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Bab 5 Aplikasi Hukum-hukum Newton
Modern parachutes also add a third term, lift, to
change the equation to
SOALAN-SOALAN
m
D  A 2 L A 2
vy 
vx
dt
2
2
the vertical velocity, and vx
dvy
 m g
10. It has been suggested that rotating cylinders
about 10 mi in length and 5 mi in diameter be
placed in space and used as colonies. The
purpose of the rotation is to simulate gravity for
the inhabitants. Explain this concept for
producing an effective imitation of gravity.
where vy is
is the
horizontal velocity. The effect of lift is clearly
seen in the “paraplane,” an ultralight airplane
made from a fan, a chair, and a parachute.
Solution
This is the same principle as the centrifuge. All
the material inside the cylinder tends to move
along a straight-line path, but the walls of the
cylinder exert an inward force to keep
everything moving around in a circular path.
MASALAH-MASALAH
14. Why does a pilot tend to black out when
pulling out of a steep dive?
Solution
Blood pressure cannot supply the force
necessary both to balance the gravitational force
and to provide the centripetal acceleration, to
keep blood flowing up to the pilot’s brain.
17. A falling sky diver reaches terminal speed
with her parachute closed. After the parachute is
opened, what parameters change to decrease this
terminal speed?
Solution
From the proportionality of the drag force to the
speed squared and from Newton’s second law,
we derive the equation that describes the motion
of the skydiver:
DA 2
m
 m g
vy
dt
2
dvy
where D is the coefficient of drag of the
parachutist, and A is the projected area of the
parachutist’s body. At terminal speed,
 2m g 
ay 
 0 and VT 
dt
 D  A 
dvy
12
.
When the parachute opens, the coefficient of
drag D and the effective area A both increase,
thus reducing the speed of the skydiver.
1.
A light string can support a
stationary hanging load of 25.0 kg before
breaking. A 3.00-kg object attached to the
string rotates on a horizontal, frictionless
table in a circle of radius 0.800 m, while the
other end of the string is held fixed. What
range of speeds can the object have before
the string breaks?
Solution
m  3.00 kg , r 0.800 m . The string will break if
the tension exceeds the weight corresponding to
25.0 kg, so Tm ax  M g  25.0 9.80  245 N .
When the 3.00 kg mass rotates in a horizontal
circle, the tension causes the centripetal
acceleration,
so
T
m v2  3.00 v2
.

r
0.800
Then
v2 
and
rT  0.800 T  0.800 Tm ax 0.800 245



 65.3 m
m
3.00
3.00
3.00
0  v  65.3 or
0  v  8.08 m s .
2
s2
The force causing the centripetal acceleration is
the frictional force f. From Newton’s second
m v2
. But the friction condition is
r
m v2
f  sn i.e.,
 sm g
r
law f  m ac 

v  srg  0.600 35.0 m  9.80 m s2

v 14.3 m s
5.
A coin placed 30.0 cm from the
center of a rotating, horizontal turntable
slips when its speed is 50.0 cm/s. (a) What
force causes the centripetal acceleration
when the coin is stationary relative to the
turntable? (b) What is the coefficient of
static friction between coin and turntable?
Solution
(a)
(b)
13.
A 40.0-kg child swings in a swing
supported by two chains, each 3.00 m long.
If the tension in each chain at the lowest
point is 350 N, find (a) the child's speed at
the lowest point and (b) the force exerted
by the seat on the child at the lowest point.
(Neglect the mass of the seat.)
static friction
 
Solution
m aˆ
i fˆ
i nˆ
j m g  ˆ
j
 Fy  0  n  m g
 Fr  m
Then  
thus n  m g and
2
v
 f  n  m g .
r
 50.0 cm s
v

 0.085 0
rg  30.0 cm  980 cm s2
2
2


7.
A crate of eggs is located in the
middle of the flat bed of a pickup truck as
the truck negotiates an unbanked curve in
the road. The curve may be regarded as an
arc of a circle of radius 35.0 m. If the
coefficient of static friction between crate
and truck is 0.600, how fast can the truck be
moving without the crate sliding?
Solution
n  m g since ay  0
M  40.0 kg , R  3.00 m , T  350 N
(a)
 F  2T  M g 
M v2
R
R
v2   2T  M g  
M 

 3.00
v2  700   40.0 9.80  
 23.1 m
 40.0
2
v  4.81 m s
(b) n  M g  F 
n  M g
M v2
R
M v2
23.1

 40.0 9.80 
  700 N

R
3.00
s2

T
T
Mg
circle if no water is to spill out?
Solution
m v2
 m g n
r
But n  0 at this minimum speed condition, so
 Fy 
Mg
child + seat
n
child alone
FIG. P6.13(b)
FIG. P6.13(a)
15.
Tarzan (m = 85.0 kg) tries to cross a
river by swinging from a vine. The vine is
10.0 m long, and his speed at the bottom of
the swing (as he just clears the water) will
be 8.00 m/s. Tarzan doesn't know that the
vine has a breaking strength of 1 000 N.
Does he make it safely across the river?
Solution
Let the tension at the lowest point be T.
mv 2
 F  ma : T  mg  mac  r

v2 
T  m g  

r 


8.00 m s 2 
T   85.0 kg  9.80 m s 2 
10.0 m 



 1.38 kN  1 000 N
He doesn’t make it across the river because
the vine breaks.
m v2
 m g  v  gr 
r
9.80 m s  1.00 m  
2
3.13 m s
.
23.
A 0.500-kg object is suspended from
the ceiling of an accelerating boxcar as in
Figure 6.13. If a = 3.00 m/s2, find (a) the
angle that the string makes with the vertical
and (b) the tension in the string.
Solution
The only forces acting on the suspended object
are the force of gravity m g and the force of
tension T, as shown in the free-body diagram.
Applying Newton’s second law in the x and y
directions,
(1)
 Fx  T sin  m a
 Fy  T cos  m g  0
or
(2)
T cos  m g
T cos 
T sin 
mg
(a)
Dividing equation (1) by (2) gives
a 3.00 m s2
tan   
 0.306 .
g 9.80 m s2
Solving for ,   17.0
FIG. P6.15
17.
A pail of water is rotated in a vertical
circle of radius 1.00 m. What is the
minimum speed of the pail at the top of the
(b)From Equation (1),
T


2
m a  0.500 kg 3.00 m s

 5.12 N .
sin 
sin 17.0
(b)At t 40.0 s
25.
A person stands on a scale in an
elevator. As the elevator starts, the scale has
a constant reading of 591 N. As the
elevator later stops, the scale reading is
391 N. Assume the magnitude of the
acceleration is the same during starting and
stopping, and determine (a) the weight of
the person, (b) the person's mass, and (c)
the acceleration of the elevator.
Solution
Fm ax  Fg  m a  591 N
Fm in  Fg  m a  391 N
(a)
Adding, 2Fg  982 N , Fg  491 N
(b)
Since Fg  m g , m 
(c)
Subtracting the above equations,
2m a 200 N
491 N
 50.1 kg
9.80 m s2
 a  2.00 m s2
39.
A motorboat cuts its engine when its
speed is 10.0 m/s and coasts to rest. The
equation describing the motion of the
motorboat during this period is v = vie–ct,
where v is the speed at time t, vi is the initial
speed, and c is a constant. At t = 20.0 s, the
speed is 5.00 m/s. (a) Find the constant c.
(b) What is the speed at t = 40.0 s? (c)
Differentiate the expression for v(t) and
thus show that the acceleration of the boat
is proportional to the speed at any time.
Solution
v t  viect
(a)
v 20.0 s  5.00  vie20.0c , vi  10.0 m s.
 1
So 5.00  10.0e20.0c and 20.0c  ln  
 2
c 
ln
 12  
20.0
3.47  102 s1
v  10.0 m s e40.0c  10.0 m s  0.250  2.50 m s
(c)
v  viect
s
dv
 cviect  cv
dt
57.
Because the Earth rotates about its
axis, a point on the equator experiences a
centripetal acceleration of 0.033 7 m/s2,
while a point at the poles experiences no
centripetal acceleration. (a) Show that at the
equator the gravitational force on an object
must exceed the normal force required to
support the object. That is, show that the
object's true weight exceeds its apparent
weight. (b) What is the apparent weight at
the equator and at the poles of a person
having a mass of 75.0 kg? (Assume the
Earth is a uniform sphere and take g =
9.800 m/s2.)
Solution
(a)
Since the centripetal acceleration of a person
is downward (toward the axis of the earth), it is
equivalent to the effect of a falling elevator.
Therefore,
m v2
or Fg  Fg
r
(b)
At the poles v  0 and
Fg  Fg  m g  75.0 9.80  735 N
Fg  Fg 
down.
65.
An amusement park ride consists of
a large vertical cylinder that spins about its
axis fast enough that any person inside is
held up against the wall when the floor
drops away (Fig. P6.65). The coefficient of
static friction between person and wall is µs,
and the radius of the cylinder is R. (a) Show
that the maximum period of revolution
necessary to keep the person from falling is
T = (42Rµs/g)1/2. (b) Obtain a numerical
value for T if R = 4.00 m and µs = 0.400. How
many revolutions per minute does the
cylinder make?
71.
A model airplane of mass 0.750 kg
flies in a horizontal circle at the end of a
60.0-m control wire, with a speed of
35.0 m/s. Compute the tension in the wire if
it makes a constant angle of 20.0 with the
horizontal. The forces exerted on the
airplane are the pull of the control wire, the
gravitational force, and aerodynamic lift,
which acts at 20.0 inward from the vertical
as shown in Figure P6.71.
Figure P6.65
Solution
(a)
n
m v2
R
f  sn
T
(b)
f m g  0
v
2 R
T
4 2R s
g
Figure P6.71
T  2.54 s
#
rev 1 rev  60 s
rev


  23.6
m in 2.54 s m in
m in
Solution
 Fy  Ly  Ty  m g  L cos20.0  T sin 20.0  7.35 N
 Fx  Lx  Tx  L sin 20.0  T cos20.0  m
v
r
 35.0 m s  16.3 N
v2
 0.750 kg
r
 60.0 m  cos20.0
2
m
 L sin 20.0  T cos20.0  16.3 N
L cos20.0  T sin 20.0  7.35 N
cos20.0
16.3 N
L T

sin 20.0 sin 20.0
sin 20.0
7.35 N
L T

cos20.0 cos20.0
16.3 N
7.35 N
T  cot20.0  tan 20.0 

sin 20.0 cos20.0
T  3.11  39.8 N
T  12.8 N
2
 m ay  0