5.1: a) The tension in the rope must be equal to each suspended weight, 25.0 N. b) If the
mass of the light pulley may be neglected, the net force on the pulley is the vector sum of
the tension in the chain and the tensions in the two parts of the rope; for the pulley to be
in equilibrium, the tension in the chain is twice the tension in the rope, or 50.0 N.
5.2: In all cases, each string is supporting a weight w against gravity, and the tension in
each string is w. Two forces act on each mass: w down and
)( wT 
up.
5.3: a) The two sides of the rope each exert a force with vertical component T
θsin
, and
the sum of these components is the hero’s weight. Solving for the tension T,
.N 1054.2
0.01sin 2
)sm(9.80kg)0.90(
sin 2
3
2





w
T
b) When the tension is at its maximum value, solving the above equation for the angle
θ
gives
.01.1
N) 1050.2(2
sm(9.80kg)(90.0

3
2
4
3
2
 θθw
ww
5.5: With the positive y-direction up and the positive x-direction to the right, the free-
body diagram of Fig. 5.4(b) will have the forces labeled n and T resolved into x- and y-
components, and setting the net force equal to zero,
.0sincos
0sincos


wTnF
nTF
y
x


Solving the first for
αTn cot 
and substituting into the second gives
w
T
TTT





4
40cos
)smkg)(9.84090(
2


θWTWθT
BB
b)
N. 1036.340sin N)1023.5(sin
44
 θTT
BA
5.8: a)
.045cos30cosand,45sin30sin , 
BACBAC
TTwTTTwT
Since
,45cos45sin 
adding the last two equations gives
,)30sin30(cos wT
A

and so
.732.0
366.1
wT
w
A


BB



5.9: The resistive force is
N. 523m)6000m200)(smkg)(9.801600(sin
2
w
.
5.10: The magnitude of the force must be equal to the component of the weight along the
incline, or
N. 3370.11sin)smkg)(9.80180(sin
2
θW
5.11: a)
,sinN,60 WθTW 
so
,sin45N)60( T
or
N. 85T
b)
N. 6045cosN85,cos
2121
 FFθTFF
5.12: If the rope makes an angle

with the vertical, then
0 110
1 51
sin 0 073

,,0 wn 
when
.0,90  n
5.14: a) In level flight, the thrust and drag are horizontal, and the lift and weight are
vertical. At constant speed, the net force is zero, and so
fF 
and
.Lw 
b) When the
plane attains the new constant speed, it is again in equilibrium and so the new values of
the thrust and drag,
F

and
f

, are related by
fF



; if
.2,2 ffFF 



c) In order to
increase the magnitude of the drag force by a factor of 2, the speed must increase by a
factor of
2











mm
mm
ga
ammgmm
c) The result of part (b) may be substituted into either of the above expressions to find
the tension
N.191T
As an alternative, the expressions may be manipulated to
eliminate a algebraically by multiplying the first by
2
m
and the second by
1
m
and adding
(with
12
aa 
) to give
N.191

m
wT




If, as in this case,
21212
2, mmmmm 
and
,2
211
mmm 
so the tension is greater
than
1
w
and less than
;
2
w
this must be the case, since the load of bricks rises and the
counterweight drops.
5.16: Use Second Law and kinematics:
,2,sin
2
vaxθga 
solve for
θ
.

so
,mgT 
because
.0a
5.18: The maximum net force on the glider combination is
N, 7000N 25002N 000,12 
so the maximum acceleration is
.sm0.5
2
kg1400
N7000
max
a
a) In terms of the runway length L and takoff speed
,,
max
2
2
aav
L
v

so
m.160
)sm0.5(2
)sm40(

2

2

gaa
g
w
mawF
a)
22
sm78.1)1N)550(N)450()(sm80.9( a
, down.
b)
,sm14.2)1N) (550N)670()(sm80.9(
22
a
up. c) If
gaF  ,0
and the
student, scale, and elevator are in free fall. The student should worry.
5.20: Similar to Exercise 5.16, the angle is
),arcsin(
2
2
gt
L
, but here the time is found in
terms of velocity along the table,
xt
v
x
,
0























gx
Lv
vxg
L
5.21:
5.22:
5.23: a) For the net force to be zero, the applied force is
N. 0.22)sm(9.80kg)2.11()20.0(
2

n

e) The applied force is enough
to either start the box moving or to keep it moving. The answer to part (d), from
Eq. (5.5), is independent of speed (as long as the box is moving), so the friction force is
8.0 N. The acceleration is
.sm45.2)(
2
k
 mfF
5.25: a) At constant speed, the net force is zero, and the magnitude of the applied force
must equal the magnitude of the kinetic friction force,
N. 7)sm(9.80kg)00.6()12.0(
2
kkk
 mgμnμfF

b)
,maf
k
F

so
N. 8)sm80.9)12.0(smkg)(0.18000.6(
)(
22
kkk


gammgμmafma

Hk











g
a
wFf
(note that the acceleration is negative), and so
.25.0
N110
N28
k


5.27: As in Example 5.17, the friction force is

cos
kk
wn 
and the component of the
weight down the skids is
.sin

)sm7.28(
22
2
2
k
22



g
v
a
v
b) The stopping distance is inversely proportional to the coefficient of friction and
proportional to the square of the speed, so to stop in the same distance the initial speed
should not exceed
.sm 16
80.0
25.0
)sm7.28(
dryk,
wetk,



v
5.29: For a given initial speed, the distance traveled is inversely proportional to the
coefficient of kinetic friction. From Table 5.1, the ratio of the distances is then
.11
04.0

.
or,,
BA
BA
ff
fTfT


F
F


Using
AA
gmf
k


and
BB
gmf
k


gives
)(
k BA
mmg F

.






where L is the distance covered before the wheel’s speed is reduced to half its original
speed. Low pressure,
.0259.0m;1.18
)smm)(9.80(18.1
s)m50.3(
8
3
2
2
L
High pressure,
.00505.0m;9.92
)smm)(9.809.92(
)sm50.3(
8
3
2
2
L
5.33: Without the dolly:
mgn 
and
0
k
 nF


m
mgF
a
mamgF


5.34: Since the speed is constant and we are neglecting air resistance, we can ignore the
2.4 m/s, and
net
F
in the horizontal direction must be zero. Therefore
 nf
rr

N200
horiz
 F
before the weight and pressure changes are made. After the changes,
,)42.1()81.0(
horiz
Fn 
because the speed is still constant and
0
net
F
. We can simply
divide the two equations:
N230N)200()42.1()81.0(
)42.1)(81.0(

s.m22.0)2(
21
 axv
(b) Solving either equation for the tension gives
N. 7.11T
5.36: a) The normal force will be
θw cos
and the component of the gravitational force
along the ramp is
θw sin
. The box begins to slip when
,cossin
s
θwθw


or
,35.0tan
s


θ
so slipping occurs at
 3.19)35.0arctan(θ
, or
19
to two figures.
b) When moving, the friction force along the ramp is
θwcos
k

k
θmgμ FF



solving for
F

gives
θμθ
mgμ
sincos
k
k


F

b) If the crate remains at rest, the above expression, with
s

instead of
k

, gives the
force that must be applied in order to start the crate moving. If
,cot
s

θ


b) Using the given values,
N, 293
)25sin)35.0(25(cos
)smkg)(9.8090)(35.0(
2



F
or 290 N to two figures.
5.39: a)
b) The blocks move with constant speed, so there is no net force on block A; the
tension in the rope connecting A and B must be equal to the frictional force on block A,
N. 9N)0.25()35.0(
k

c) The weight of block C will be the tension in the rope
connecting B and C; this is found by considering the forces on block B. The components
of force along the ramp are the tension in the first rope (9 N, from part (a)), the
component of the weight along the ramp, the friction on block B and the tension in the
second rope. Thus, the weight of block C is
N,31.0)36.9(0.35)cos36.9N)(sin (25.0N9
)9.36cos9.36(sinN9
k



BC
ww






where Eq. (5.9),
kmgv 
t
has been used.
Integrating Eq. (5.10) with respect to time with
0
0
y
gives
 
.1
]1[
)(
t
t
)(
t
0
)(
t






tmk
e
k
m
tv
k
m
ve
k
m
tv
dtevy
5.41: a) Solving for D in terms of
t
v
,
.mkg44.0
)sm42(
)sm(9.80kg)80(
2
2
2

t
v
mg
D
b)
.sm42
m)kg25.0(

.290.0
)smm)(9.80(220
s)m0.25(
2
22
s

Rg
v

5.45: a) The magnitude of the force F is given to be equal to 3.8w. “Level flight” means
that the net vertical force is zero, so
,cos)8.3(cos wwβF 

, and
 75)8.31arccos(

.
(b) The angle does not depend on speed.
5.46: a) The analysis of Example 5.22 may be used to obtain
),(tan
2
gRv
but the
subsequent algebra expressing R in terms of L is not valid. Denoting the length of the
horizontal arm as r and the length of the cable as
.sin, βlrRl 
The relation
T
R








g
lr
T
Note that in the analysis of Example 5.22,
β
is the angle that the support (string or cable)
makes with the vertical (see Figure 5.30(b)). b) To the extent that the cable can be
considered massless, the angle will be independent of the rider’s weight. The tension in
the cable will depend on the rider’s mass.
5.47: This is the same situation as Example 5.22, with the lift force replacing the tension
in the string. As in that example, the angle
β
is related to the speed and the turning
radius by
.tan
2
gR
v


Solving for
β
,

β
5.48: a) This situation is equivalent to that of Example 5.23 and Problem 5.44, so
Rg
v
2
s


. Expressing v in terms of the period T,
,
2
T
R
v


so
.
2
2
4
S
gT
R



A platform speed
of 40.0 rev/min corresponds to a period of 1.50 s, so
.269.0

 π
g
R
πT
so the number of revolutions per minute is
minrev1.5s)1.40(min)s60( 
.
b) The lower acceleration corresponds to a longer period, and hence a lower rotation
rate, by a factor of the square root of the ratio of the accelerations,
min.rev92.09.83.70min)rev5.1( 

T
.
5.50: a)
.sm5.24s)(60.0m)0.50(22  TR
b) The magnitude of the radial force
is
N49)4(4
22222
 gTRwTRmRmv

(to the nearest Newton), so the apparent
weight at the top is
N,833N49N882 
and at the bottom is
N931N49N882 
.
c) For apparent weightlessness, the radial acceleration at the top is equal to g in
magnitude. Using this in Eq. (5.16) and solving for T gives
.s14



















g
a
wmaw
or 3580 N to three places.
5.52: a) Solving Eq. (5.14) for R,
m.230)sm80.94(s)m0.95(4
2222
 gvavR
b) The apparent weight will be five times the actual weight,
N 2450)sm(9.80kg)0.50(55
2
mg
















g
a
wgamT
5.55: a)

1
T
is more vertical so supports more
of the weight and is larger.
You can also see this from
:
xx
maF 

221


N6400
40sin60sin
21



w
wTT
maF
yy
5.56:
The tension in the lower chain balances the weight and so is equal to w. The lower
pulley must have no net force on it, so twice the tension in the rope must be equal to w,
and so the tension in the rope is
2w
. Then, the downward force on the upper pulley due
to the rope is also w, and so the upper chain exerts a force w on the upper pulley, and the
tension in the upper chain is also w.
5.57: In the absence of friction, the only forces along the ramp are the component of the
weight along the ramp,

sinw
, and the component of
F

along the ramp,
 coscos FF

. These forces must sum to zero, so

,cos
middleend
TθT 
so
.)tan 2()sin2(cos
middle
θMgθθMgT 
(c) Mathematically speaking,
0


because this would cause a division by zero in the
equation for
end
T
or
middle
T
. Physically speaking, we would need an infinite tension to
keep a non-massless rope perfectly straight.
5.59: Consider a point a distance x from the top of the rope. The forces acting in this
point are T up and
 
gM
L
xLm
)( 

downwards. Newton’s Second Law becomes
   

mM
MF


as expected.
5.60: a) The tension in the cord must be
2
m g
in order that the hanging block move at
constant speed. This tension must overcome friction and the component of the
gravitational force along the incline, so
 
 cossin
112
gmμgmgm
k
and
)cos(sin
12

k
μmm 
.
b) In this case, the friction force acts in the same direction as the tension on the block
of mass
1
m
, so
)cossin(
1k12

be the tension in the horizontal wire, and hence the friction force on block A is also 12.0
N
. b) The maximum frictional force is
N 15N)0.60)(25.0(
s

A
wμ
; this will be the
tension in both the horizontal and vertical parts of the wire, so the maximum weight is 15
N.
5.62: a) The most direct way to do part (a) is to consider the blocks as a unit, with total
weight 4.80 N. Then the normal force between block B and the lower surface is 4.80 N,
and the friction force that must be overcome by the force F is
N, 1.44or N, 1.440N) 80.4)(30.0(
k
nμ
to three figures. b) The normal force
between block B and the lower surface is still 4.80 N, but since block A is moving
relative to block B, there is a friction force between the blocks, of magnitude
N,0.360N) 20.1)(30.0( 
so the total friction force that the force F must overcome is
N 1.80N0.360N440.1 
. (An extra figure was kept in these calculations for clarity.)
5.63: (Denote
F

by F.) a) The force normal to the surface is
θFn cos
; the vertical



 mggmmaF
This force is 62.5 times the flea’s weight.
b)
N 102.9dynes29
140)140(
4
maxmax


 mggmmaF
Occurs at approximately 1.2 ms.
c)
 vvvvv 0
0
area under a-t graph. Approximate area as shown:
sm2.1scm120
g)ms)(14005.0(
2
1

g)ms)(62.5(1.2g)ms)(77.52.1(
2
1
)3()2()1(



 AAAA

. Let F be the thrust of the
rocket engines.
N1072.5)sm07.13sm (9.80kg)000,25()(
522


agmF
mamgF
b)
m.4170gives
0
2
2
1
00
 yytatvyy
yy
5.66: The elevator’s acceleration is:
tt
dt
tdv
a
)sm40.0(sm0.3)sm20.0(2sm0.3
)(
32
3
2

At
232

2
0


yyyy
yyy
vyyavv
vvayy
5.68: (a) Choosing upslope as the positive direction:
mamgmgfmgF  37cos37sin37sin
kknet

and
22
sm25.8))799.0)(30.0(602.0()sm8.9( a
Since we know the length of the slope, we can use
)(2
0
2
0
2
xxavv 
with
0
0
x
and
0v
at the top.
sm11or sm5.11sm132

0
2
00




v
xxavv
xxv