Transcript File - Phy 2048-0002

Chapter 6 – Force and Motion II
•
Drag forces and terminal speed.
•
Uniform circular motion.
•
Other Applications of Newton’s Laws
Uniform Circular Motion
• A force, Fr , is directed
toward the center of
the circle
• This force is associated
with an acceleration, ac
• Applying Newton’s
Second Law along the
radial direction gives
2
v
 F  mac  m r
Uniform Circular Motion
• A force causing a
centripetal acceleration
acts toward the center of
the circle
• It causes a change in the
direction of the velocity
vector
• If the force vanishes, the
object would move in a
straight-line path tangent to
the circle
Centripetal Force
• The force causing the centripetal acceleration
is sometimes called the centripetal force
v2
 F  mac  m r
• This is not a new force, it is a new role for a
force
• It is a force acting in the role of a force that
causes a circular motion
Conical Pendulum
• The object is in equilibrium in
the vertical direction and
undergoes uniform circular
motion in the horizontal
direction
(1) Tcosθ = mg
(2) Tsinθ = mac= mv2/r
Conical Pendulum
• Dividing (2) by (1) and using
sinθ/cosθ = tanθ
we eliminate T and find:
v2
tan  
rg

v  rg tan 
Using r = Lsinθ
v
Lg sin  tan 
• v is independent of m
Consider a conical pendulum with an 80.0-kg bob on a
10.0-m wire making an angle of 5.00 with the vertical.
Determine (a) the horizontal and vertical components of the
force exerted by the wire on the pendulum and (b) the
radial acceleration of the bob.
Consider a conical pendulum with an 80.0-kg bob on a
10.0-m wire making an angle of 5.00 with the vertical.
Determine (a) the horizontal and vertical components of the
force exerted by the wire on the pendulum and (b) the
radial acceleration of the bob.

T cos5.00  m g   80.0 kg 9.80 m s2
(a)
T  787 N
(b)
T sin 5.00  m ac
T

 68.6 N  iˆ  784 N  ˆj
:
toward the center of the
circle.
ac= (T sin5.00/)m = .857 m/s2
The length of the wire is unnecessary information. We could,
on the other hand, use it to find the radius of the circle, the
speed of the bob, and the period of the motion
A 4.00-kg object is attached to a vertical rod by two strings,
as in Figure. The object rotates in a horizontal circle at
constant speed 6.00 m/s. Find the tension in (a) the upper
string and (b) the lower string.
A 4.00-kg object is attached to a vertical rod by two strings,
as in Figure. The object rotates in a horizontal circle at
constant speed 6.00 m/s. Find the tension in (a) the upper
string and (b) the lower string.
Ta
2


Fg  m g   4 kg 9.8 m s  39.2 N

1.5 m
sin  
2m
  48.6
r  2 m  cos48.6  1.32 m
39.2 N
Tb
forces
m v2
 Fx  m ax  r
Ta cos48.6  Tb
Ta  Tb 
4 kg 6 m s

cos48.6 
109 N
 165 N
cos48.6
ac
v
2
1.32 m
motion
A 4.00-kg object is attached to a vertical rod by two strings,
as in Figure. The object rotates in a horizontal circle at
constant speed 6.00 m/s. Find the tension in (a) the upper
string and (b) the lower string.
Ta
2
 Fx  m ax 
mv
r
Ta cos48.6  Tb
Ta  Tb 
4 kg 6 m s

cos48.6 

2
39.2 N
1.32 m
109 N
 165 N
cos48.6
 Fy  m ay
Ta sin 48.6  Tb sin 48.6  39.2 N  0
39.2 N
Ta  Tb 
 52.3 N
sin 48.6
Tb
forces
ac
v
motion
A 4.00-kg object is attached to a vertical rod by two strings,
as in Figure. The object rotates in a horizontal circle at
constant speed 6.00 m/s. Find the tension in (a) the upper
string and (b) the lower string.
Ta
(a)To solve simultaneously, we add the
equations in Ta and Tb :

Ta  Tb  Ta  Tb  165 N  52.3 N
217 N
Ta 
 108 N
2
(b)
39.2 N
Tb
forces
ac
v
Tb  165 N  Ta  165 N  108 N  56.2 N
motion
How fast can it spin?
• The speed at which the
object moves depends
on the mass of the object
and the tension in the
cord. The centripetal
force is supplied by the
tension:
v2
T m
r
Tr
v
m
This shows tat v increases with T
and decreases with larger m.
Horizontal (Flat) Curve
• The force of static friction
supplies the centripetal force
2
v
fS  m
r
• The maximum speed at which
the car can negotiate the curve
is
v   gr
• Note, this does not depend on
the mass of the car
Fr
Banked Curve
• These are designed with
friction equaling zero
• There is a component of the
normal force that supplies the
centripetal force (1), and
component of the normal
force that supplies the
gravitational force (2).
• Dividing (1) by (2) gives:
2
v
tan  
rg
mv2
n sin  
r
(1)
n cos   mg
(2)
Loop-the-Loop
• A pilot in a jet aircraft
executes a loop-the loop.
This is an example of a
vertical circle
• At the bottom of the loop,
(b), the upward constant
force, that keeps the pilot
moving in a circular path
at a constant speed, is
greater than its weight,
because the normal and
gravitational forces act in
opposite direction:
2
v
 F  nbot  mg  m r
 v 
nbottom  mg 1  
 rg 
2
Loop-the-Loop
• At the top of the circle,
(c), the force exerted on
the object is less than
its weight, because
both the gravitational
force and the normal
force, ntop, exerted on
the pilot by the seat act
in same direction:
v2
 F  ntop  mg  m r
ntop
 v2

 mg   1
 rg 
A 0.400-kg object is swung in a vertical circular path on a
string 0.500 m long. If its speed is 4.00 m/s at the top of the
circle, what is the tension in the string there?
A 0.400-kg object is swung in a vertical circular path on a
string 0.500 m long. If its speed is 4.00 m/s at the top of the
circle, what is the tension in the string there?
At the top of the vertical circle,
T   0.400
 4.00
2
0.500
v2
Tm
mg
R
  0.400 9.80  8.88 N
Non-Uniform Circular Motion
• When the force acting on a
particle moving in a circular
path has a tangential
component
Ft , the
particles speed changes.
• The acceleration has a
tangential components
• Fr produces the centripetal
acceleration
• Ft produces the tangential
acceleration.
• The total force is the vector
sum or the radial force and
tangential force:

F F F
r
t
Vertical Circle with Non-Uniform Speed
• The gravitational force
exerts a tangential force on
the object
– Look at the components of Fg
• The gravitational force
resolve into a tangential
component mg sinθ and a
radial component mg cosθ.
Vertical Circle with Non-Uniform Speed
•Applying II NL to the tangential and
radial directions:
 F  mg sin   ma
t
t
at  g sin 
mv2
 Fr  T  mg cos  r
The tension at any point
can be found:
v

T  m   g cos  
R

2
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?
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?
Let the tension at the lowest point be T.
m v2
 F  m a : T  m g  m ac  r

v2 
T  m  g 
r

2

00 m s
2  8.

  1.38 kN  1000 N
T   85.0 kg 9.80 m s 
10.0 m




H e doesn’tm ake itacross the riverbecause the vine breaks.
Motion in accelerated Frames
When Newton’s laws of motion was
introduced in Chapter 5, we emphasized that
they are valid only for inertial frames of
reference.
In this section, we will analyze the noninertial
frames, that is, on that is accelerating.
Example: Let’s consider a hockey puck on at
table in a moving train. The train moving with a
constant velocity represents an inertial frame.
The puck at rest remains at rest, and Newton’s I
low is obeyed.
Motion in accelerated Frames
Example: Let’s consider a hockey puck on at table in a
moving train. The train moving with a constant velocity
represents an inertial frame. The puck at rest remains at
rest, and Newton’s I low is obeyed.
The accelerating train is not an inertial frame. For the
observer on the train, there appears to be no visible force
on the puck, but it will accelerate from rest toward the back
of the train, as the train start to accelerate. The Newton’s I
law is violated.
The observer on the accelerating train, if he applied the
N II law to the puck, might conclude that a force has acted
on the puck to cause it to accelerate.
We call an apparent force such as this a fictitious
force.
Motion in Accelerated Frames
• A fictitious force results from an
accelerated frame of reference
– A fictitious force appears to act on an object in
the same way as a real force, but you cannot
identify a second object for the fictitious force
“Centrifugal” Force
• From the frame of the passenger
(b), a force appears to push her
toward the door
• From the frame of the Earth, the car
applies a leftward force on the
passenger
• The outward force is often called a
centrifugal force
– It is a fictitious force due to the
acceleration associated with the car’s
change in direction
If the coefficient of static friction between your coffee cup
and the horizontal dashboard of your car is μs = 0.800, how
fast can you drive on a horizontal roadway around a right
turn of radius 30.0 m before the cup starts to slide? If you
go too fast, in what direction will the cup slide relative to the
dashboard?
If the coefficient of static friction between your coffee cup
and the horizontal dashboard of your car is μs = 0.800, how
fast can you drive on a horizontal roadway around a right
turn of radius 30.0 m before the cup starts to slide? If you
go too fast, in what direction will the cup slide relative to the
dashboard?
We adopt the view of an inertial observer. If it is on the verge
of sliding, the cup is moving on a circle with its centripetal
acceleration caused by friction.
 Fy  m ay :
n  m g  0
m v2
 Fx  m ax : f  r  sn  sm g


v  sgr  0.8 9.8 m s2  30 m   15.3 m s
If you go too fast the cup will begin sliding
straightacross the dashboard to the left.
•
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.
•
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.
Fm ax  Fg  m a  591 N
+
Fm in  Fg  m a  391 N
2Fg  982 N
(b) m 
491 N
 50.1 kg
2
9.80 m s
(a) Fg  491 N
•
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.
(c)
-
Fm ax  Fg  m a  591 N
Fm in  Fg  m a  391 N
2m a 200 N
 a  2.00 m s2
“Coriolis Force”
• This is an apparent
force caused by
changing the radial
position of an object
in a rotating
coordinate system
• The result of the
rotation is the curved
path of the ball
The Earth rotates about its axis with a period of 24.0 h.
Imagine that the rotational speed can be increased. If an
object at the equator is to have zero apparent weight, (a)
what must the new period be? (b) By what factor would
the speed of the object be increased when the planet is
rotating at the higher speed?
Note that the apparent weight of the object becomes
zero when the normal force exerted on it is zero.
The Earth rotates about its axis with a period of 24.0 h.
Imagine that the rotational speed can be increased. If an object
at the equator is to have zero apparent weight, (a) what must
the new period be? (b) By what factor would the speed of the
object be increased when the planet is rotating at the higher
speed? Note that the apparent weight of the object becomes
zero when the normal force exerted on it is zero.
(a)
 Fr  m ar
m v2 m
mg

R
R
2
4 2R
g
T2
T
(b)
 2 R 


T 
4 2R
6.37  106 m
3
 2

5.
07

10
s  1.41 h
2
g
9.80 m s
speed increase factor
vnew
2 R  Tcurrent  Tcurrent 24.0 h



 17.1

 
vcurrent Tnew
2 R
Tnew
1.41 h
Fictitious Forces, examples
• Although fictitious forces are not real
forces, they can have real effects
• Examples:
– Objects in the car do slide
– You feel pushed to the outside of a rotating
platform
– The Coriolis force is responsible for the
rotation of weather systems and ocean
currents
Fictitious Forces in Linear Systems
• The inertial observer (a)
sees
F
F
x
 T sin   ma
y
 T cos  mg  0
• The noninertial observer
(b) sees
F '
F '
x
 T sin   F fictitious  ma
y
 T cos  mg  0
Fictitious Forces in a Rotating System
• According to the inertial observer (a), the tension is the
2
centripetal force
mv
T
r
• The noninertial observer (b) sees
T  Ffictitious
mv 2
T 
0
r
Motion with Resistive Forces
• Motion can be through a medium
– Either a liquid or a gas
• The medium exerts a resistive force, R, on an
object moving through the medium
• The magnitude of R depends on the medium
• The direction of R is opposite the direction of
motion of the object relative to the medium
• R nearly always increases with increasing speed
Motion with Resistive Forces
• The magnitude of R can depend on the
speed in complex ways
• We will discuss only two
– R is proportional to v
• Good approximation for slow motions or small
objects
– R is proportional to v2
• Good approximation for large objects
R Proportional To v
• The resistive force can be expressed as
R=-bv
• b depends on the property of the medium,
and on the shape and dimensions of the
object
• The negative sign indicates R is in the
opposite direction to v
R Proportional to v, Example
• Analyzing the motion
results in
dv
mg  bv  ma  m
dt
dv
b
a
g v
dt
m
R Proportional to v
• Initially, v = 0 and dv/dt = g
• As t increases, R increases
and a decreases
• The acceleration approaches
0 when R → mg
• At this point, v approaches
the terminal speed of the
object
Terminal Speed
 To find the terminal speed, let
a=0
mg
vT 
b
 Solving the differential
equation gives
mg
 bt m
t t
v
1

e

v
1

e



T 
b
 t is the time constant and
t = m/b
R Proportional to v2
• For objects moving at high speeds through air,
the resistive force is approximately equal to the
square of the speed
R = ½ DrAv2
– D is a dimensionless empirical quantity that called the
drag coefficient
 r is the density of air
– A is the cross-sectional area of the object
– v is the speed of the object
R Proportional to v2
• Analysis of an object falling
through air accounting for
air resistance
1
2
F

mg

D
r
Av
 ma

2
 Dr A  2
a  g 
v
 2m 
R Proportional to v2, Terminal Speed
• The terminal speed will
occur when the
acceleration goes to zero
• Solving the equation
gives
vT 
2mg
Dr A
Some Terminal Speeds
Process for Problem-Solving
• Analytical Method
– The process used so far involves the
identification of well-behaved functional
expressions generated from algebraic
manipulation or techniques of calculus
Analytical Method
• Apply the method using this procedure:
– Sum all the forces acting on the particle to
find the net force, SF.
– Use this net force to determine the
acceleration from the relationship a =SF/m
– Use this acceleration to determine the velocity
from the relationship dv/dt = a
– Use this velocity to determine the position
from the relationship dx/dt = v
Analytic Method, Example
• Applying the procedure:
Fg = may = - mg
ay = -g and dvy/dt = -g
vy(t) = vyi – gt
y(t) = yi + vyi t – ½ gt2
Numerical Modeling
• In many cases, the analytic method is not
sufficient for solving “real” problems
• Numerical modeling can be used in place
of the analytic method for these more
complicated situations
• The Euler method is one of the simplest
numerical modeling techniques
Euler Method

In the Euler Method, derivatives are
approximated as ratios of finite differences
 Dt is assumed to be very small, such
that the change in acceleration during the
time interval is also very small
Equations for the Euler Method
Dv v (t  Dt )  v (t )
a t  

Dt
Dt
v (t  Dt )  v (t )  a (t ) Dt
and
Dx x (t  Dt )  x (t )
v (t ) 

Dt
Dt
x  t  Dt   x (t )  v (t ) Dt
Euler Method
• It is convenient to set up the numerical
solution to this kind of problem by
numbering the steps and entering the
calculations into a table
• Many small increments can be taken, and
accurate results can be obtained by a
computer
Euler Method Set Up
Euler Method Final
• One advantage of the method is that the dynamics
are not obscured
– The relationships among acceleration, force, velocity
and position are clearly shown
• The time interval must be small
– The method is completely reliable for infinitesimally small
time increments
– For practical reasons a finite increment must be chosen
– A time increment can be chosen based on the initial
conditions and used throughout the problem
• In certain cases, the time increment may need to
be changed within the problem
Accuracy of the Euler Method
• The size of the time increment influences the
accuracy of the results
• It is difficult to determine the accuracy of the
result without knowing the analytical solution
• One method of determining the accuracy of the
numerical solution is to repeat the solution with a
smaller time increment and compare the results
– If the results agree, the results are correct to the
precision of the number of significant figures of
agreement
Euler Method, Numerical Example
Euler Method, Numerical Example
II. Drag force and terminal speed
-Fluid: anything that can flow. Example: gas, liquid.
-Drag force: D
- Appears when there is a relative velocity between a
fluid and a body.
Opposes the relative motion of a body in a fluid.
- Points in the direction in which the fluid flows.
Assumptions:
* Fluid = air.
* Body is blunt (baseball).
* Fast relative motion  turbulent air.
1
2
D  CrAv
2
C = drag coefficient (0.4-1).
ρ = air density (mass/volume).
A = effective body’s cross sectional area  area
perpendicular to v
-Terminal speed: vt
- Reached when the acceleration of an object that experiences
a vertical movement through the air becomes zero  Fg=D
1
2
D  Fg  ma  if a  0  CrAv  Fg  0
2
vt 
2 Fg
C rA
III. Uniform circular motion
-Centripetal acceleration:
2
v
a
r
v, a = const, but direction changes during motion.
A centripetal force accelerates a body by changing the
direction of the body’s velocity without changing its speed.
-Centripetal force:
v2
F m
R
a, F are directed toward the center
of curvature of the particle’s path.
49. A puck of mass m slides on a frictionless table while
attached to a hanging cylinder of mass M by a cord
through a hole in the table. What speed
keeps the cylinder at rest?
N
T
mg
T
Mg
49. A puck of mass m slides on a frictionless table while
attached to a hanging cylinder of mass M by a cord
through a hole in the table. What speed
keeps the cylinder at rest?
N
T
mg
T
T  Mg
For M :
Mg
2
For m :
v
T m
r
2
v
Mgr
Mg  m  v 
r
m
33E. Calculate the drag force on a missile 53cm in diameter
cruising with a speed of 250m/s at low altitude, where the
density of air is 1.2kg/m3. Assume C=0.75
1
D  CrAv 2  0.5  0.75  (1.2 kg / m 3 )    (0.53m / 2) 2 250 m / s 2  6.2 kN
2
32. The terminal speed of a ski diver is 160 km/h in the spread
eagle position and 310 km/h in the nose-dive position.
Assuming that the diver’s drag coefficient C does not change
from one point to another, find the ratio of the effective cross
sectional area A in the slower position to that of the
faster position.
2 Fg
vt 
2 Fg
160km / h


C rA
310km / h
CrAE
2 Fg
CrAD

AD
AE

AE
 3 .7
AD
32. The terminal speed of a ski diver is 160 km/h in the spread
eagle position and 310 km/h in the nose-dive position.
Assuming that the diver’s drag coefficient C does not change
from one point to another, find the ratio of the effective cross
sectional area A in the slower position to that of the faster
position.
2 Fg
2 Fg
160km / h
vt 


CrA
310km / h
CrAE
2 Fg
CrAD

AD
AE

 3.7
AD
AE
Block B weighs 711N. The coefficient of static friction between
the block and the table is 0.25;
assume that the cord between B and
N
the knot is horizontal. Find the maximum
T
f
weight of block A for which the system
T
T T
T
will be stationary.
2
1
1
3
3
System stationary  f s ,max   s N
FgB
FgA
Block B  N  mB g
T1  f s ,max  0  T1  0.25  711N  177.75 N
Knot  T1  T2 x  T2 cos 30  T2 
177.75 N
 205.25 N

cos 30
T2 y  T2 sin 30  T3
Block A  T3  mA g  T2 sin 30  0.5  205.25 N  102.62 N
Two blocks of weights 3.6N and 7.2N, are connected by a
massless string and slide down a 30º inclined plane. The
coefficient of kinetic friction between the lighter block and the
plane is 0.10; that between the heavier block and the plane
is 0.20. Assuming that the lighter block leads, find (a) the
magnitude of the acceleration of the blocks and (b) the
tension in the string. (c) Describe the motion if, instead, the
heavier block leads.
Block A
NA
T
FgxA
fkA
Block B
NB
NB
FgxB
fkB
T
fk,A
NA
T
A
FgyA
Light block A leads
FgyB
T
FgA
fk,B
B
FgB
Light block A leads
Block A  N A  FgyA  m A g cos 30  3.12 N
f kA   kA N A  (0.1)(3.12 N )  0.312 N
FgxA  f kA  T  m A a  (3.6 N ) sin 30  0.312 N  T  0.37 a  1.49  T  0.37 a
Block B  N B  FgyB  mB g cos 30  6.23 N

f kB   kB N B  (0.2)(6.23 N )  1.25 N
a  3.49m / s 2
T  0.2 N
FgxB  T  f kB  mB a  (7.2 N ) sin 30  T  1.25 N  0.73a  2.35  T  0.73a
 WAWB
T  
 WA  WB

kB  kA  cos   0.2 N

Heavy block B leads
Reversing the blocks is equivalent to switching the labels. This would
give T~(μkA-μkB)<0 impossible!!!
The above set of equations is not valid in this circumstance  aA≠ aB 
The blocks move independently from each other.
74. A block weighing 22N is held against a vertical wall by a horizontal force
F of magnitude 60N. The coefficient of static friction between the wall
and the block is 0.55 and the coefficient of kinetic friction between them
is 0.38. A second force P acting parallel to the wall is applied to the
block. For the following magnitudes and directions of P,
N
determine whether the block moves, the direction of motion,
and the magnitude and direction of the
F=60N
frictional force acting on the block:
P
mg=22N
(a) 34N up, (b) 12N up, (c) 48N up,
N=F=60N
(d) 62N up, (e) 10N down, (f) 18N down.
22N
f
Without P, the block is at rest 
(a) P=34N, up
P
f s ,max   s N  0.55(60 N )  33N
f k   k N  0.38(60 N )  22.8 N
P  mg  f  ma
If we assume f  f s  a  0
34 N  22 N  f  f  12 N down
f  f s ,max  33 N  Block does not move
74.
A block weighing 22N is held against a vertical wall by a
horizontal force F of magnitude 60N. The coefficient of static
friction between the wall and the block is 0.55 and the
coefficient of kinetic friction between them is 0.38. A second P
acting parallel to the wall is applied to the block. For the
following magnitudes and directions of P, determine whether
the block moves, the direction of motion, and the magnitude
P direction of the frictional force acting on the block: (a) 34N
and
up (b) 12N up, (c) 48N up, (d) 62N up, (e) 10N down, (f) 18N
down.
22N
P
P  f  mg  ma  0
f  22 N  12 N  10 N up
f  f s,max  33N  Not moving
(d) P=62N, up
P
f
22N
F=60N
mg=22N
N=F=60N
(c) P=48N, up
(b) P=12N, up
f
N
P  f  mg  ma  0
P f  48 N  22 N  26 N down
22N
f
f  f s ,max  33N  Not moving
P  f  mg  0 (*)  f  62 N  22 N  40 N up
f  f s ,max  33N  Block moves up  Assumption (*) wrong
 P  f  mg  ma with f  f k  22.8 N down
(e) P=10N, down
f
(f) P=18N, down
f
P
22N
22N
P
f  P  mg  ma  0
f  P  mg  ma  0
f  22 N  10 N  32 N up
f  18 N  22 N  40 N up
f  f s, max  33N  Not moving
f  f s ,max  33 N  moves
f  f k  22.8 N up
WA=44N
28. Blocks A and B have weights of 44N and 22N, respectively. (a)
Determine the minimum weight of block C to keep A from sliding
if μPs between A and the table is 0.2. (b) Block C suddenly is lifted
of A. What is the acceleration of block A if μk between A and the
table is 0.15?
(a)
f  f s ,max   s N
Block A  a  0  T  f s ,max  0  T   s N
(1)
N
f
T
Block B  T  mB g  0  T  22 N
Wc
WA=44N
T
T
WB=22N
22 N
(1)  (2) N 

 110 N
s
0.2
Blocks A, B  N  WA  WC  WC  110 N  44 N  66 N
(2)
28. Blocks A and B have weights of 44N and 22N, respectively. (a)
Determine the minimum weight of block C to keep A from sliding
if μs between A and the table is 0.2. (b) Block C suddenly is lifted
of A. What is the acceleration of block A if μk between A and the
table is 0.15?
C disappears  N  m A g  44 N
(b)
T  k N  mAa
mB g  T  mB a
N
f
T
Wc
WA=44N
T
WB=22N
T  6.6  4.5a
 a  2.3m / s 2
22  T  2.2a
 T  17 N
29. The two blocks (with m=16kg and m=88kg) shown in the figure below
are not attached. The coefficient of static friction between the blocks
is: μs=0.38 but the surface beneath the larger block is frictionless.
What is the minimum value of the
N
f
horizontal force F required to keep the
F’
F’
smaller block from slipping
down the larger block?
mg
Fmin required to keep m from sliding down?
Mg
Treat both blocks as a single system sliding
across a frictionless floor
Movement
F
F  mtotala  a 
mM
 F 
Small block  F  F '  ma  m

m

M


f s  mg  0   s F ' mg  0
(1)  ( 2)
(1)
( 2)
F 
mg  m  M 
  mg  F 

  488 N
s  M 
mM 

s M 
44. An
amusement park ride consists of a car moving in a vertical circle
on the end of a rigid boom of negligible mass. The combined weigh
of the car and riders is 5kN, and the radius of the circle is 10m.
What are the magnitude and the direction of the force of the boom
on the car at the top of the circle if the car’s speed is (a) 5m/s (b)
12m/s?
The force of the boom on the car is capable of pointing any direction
mg
 v2
FB  W  m 
 R


v2 
  FB  W 1 



Rg 


(a) v  5m / s  FB  3.7 N up
y
FB
W
(b) v  12m / s  FB  2.3 down