone is
5
5
3
 5d 
FB  pavgB AB   g   d 2   w gd 3 
1.0 103 kg m3 9.8 m s 2  5.0 m 
2
2
 2 
 3.1106 N.


Adding the contribution from the atmospheric pressure,
F0= (1.0  105 Pa)(5.0 m)2 = 2.5  106 N,
we have
FB '  F0  FB  2.5 106 N  3.1106 N  5.6 106 N.

19. (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

F 
D
0
 gyW dy 
1
 gWD 2
2



1
2
1.00 103 kg m3 9.80 m s 2  314 m  35.0 m   1.88 109 N.
2

(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
 


D
0
1

1


1
 gyW  D  y  dy   gW  D 3  D 3    gWD3
2
3
6



1
3
1.00 103 kg m3 9.80 m s 2  314 m  35.0 m   2.20 1010 N  m.
6
(c) We write  = rF, where r is the effective moment arm. Then,
r

F

1
6
1
2
 gWD3 D 35.0 m
 
 11.7 m.
 gWD 2 3
3
23. Eq. 14-13 combined with Eq. 5-8 and Eq. 7-21 (in absolute value) gives
mg = kx
A1
.
A2
With A2 = 18A1 (and the other values given in the problem) we find m = 8.50 kg.
25. (a) The anchor is completely submerged in water of density w. Its effective
weight is Weff = W – w gV, where W is its actual weight (mg). Thus,
V 
W  Weff
200 N

 2.04  102 m3 .
3
2
w g
1000 kg/m   9.8 m/s 
(b) The mass of the anchor is m = V, where  is the density of iron (found in Table
14-1). Its weight in air is
W  mg  Vg   7870 kg/m3  (2.04  102 m3 ) 9.80 m/s2   1.57  103 N .
32. (a) An object of the same density as the surrounding liquid (in which case the
“object” could just be a packet of the liquid itself) is not going to accelerate up or
down (and thus won’t gain any kinetic energy). Thus, the point corresponding to
zero K in the graph must correspond to the case where the density of the object equals
liquid. Therefore, ball = 1.5 g/cm3 (or 1500 kg/m3).
(b) Consider the liquid = 0 point (where Kgained = 1.6 J). In this case, the ball is
falling through perfect vacuum, so that v2 = 2gh (see Eq. 2-16) which means that K =
1
2
mv2 = 1.6 J can be used to solve for the mass. We obtain mball = 4.082 kg. The
volume of the ball is then given by mball/ball = 2.72  103 m3.
35. The volume Vcav of the cavities is the difference between the volume Vcast of the
casting as a whole and the volume Viron contained: Vcav = Vcast – Viron. The volume of
the iron is given by Viron = W/giron, where W is the weight of the casting and iron is
the density of iron. The effective weight in water (of density w) is Weff = W – gw
Vcast. Thus, Vcast = (W – Weff)/gw and
Vcav 
W  Weff
W
6000 N  4000 N
6000 N



2
3
2
g w
g iron (9.8 m/s ) (1000 kg/m ) (9.8 m/s ) (7.87  103 kg/m3 )
 0.126 m3 .
38. (a) Since the lead is not displacing any water (of density w), the lead’s volume is
not contributing to the buoyant force Fb. If the immersed volume of wood is Vi, then
m

Fb   wVi g  0.900  wVwood g  0.900  w g  wood  ,
  wood 
which, when floating, equals the weights of the wood and lead:
m

Fb  0.900  w g  wood   (mwood  mlead ) g .
  wood 
Thus,
(0.900) (1000 kg/m3 ) (3.67 kg)
m

mlead  0.900  w  wood   mwood 
 3.67 kg  1.84 kg .
600 kg/m3
  wood 
(b) In this case, the volume Vlead = mlead/lead also contributes to Fb. Consequently,
m
   
Fb  0.900  w g  wood    w  mlead g  (mwood  mlead ) g ,
  wood   lead 
which leads to
mlead 
0.900 (  w /  wood )mwood  mwood
1   w / lead

1.84 kg
1  (1.00  10 kg/m3 /1.13  10 4 kg/m 3 )
3
 2.01 kg.
47. (a) The equation of continuity leads to
v2 A2  v1 A1
 r12 
 v2  v1  2 
 r2 
which gives v2 = 3.9 m/s.
(b) With h = 7.6 m and p1 = 1.7  105 Pa, Bernoulli’s equation reduces to
p2  p1   gh 


1
 v12  v22  8.8 104 Pa.
2
55. (a) Since Sample Problem 14-8 deals with a similar situation, we use the final
equation (labeled “Answer”) from it:
v  2gh  v  v0 for the projectile motion.
The stream of water emerges horizontally (0 = 0° in the notation of Chapter 4), and
setting y – y0 = –(H – h) in Eq. 4-22, we obtain the “time-of-flight”
t
2( H  h)

g
2
( H  h).
g
Using this in Eq. 4-21, where x0 = 0 by choice of coordinate origin, we find
x  v0t  2 gh
2( H  h)
 2 h( H  h)  2 (10 cm)(40 cm  10 cm)  35 cm.
g
(b) The result of part (a) (which, when squared, reads x2 = 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:
h2  Hh 
x2
H  H 2  x2
0h
4
2
which permits us to see that both roots are physically possible, so long as x  H.
Labeling the larger root h1 (where the plus sign is chosen) and the smaller root as h2
(where the minus sign is chosen), then we note that their sum is simply
h1  h2 
H  H 2  x2 H  H 2  x2

 H.
2
2
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 = x2 = 4h(H – h). We differentiate with
respect to h and set equal to zero to obtain
df
H
 4 H  8h  0  h 
dh
2
or h = (40 cm)/2 = 20 cm, as the depth from which an emerging stream of water will
travel the maximum horizontal distance.
59. (a) The continuity equation yields Av = aV, and Bernoulli’s equation yields
p  12  v 2  12 V 2 , where p = p1 – p2. The first equation gives V = (A/a)v. We use
this to substitute for V in the second equation, and obtain p  12  v 2  12   A a  v 2 .
2
We solve for v. The result is
2p

 A2 
  2  1
a

v
2a 2 p
.
  A2  a 2 
(b) We substitute values to obtain
2(32  104 m 2 ) 2 (55  103 Pa  41  103 Pa)
v
 3.06 m/s.
(1000 kg / m3 ) (64  104 m 2 ) 2  (32  10 4 m 2 ) 2


Consequently, the flow rate is
Av  (64  104 m2 ) (3.06 m/s)  2.0  102 m3 / s.
Chap 15
5. (a) The motion repeats every 0.500 s so the period must be T = 0.500 s.
(b) The frequency is the reciprocal of the period: f = 1/T = 1/(0.500 s) = 2.00 Hz.
(c) The angular frequency  is  = 2f = 2(2.00 Hz) = 12.6 rad/s.
(d) The angular frequency is related to the spring constant k and the mass m by
  k m . We solve for k: k = m2 = (0.500 kg)(12.6 rad/s)2 = 79.0 N/m.
(e) Let xm be the amplitude. The maximum speed is vm = xm = (12.6 rad/s)(0.350 m)
= 4.40 m/s.
(f) The maximum force is exerted when the displacement is a maximum and its
magnitude is given by Fm = kxm = (79.0 N/m)(0.350 m) = 27.6 N.
15. The maximum force that can be exerted by the surface must be less than sFN or
else the block will not follow the surface in its motion. Here, µs is the coefficient of
static friction and FN is the normal force exerted by the surface on the block. Since the
block does not accelerate vertically, we know that FN = mg, where m is the mass of
the block. If the block follows the table and moves in simple harmonic motion, the
magnitude of the maximum force exerted on it is given by F = mam = m2xm =
m(2f)2xm, where am is the magnitude of the maximum acceleration,  is the angular
frequency, and f is the frequency. The relationship  = 2f was used to obtain the last
form. We substitute F = m(2f)2xm and FN = mg into F  µsFN to obtain m(2f)2xm 
µsmg. The largest amplitude for which the block does not slip is
xm
b0.50gc9.8 m / s h 0.031 m.
=
=
b2f g b2  2.0 Hzg
sg
2
2
2
A larger amplitude requires a larger force at the end points of the motion. The surface
cannot supply the larger force and the block slips.
20. Both parts of this problem deal with the critical case when the maximum
acceleration becomes equal to that of free fall. The textbook notes (in the discussion
immediately after Eq. 15-7) that the acceleration amplitude is am = 2xm, where  is
the angular frequency; this is the expression we set equal to g = 9.8 m/s2.
(a) Using Eq. 15-5 and T = 1.0 s, we have
2 I
F
G
HT J
Kx
2
gT 2
= 0.25 m.
m = g  xm =
4 2
(b) Since  = 2f, and xm = 0.050 m is given, we find
 2 f 
2
xm = g 
f=
1
2
g
 2.2 Hz.
xm
22. They pass each other at time t, at x1  x2  12 xm where
x1  xm cos(t   1 )
and
x2  xm cos(t   2 ).
From this, we conclude that cos(t  1 )  cos(t   2 )  12 , and therefore that the
phases (the arguments of the cosines) are either both equal to /3 or one is /3 while
the other is –/3. Also at this instant, we have v1 = –v2≠0where
v1   xm sin(t   1 )
and
v2   xm sin(t   2 ).
This leads to sin(t + 1) = – sin(t +  2). This leads us to conclude that the phases
have opposite sign. Thus, one phase is /3 and the other phase is – /3; the wt term
cancels if we take the phase difference, which is seen to be  /3 – (– /3) = 2 /3.
25. To be on the verge of slipping means that the force exerted on the smaller block
(at the point of maximum acceleration) is fmax = µs mg. The textbook notes (in the
discussion immediately after Eq. 15-7) that the acceleration amplitude is am =2xm,
where   k / (m  M) is the angular frequency (from Eq. 15-12). Therefore, using
Newton’s second law, we have
mam =  s mg 
which leads to xm = 0.22 m.
k
xm =  s g
m+ M
26. We wish to find the effective spring constant for the combination of springs
shown in Fig. 15-35. 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 xr.
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 kxr on the left-hand spring. By
Newton’s third law these must be equal, so x  xr . 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
keff  kx / 2xr  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 keff
in f  1 / 2 keff / m with k/2 to obtain
a f
f=
1
k
.
2 2m
With m = 0.245 kg and k = 6430 N/m, the frequency is f = 18.2 Hz.
33. (a) Eq. 15-12 (divided by 2) yields
f=
1
2
k
1 1000 N / m

 2.25 Hz.
m 2
5.00 kg
(b) With xo = 0.500 m, we have U0  12 kx02  125 J .
(c) With vo = 10.0 m/s, the initial kinetic energy is K0  12 mv02  250 J .
(d) Since the total energy E = Ko + Uo = 375 J is conserved, then consideration of the
energy at the turning point leads to
1
2E
E = kxm2  xm 
= 0.866 m.
2
k
39. (a) We take the angular displacement of the wheel to be  = m cos(2t/T), where
m is the amplitude and T is the period. We differentiate with respect to time to find
the angular velocity:  = –(2/T)msin(2t/T). The symbol  is used for the
angular velocity of the wheel so it is not confused with the angular frequency. The
maximum angular velocity is
m =
a fa f
2 m
2  rad
=
= 39.5 rad / s.
T
0.500 s
(b) When  = /2, then /m = 1/2, cos(2t/T) = 1/2, and
sin  2 t T   1  cos2  2 t T   1  1 2   3 2
2
where the trigonometric identity cos2+sin2= 1 is used. Thus,
=
3I
Ia fF
G
K H2 J
K= 34.2 rad / s.
F I F
HKH
2
2 t
2
 msin
=
 rad
T
T
0.500 s
During another portion of the cycle its angular speed is +34.2 rad/s when its angular
displacement is /2 rad.
(c) The angular acceleration is
d 2
 2 
 2 
 
  m cos  2 t / T    
 .
2
dt
 T 
 T 
2

2
When  = /4,
 2    
2
  
   = 124 rad/s ,
0.500
s
4

  
2
or |  |  124 rad/s 2 .
51. This is similar to the situation treated in Sample Problem 15-5, except that O is no
longer at the end of the stick. Referring to the center of mass as C (assumed to be the
geometric center of the stick), we see that the distance between O and C is h = x. The
parallel axis theorem (see Eq. 15-30) leads to
I=
F
IJ
G
H K
1
L2
mL2 + mh 2 = m
+ x2 .
12
12
And Eq. 15-29 gives
I
T  2
 2
mgh
c  x h 2 cL  12 x h.
L2
12
2
2
gx
2
12 gx
(a) Minimizing T by graphing (or special calculator functions) is straightforward, but
the standard calculus method (setting the derivative equal to zero and solving) is
somewhat awkward. We pursue the calculus method but choose to work with
12gT2/2 instead of T (it should be clear that 12gT2/2 is a minimum whenever T is a
minimum).
d
e j  0  d d  12 xi   L  12
12 gT 2
2
L2
x
dx
2
dx
x2
which yields x  L / 12  (1.85 m)/ 12  0.53 m as the value of x which should
produce the smallest possible value of T.
(b) With L = 1.85 m and x = 0.53 m, we obtain T = 2.1 s from the expression derived
in part (a).
60. (a) From Hooke’s law, we have
k=
 500 kg   9.8 m/s2 
10cm
 4.9 102 N/cm.
(b) The amplitude decreasing by 50% during one period of the motion implies
e  bT 2 m 
1
2
.
where T 
2

Since the problem asks us to estimate, we let      k / m . That is, we let
 
49000 N / m
 9.9 rad / s,
500 kg
so that T  0.63 s. Taking the (natural) log of both sides of the above equation, and
rearranging, we find
b=
b gb g
2 500
2m
ln2 
0.69 = 1.1  103 kg / s.
T
0.63
Note: if one worries about the ´   approximation, it is quite possible (though
messy) to use Eq. 15-43 in its full form and solve for b. The result would be (quoting
more figures than are significant)
b=
2 ln 2 mk
(ln 2)2 + 4  2
= 1086 kg / s
which is in good agreement with the value gotten “the easy way” above.
63. With M = 1000 kg and m = 82 kg, we adapt Eq. 15-12 to this situation by writing

2
k
.

T
M  4m
If d = 4.0 m is the distance traveled (at constant car speed v) between impulses, then
we may write T = v/d, in which case the above equation may be solved for the spring
constant:
2 v
k
 2 v 
=
 k   M  4m  
 .
d
M  4m
 d 
2
Before the people got out, the equilibrium compression is xi = (M + 4m)g/k, and
afterward it is xf = Mg/k. Therefore, with v = 16000/3600 = 4.44 m/s, we find the rise
of the car body on its suspension is
F I = 0.050 m.
HK
4mg
4mg
d
xi  x f =
=
k
M + 4m 2v
2
Chap 16
6. (a) The amplitude is ym = 6.0 cm.
(b) We find  from 2/ = 0.020:  = 1.0×102 cm.
(c) Solving 2f =  = 4.0, we obtain f = 2.0 Hz.
(d) The wave speed is v = f = (100 cm) (2.0 Hz) = 2.0×102 cm/s.
(e) The wave propagates in the –x direction, since the argument of the trig function is
kx + t instead of kx – t (as in Eq. 16-2).
(f) The maximum transverse speed (found from the time derivative of y) is


umax  2 fym  4.0  s 1  6.0 cm   75cm s.
(g) y(3.5 cm, 0.26 s) = (6.0 cm) sin[0.020(3.5) + 4.0(0.26)] = –2.0 cm.
8. With length in centimeters and time in seconds, we have
u
=
du
dt
= 225 sin (x  15t)
.
Squaring this and adding it to the square of 15y, we have
u2 + (15y)2
so that
= (225 )2 [sin2 (x  15 t) + cos2 (x  15 t)]
u
=
(225)2 - (15y)2
= 15
152 - y2
.
Therefore, where y = 12, u must be  135. Consequently, the speed there is 424
cm/s = 4.24 m/s.
19. (a) We read the amplitude from the graph. It is about 5.0 cm.
(b) We read the wavelength from the graph. The curve crosses y = 0 at about x = 15
cm and again with the same slope at about x = 55 cm, so
 = (55 cm – 15 cm) = 40 cm = 0.40 m.
(c) The wave speed is v   /  , where  is the tension in the string and  is the
linear mass density of the string. Thus,
v
3.6 N
 12 m/s.
25 103 kg/m
(d) The frequency is f = v/ = (12 m/s)/(0.40 m) = 30 Hz and the period is
T = 1/f = 1/(30 Hz) = 0.033 s.
(e) The maximum string speed is
um = ym = 2fym = 2(30 Hz) (5.0 cm) = 940 cm/s = 9.4 m/s.
(f) The angular wave number is k = 2/ = 2/(0.40 m) = 16 m–1.
(g) The angular frequency is  = 2f = 2(30 Hz) = 1.9×102 rad/s
(h) According to the graph, the displacement at x = 0 and t = 0 is 4.0  10–2 m. The
formula for the displacement gives y(0, 0) = ym sin . We wish to select  so that 5.0 
10–2 sin  = 4.0  10–2. The solution is either 0.93 rad or 2.21 rad. In the first case the
function has a positive slope at x = 0 and matches the graph. In the second case it has
negative slope and does not match the graph. We select  = 0.93 rad.
(i) The string displacement has the form y (x, t) = ym sin(kx + t + ). A plus sign
appears in the argument of the trigonometric function because the wave is moving in
the negative x direction. Using the results obtained above, the expression for the
displacement is
y ( x, t )   5.0  102 m  sin (16 m 1 ) x  (190s 1 )t  0.93 .
21. The pulses have the same speed v. Suppose one pulse starts from the left end of
the wire at time t = 0. Its coordinate at time t is x1 = vt. The other pulse starts from the
right end, at x = L, where L is the length of the wire, at time t = 30 ms. If this time is
denoted by t0 then the coordinate of this wave at time t is x2 = L – v(t – t0). They meet
when x1 = x2, or, what is the same, when vt = L – v(t – t0). We solve for the time they
meet: t = (L + vt0)/2v and the coordinate of the meeting point is x = vt = (L + vt0)/2.
Now, we calculate the wave speed:
v
L
m

(250 N) (10.0 m)
 158m/s.
0.100 kg
Here  is the tension in the wire and L/m is the linear mass density of the wire. The
coordinate of the meeting point is
10.0 m  (158m/s) (30.0 103 s)
x
 7.37 m.
2
This is the distance from the left end of the wire. The distance from the right end is
L – x = (10.0 m – 7.37 m ) = 2.63 m.
22. (a) The tension in each string is given by  = Mg/2. Thus, the wave speed in string
1 is
v1 

Mg
(500 g) (9.80 m/s 2 )


 28.6 m/s.
1
21
2(3.00 g/m)
(b) And the wave speed in string 2 is
v2 
Mg
(500g) (9.80 m/s 2 )

 22.1m/s.
2 2
2(5.00g/m)
(c) Let v1  M1 g /(21 )  v2  M 2 g /(22 ) and M1 + M2 = M. We solve for M1 and
obtain
M1 
M
1   2 / 1

500 g
 187.5g  188g.
1  5.00 / 3.00
(d) And we solve for the second mass: M2 = M – M1 = (500 g – 187.5 g)  313 g.
29. The displacement of the string is given by
y  ym sin(kx  t )  ym sin(kx  t   )  2 ym cos  12   sin  kx  t  12   ,
where  = /2. The amplitude is
A  2 ym cos  12    2 ym cos( / 4)  1.41ym .
34. The phasor diagram is shown below. We use the cosine theorem:
ym2  ym2 1  ym2 2  2 ym1 ym 2 cos   ym2 1  ym2 2  2 ym1 ym 2 cos .
We solve for cos :
cos  
ym2  ym2 1  ym2 2 (9.0 mm) 2  (5.0 mm) 2  (7.0 mm) 2

 0.10.
2 ym1 ym 2
2(5.0 mm) (7.0 mm)
The phase constant is therefore  = 84°.
43. (a) Eq. 16–26 gives the speed of the wave:
v

150 N

 144.34 m/s  1.44 102 m/s.
3

7.20 10 kg/m
(b) From the Figure, we find the wavelength of the standing wave to be  = (2/3)(90.0
cm) = 60.0 cm.
(c) The frequency is
f 
v 1.44 102 m/s

 241Hz.

0.600 m
47. (a) The amplitude of each of the traveling waves is half the maximum
displacement of the string when the standing wave is present, or 0.25 cm.
(b) Each traveling wave has an angular frequency of  = 40 rad/s and an angular
wave number of k = /3 cm–1. The wave speed is
v = /k = (40 rad/s)/(/3 cm–1) = 1.2×102 cm/s.
(c) The distance between nodes is half a wavelength: d = /2 = /k = /(/3 cm–1) =
3.0 cm. Here 2/k was substituted for .
(d) The string speed is given by u(x, t) = y/t = –ymsin(kx)sin(t). For the given
coordinate and time,
 



9
u  (40 rad/s) (0.50cm) sin  cm 1  (1.5cm)  sin  40 s 1 

8
 3





s   0.

53. (a) The angular frequency is  = 8.00/2 = 4.00 rad/s, so the frequency is
f = /2 = (4.00 rad/s)/2 = 2.00 Hz.
(b) The angular wave number is k = 2.00/2 = 1.00 m–1, so the wavelength is
 = 2/k = 2/(1.00 m–1) = 2.00 m.
(c) The wave speed is
v  f  (2.00 m) (2.00 Hz) = 4.00 m/s.
(d) We need to add two cosine functions. First convert them to sine functions using
cos  = sin ( + /2), then apply




     
   
cos   cos   sin      sin      2sin 
 cos 

2
2
2




 2 
   
   
 2 cos 
 cos 

 2 
 2 
Letting  = kx and  = t, we find
ym cos(kx  t )  ym cos(kx  t )  2 ym cos(kx) cos(t ).
Nodes occur where cos(kx) = 0 or kx = n + /2, where n is an integer (including
zero). Since k = 1.0 m–1, this means x   n  12  (1.00m) . Thus, the smallest value of
x which corresponds to a node is x = 0.500 m (n=0).
(e) The second smallest value of x which corresponds to a node is x = 1.50 m (n=1).
(f) The third smallest value of x which corresponds to a node is x = 2.50 m (n=2).
(g) The displacement is a maximum where cos(kx) = 1. This means kx = n, where n
is an integer. Thus, x = n(1.00 m). The smallest value of x which corresponds to an
anti-node (maximum) is x = 0 (n=0).
(h) The second smallest value of x which corresponds to an anti-node (maximum) is
x  1.00 m (n=1).
(i) The third smallest value of x which corresponds to an anti-node (maximum) is
x  2.00 m (n=2).
55. (a) The frequency of the wave is the same for both sections of the wire. The wave
speed and wavelength, however, are both different in different sections. Suppose there
are n1 loops in the aluminum section of the wire. Then,
L1 = n11/2 = n1v1/2f,
where 1 is the wavelength and v1 is the wave speed in that section. In this
consideration, we have substituted 1 = v1/f, where f is the frequency. Thus f =
n1v1/2L1. A similar expression holds for the steel section: f = n2v2/2L2. Since the
frequency is the same for the two sections, n1v1/L1 = n2v2/L2. Now the wave speed in
the aluminum section is given by 1   / 1 , where 1 is the linear mass density of
the aluminum wire. The mass of aluminum in the wire is given by m1 = 1AL1, where
1 is the mass density (mass per unit volume) for aluminum and A is the
cross-sectional area of the wire. Thus
1 = 1AL1/L1 = 1A
and 1   / 1 A. A similar expression holds for the wave speed in the steel section:
v2   / 2 A. We note that the cross-sectional area and the tension are the same for
the two sections. The equality of the frequencies for the two sections now leads to
n1 / L1 1  n2 / L2 2 , where A has been canceled from both sides. The ratio of the
integers is
3
3
n2 L2 2  0.866 m  7.80 10 kg/m


 2.50.
n1 L1 1  0.600 m  2.60 103 kg/m3
The smallest integers that have this ratio are n1 = 2 and n2 = 5. The frequency is
f  n1v1 / 2L1   n1 / 2L1   / 1 A.
The tension is provided by the hanging block and is  = mg, where m is the mass of
the block. Thus
n
f  1
2 L1
mg
2

1 A 2  0.600 m 
10.0 kg   9.80 m/s2 
 2.60 10 kg/m 1.00 10
3
3
6
m2

 324 Hz.
(b) The standing wave pattern has two loops in the aluminum section and five loops in
the steel section, or seven loops in all. There are eight nodes, counting the end points.