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3.
Force
Chapter 3
Force
Basic Requirements
1. Master the Newton's first law, second law and Newton's third law;
2. Know about some particular forces: the gravitational force, the normal force and the
frictional force;
3. Learn to apply the Newton's laws of motion in question solving with the help of the
free-body diagram;
4.Understand the concept of Inertial reference frame and noninertial reference frame;
5. Understand the Inertial forces.
6. Master the properties of the frictional force;
7. Understand the drag force and terminal speed;
8. Understand the concept of the centripetal acceleration and the centripetal force in
uniform circular motion.
Supplementary Material
Non-inertial reference frames and the inertial forces
Based on the discussion of inertial reference frames we have that Newton’s laws
only apply to inertial reference frames, and they do not apply in accelerated reference
frames with relative motion with respect to inertial reference frames. Such accelerated
reference frames are non-inertial reference frames. Trains accelerating with respect to
the ground, accelerated elevators, and rotating circular plates are all non-inertial
reference frames.
Fig. 3-1
3.
Force
In order to conveniently solve mechanical problems in non-inertial reference
frames, we introduce inertial forces. Shown in Fig. 3-1 is a train with an acceleration
a0 along the Ox axis moving with respect to the ground, as seen by an observer in
that there is an inertial force acting on the ball with mass m and take it as F1   ma0 ,
then Newton’s laws can be applied to this non-inertial reference frame. That is to say,
for the observer inside the train moving with acceleration a0 with respect to the ground,
he thinks that there is an inertial force with the magnitude equal to ma0 but in opposite
direction as a0 acting on the ball.
Generally speaking, if the forces acting on a body include an inertial force Fi , then
the mathematical expression of Newton’s second law is:
or
F  Fi  ma
F  (ma0 )  ma
where a0 is the acceleration of the non-inertial frame with respect to the inertial frame,
a is the acceleration of the object with respect to the non-inertial frame, and F is the
combined force of all forces except the inertial force.
We now introduce the concept of inertial centrifugal forces, as shown in Fig. 3-2,
Fig. 3-2 Illustration of the inertial centrifugal force
in a horizontally placed rotating platform there is a light spring tied in the middle of a
thin rope, one end of the thin rope is tied at the center of the rotating platform, and a
small ball with mass m is tied at the other end.
Assume that the surface of the rotating platform is very smooth, the frictional
force between the small ball and the surface is negligible, and the frictional force
between the light spring and the surface of the platform is also negligible. The
platform rotates with a constant angular velocity w with respect to the vertical central
axis of the platform. There are two observers, one standing on the ground (in the
inertial reference frame) and the other staying fixed with respect to the platform and
rotating with it (in the non-inertial frame). When the platform rotates, the observer on
the ground sees that the spring is stretched to be longer. Now, the force acted by the
3.
Force
rope on the small ball is the centripetal force F . The magnitude of F is
ml 2 . This is understandable based on Newton's second law since the small ball is
carrying out a circular motion with a constant speed. However, the other observer
fixed on the platform also sees the elongation of the rope and the force F acting on
the ball along the radial direction, but the ball is not moving with respect to the
platform, this is hard
to understand. Why would the ball not move when there is a force acting on it?
Therefore, this observers believes that to reconcile the fact the ball is still at rest with
Newton's second law we must imagine that there is a force with equal magnitude but
in opposite direction acting on the ball, this force Fi is termed inertial centrifugal
force. We must notice that the centripetal and the inertial centrifugal forces both act on
the same small ball and they are not action and reaction, thus they do not obey
Newton's third law.
The Coriolis Effect
In a reference flame that rotates at a constant angular speed
 (relative to an
inertial system), there exists another pseudoforce known as the Coriolis force. It
appears to act on a body in a rotating reference frame only if the body is moving
relative to that reference flame, and it acts to deflect the body sideways. It, too, is an
effect of the reference being noninertial and hence is referred to as an inertial force. To
see how the Coriolis force arises, consider two people, A and B , at rest on a platform
rotating with angular speed  , as shown in Fig. 3-3 a. They are situated at distances
rA and rB from the axis of rotation (at O ). The woman at A throws a ball with a
horizontal velocity v (in her reference flame) radially outward toward the man at B on
the outer edge platform. In Fig. 3-3 a, we view the situation from an inertial
reference flame. The ball initially has not only the velocity v radial outward, but also a
tangential velocity v A due to the rotation of platform. We know that v A  rA ,
where rA is the woman’s radial distance from the axis of rotation at O . If the man
at B had this same velocity v A , the ball would reach him perfectly. But his speed
is vB  rB , which is greater than v A because rB  rA . Thus, when the ball reaches
the outer edge of the platform, it passes a point that the man at B has already passed
3.
Force
because his speed in that direction greater than the ball's. So the ball passes
Fig. 3-3
behind him.
Figure 3-3b shows the situation as seen from the rotating form as frame of
reference. Both A and B are at rest, and the ball is thrown with velocity v toward B ,
but the ball deflects to the right as shown and passes behind B as previously described.
This is not a centrifugal-force effect, for the latter acts radially outward. Instead, this
effect acts sideways, perpendicular to v , and is called a Coriolis acceleration; it is said
to be due to the Coriolis force, which is a fictitious, inertial force. Its explanation as
seen from an inertial system was given above: it is an effect of being in a rotating
system, wherein points that are farther from the rotation axis have higher linear
speeds. On the other hand, when viewed from the rotating system, we can describe the
motion using Newton’s second law,
 F  ma , if we add a “pseudoforce” term
corresponding to this Coriolis effect.
Let us determine the magnitude of the Coriolis acceleration for the simple case
described above. We do the calculation from the inertial reference frame. The ball
moves radially outward a distance rB  rA at speed v in a time t given by
rB  rA  vt
During this time, the ball moves to the side a distance s A given by
s A  v At
The man at B , in this time, moves a distance
s B  vB t
the ball therefore passes behind him a distance s given by
s  sB  sA  (vB  vA )t
3.
Force
We saw earlier v A  rA and vB  rB , so
We substitute rB  rA  vt and get
s  (rB  rA )t
s   vt 2
This same s equals the sideways displacement as seen from the noninertial rotating
system.
We see immediately that Equation above corresponding to motion at constant
acceleration. Thus, if we write it in the form s 
1
acor t 2 , we see that the Coriolis
2
acceleration acor is
acor  2v
This relation is valid for any velocity in the plane of rotation—that is, in the plane
perpendicular to the axis of rotation.
Fig. 3-4
Because the Earth rotates, the Coriolis effect has some interesting manifestations
on the Earth. It affects the movement of air masses and thus has an influence on
weather. In the absence of the Coriolis effect, air would rush directly into a region of
3.
Force
low pressure, as shown in Fig. 3-3a. But because of the Coriolis effect, the winds
are deflected to the right in the Northern Hemisphere (Fig. 3-3b), since the Earth
rotates from west to east. So there tends to be a counterclockwise wind pattern around
a low-pressure area. The reverse is true in the Southern Hemisphere. Thus cyclones
rotate counterclockwise in the Northern Hemisphere and clockwise in the Southern
Hemisphere. The same effect explains the easterly trade winds near the equator: any
winds heading south toward the west (that is, as if coming from the east).
The Coriolis effect also acts on a falling body. A body released from the top of a
high tower will not hit the ground directly below the release point, but will be
deflected slightly to the east. A velocity, or component of velocity, parallel to the axis
of rotation will not experience a Coriolis acceleration since r (distance from axis)
doesn't change. From Fig. 3-4 we see that this is v cos  for a vertically falling body,
where  is the latitude of the place on the Earth.
Review & Summary
Newtonian Mechanics The velocity of a particle or a particle-like body can change
(the particle can accelerate) when the particle is acted on by one or more forces
(pushes or pulls) from other objects. Newtonian mechanics relates accelerations and
forces.
Force Forces are vector quantities. Their magnitudes are defined in terms of the
acceleration they would give the standard kilogram. A force that accelerates that
standard body by exactly 1m / s
2
is defined to have a magnitude of 1N . The
direction of a force is the direction of the acceleration it causes. Forces are combined
according to the rules of vector algebra. The net force on a body is the vector sum of
all the forces acting on it.
Mass The mass of a body is the characteristic of that body that relates the body’s
acceleration to the force (or net force) causing the acceleration. Mass are scalar
quantities.
Inertial Reference Frames Reference frames in which Newtonian mechanics holds
are called inertial reference frames or simply inertial frames. We can approximate the
ground as an inertial frame if Earth’s motions can be neglected. Reference frames in
which Newtonian mechanics does not hold are called noninertial reference frames or
simply noninertial frames. An elevator accelerating relative to the ground is a
3.
Force
noninertial frame.
Newton’s Second Law The net force Fnet on a body with mass m is related to the
body’s acceleration a by
Fnet  ma
(3-1)
which may be written in the component versions
Fnet , x  max Fnet , y  ma y and Fnet , z  maz
(3-2)
The second law indicates that in SI units
1N  1kg  m / s 2
A free-body diagram is helpful in solving problems with the second law: It is a
stripped-down diagram in which only one body is considered. That body is
represented by a sketch or simply a dot. The external forces on the body are drawn,
and a coordinate system is superimposed, oriented so as to simplify the solution.
Some Particular Forces A gravitational force Fg on a body is a pull by another
body. In most situations in this book, the other body is Earth or some other
astronomical body. For Earth, the force is directed down toward the ground, which is
assumed to be an inertial frame. With that assumption, the magnitude of the force is
Fg  mg
(3-14)
where m is the body’s mass and g is the magnitude of the free-fall acceleration.
The weight W of a body is the magnitude of the upward force needed to balance
the gravitational force on the body due to Earth (or another astronomical body). It is
related to the body’s mass by
W  mg
(3-18)
A normal force N is the force on a body from surface against which the body
press. The normal force is always perpendicular to the surface.
A frictional force f is the force on a body when the body slides or attempts to slid
along a surface. The force is always parallel to the surface and directed so as to oppose
the motion of the body. On a frictionless surface, the frictional force is negligible.
When a cord is under tension, it pulls on a body at each of its ends. The pull is
directed along the cord, away from the point of attachment to each body. For a
massless cord (a cord with negligible mass), the pulls at both ends of the cord have the
same magnitude T , even if the cord runs around a massless, frictionless pulley (a
pulley with negligible mass and negligible friction on its axle to oppose its rotation).
Newton’s Third Law
If a force FBC acts on a body B due to body C, then there is a
force FCB on body C due to body B. The forces are equal in magnitude and opposite
3.
Force
in direction:
FBC   FCB

Friction When a force F tends to slide a body along a surface, a frictional force
from the surface acts on the body. The frictional force is parallel to the surface and
directed so a to oppose the sliding. It is due to bonding between the body and the
surface.

If the body does not slide, the frictional force is a static frictional force f s . If there

is sliding, the frictional force is a kinetic frictional force f k .
Three Properties of Friction

1. If the body does not move, then the static frictional force f s and the component of


F that is parallel to the surface are equal in magnitude, and f s is directed opposite
that component. If that parallel component increase, magnitude f s also increase.

2. The magnitude of f s has a maximum value f s ,max that is given by
f s ,max   s N
(3-21)
where  s is the coefficient of static friction and N is the magnitude of the normal

force. If the component of F that is parallel to the surface exceeds f s ,max , then the
body slides on the surface.
3. If the body begins to slide on the surface, the magnitude of the frictional force
rapidly decrease to a constant value f k given by
f k  k N
where
(3-22)
 k is the coefficient of kinetic friction.
Drag Force When there is a relative motion between air (or some other fluid) and a

body, the body experiences a drag force D that opposes the relative motion and points

in the direction in which the fluid flows relative to the body. The magnitude of D is
related to the relative speed v by an experimentally determined drag coefficient C
according to
D
1
CAv 2
2
(3-28)
Where  is the fluid density (mass per volume) and A is the effective cross-sectional
area of the body (the area of a cross section taken perpendicular to the relative

velocity v ).
Terminal speed
When a blunt object has fallen far enough through air, the


magnitudes of the drag force D and the gravitational force Fg on the body become
equal. The body then has falls at a constant terminal speed v t given by
3.
vt 
Force
2 Fg
(3-30)
CA
Uniform Circular Motion If a particle moves in a circle or a circular arc with
radius R at constant speed, it is said to be in uniform circular motion. It then has a

centripetal acceleration a with magnitude given by
v2
a
R
(3-31)
This acceleration is due to a net centripetal force on the particle, with magnitude
given by
F
mv 2
R
(3-32)


where m is the particle’s mass. The vector quantities a and F are directed toward the
center of curvature of the particle’s path.
Examples
Example A box of mass m1  10.0kg rests on a surface inclined at   37 to
the horizontal. It is connected by a lightweight cord, which passes over a massless and
frictionless pulley, to a second box of mass m2 , which hangs freely as shown in Fig.
3-5 a. (a) If the coefficient of static friction is
values for mass m
2
 s  0.40 , determine what range of
will keep the system at rest. (b) If the coefficient of kinetic
friction is  k =0.30, and m2 =10.0 kg , determine the acceleration of the system.
Fig. 3-5 Example 1
3.
Force
Solution: (a) Figure 3-5b shows to free-body diagrams for box m1 .The tension
force exert by the cord is labeled FT .The force of friction can be either up or down
the slope ,and we show both possibilities in Fig.5-9b: (i) If m2 =0 or is sufficiently
small, m1 would tend to slide down the incline, so F fr would be directed up the
incline; (ii) If m2 is large enough, m1 will tend to be pulled up the plane, so F fr
would point down the plane. For both cases, Newton’s second law for the y
direction (perpendicular to the plane ) is the same :
FN  cos   m1  a y  0
Since there is no y motion. So
FN  m1 g cos  .
Now for the x motion. We consider case (i) first for which
 F  ma
gives
m1 g sin   FT  F fr  m1 a x .
We want a x  0 and we solve for FT is related to m2 (whose value we are
seeking ) by FT  m2 g (see Fig. 3-5 c).Thus
m1 g sin   F fr  FT  m2 g .
Since F fr can be at most
 s FT   S m1 g cos ,the minimum value that m2 can
have to prevent motion ( a x  0 ) is, after dividing through by g ,
m2  m1 sin   s m1 cos 
 (10.0kg )(sin 37  0.40cos 37 )  2.8kg
Thus if m2  2.8kg ,then box 1 will slide down the incline.
3.
Force
Now for case (ii).Newton’s second law is
m1 g sin   Frt  FT  max  0 .
Then the maximum value m2 can have without causing acceleration is given by
FT  m2 g  m1 g sin    s m1 g cos 
or
m2  m1 sin   s m1 cos   (10.0kg )(sin 37  0.40 cos 37 )  9.2kg
Thus, to prevent motion , we have the condition
2.8kg  m2  9.2kg .
(b) If m2  10.0kg and
(Answer)
 k  0.30 , then m2 will fall and m1 will rise up the
plane with an acceleration a given by
m1a  FT  m1 g sin    k FN
Since m2 also accelerates, FT  m2 g  m2 a (see Fig. 3-5c) and we substitute this
into the equation above:
m1a  m2 g  m2 a  m1 g sin    k FN .
We solve for the acceleration a and substitute FN  m1 g cos  ,and then
m1  m2  10.0kg , to find
a
m2 g  m1 g sin   k m1 g cos 
m1  m2
(9.8m / s 2 )(10.0kg )(1  sin 37  0.30 cos 37 )

20.0kg
(Answer)
 0.079 g  0.78m / s 2
Example 2 Determine the velocity as a function of time for a body failing vertically
3.
Force
from rest when there is a resistive force linearly proportional to v .
Solution：We write
dv
b
 g v.
dt
m
There are two variables, v and t .We collect variables of the same type on one or the
other side of the equation:
dv
 dt
b
g v
m
or
dv
b
  dt .
mg
m
v
b
Now we can integrate , remembering v  0 at t  0 :

v
0
dv
b t
   dt
mg
m 0
v
b
In(v 
mg
mg
b
)  In(
) t
b
b
m
or
ln
v  mg / b
b
 t.
 mg / b
m
We raise each side to the exponential [note that the natural log and the exponential are
inverse operations of each other: e
b
ln x
 x ,or ln( e x )  x ] and obtain
mg
mg  m t
v

e
or finally,
b
b
b
 t
mg
v
(1  e m ) .
b
(Answer)
This relation gives the velocity v as a function of time. As a check , note that at
t  0 , v  0 and
b
a(t  0) 
As expected. At large t , e
 t
dv mg d
mg b

(1  e m ) 
( )g,
dt
b dt
b m

b
t
m
(Answer)
approaches zero, so v approaches mg / b , which is
the terminal velocity, vT , as we saw earlier. If we set
  m / b , then
3.
Force
v  vT (1  e t /  ) . So   m / b is the time required for the velocity to reach
63 percent of the terminal velocity (since e 1  0.37 ).
Example 3 A 1000 kg car rounds a curve on a flat road of radius 50m at a speed
of 50km/ h ( 14m/ s ). Will the car make the turn ,or will it skid, if: (a) the pavement
is dry and the coefficient of static friction is
 s  0.25 ?
 s  0.60 ; (b) the pavement is icy and
Solution: Fig. 3-6 shows the free-body diagram for a car. The normal force, FN , on
the car is equal to the weight since the road is flat and there is no vertical acceleration:
FN  mg  (1000kg)(9.8m / s 2 )  9800 N .
In the horizontal direction the only force is friction, and we must compare it to the
Fig. 3-6 Example 3
force needed to produce the centripetal acceleration to see if it is sufficient. The net
horizontal force required to keep the car moving in a circle around the curve is
v2
(14m / s) 2
 FR  maR  m r  (1000kg) 50m  3900 N .
Naturally we hope the maximum total friction force (the sum of the friction forces
acting on each of the four tires ) will be at least this large. For (a),  s  0.60 ,and the
maximum friction force attainable (recall from Section 5-1 that F fr   s FN )is
3.
Force
( F fr ) max   s FN  (0.60)(9800 N )  5900 N .
Since a force of only 3900N is needed, and that is, in fact, how much will be
exerted by the road as a static friction force, the car can make the turn fine. But in (b)
the maximum friction force possible is
( F fr ) max   s FN  (0.25)(9800 N )  2500 N .
(Answer)
The car will skid because the ground cannot exert sufficient force ( 3900N is needed )
to keep it moving in a curve of radius 50m .
Example 4 (1) As shown in Fig.3-7(a), a thin rope is hung over a fixed pulley, on the
two ends of the rope are hung two weights with masses of m1 and m2 ( m1  m 2 ).
Assume that the mass of the pulley and the mass of the rope are both negligible, and
the frictional force between the pulley and the rope as well as the frictional force
between the axes of the pulley are all negligible, when the weights are released for free,
what are the accelerations and the tension of the rope ?
(1) If we fix the above equipment on the ceiling of an elevator as shown in Fig.3-7(b),
Fig. 3-7 Example 4
3.
Force
when the elevator moves vertically upward with acceleration a , what are the
accelerations of the weights relative to the elevator and the tension of the rope ?
Solution: (1) choose the ground as our inertial reference frame and plot the force
diagram as shown in Fig.3-4(a) .Since we neglect the masses of the rope and the pulley,
the forces FT 1  FT 2  FT . Referring to the illustrated acceleration a and Newton’s
second law, we have
m g  FT  m1a
{ 1
FT  m2 g  m2 a
Solving the above two equations simultaneously ,we obtain the acceleration of the two
weights and the tension of the rope as follows;
a
m1  m2
2m1m2
g , FT 
g
m1  m2
m1  m2
(Answer)
(2) We still choose the ground as our inertial reference frame , the acceleration of the
elevator with respect to the ground is a as shown in Fig.3-7(b). Let ar be the
acceleration of object one with respect to the elevator ,then the acceleration of object
one with respect to the ground is a1  ar  a .We have from Newton’s second law:
p1  FT 1  m1a1
According to the coordinate system of the figure ,consider object one is constrained to
moving along the y axis ,and
a1  ar  a . So the above becomes
m1 g  FT  m1a1  m(ar  a)
(1)
Since the length of the rope is constant , the acceleration of object two with respect to
the elevator is also ar , and the acceleration of object two with respect to the ground
is a2 . According to the coordinate system as shown in the figure, a2  ar  a .
Therefore , the equation of motion of object two is
FT  m2 g  m2 a2  m2 (ar  a)
(2)
3.
Force
According to Eqs. (1) and (2), we obtain the magnitude of the acceleration of
objects one and two with respect to the elevator as follows:
ar 
m1  m2
( g  a)
m1  m2
Substitute the above into Eq. (1), the tension of the rope is
FT 
2m1m2
( g  a)
m1  m2
(Answer)
Example 5 As shown in Fig.3-8 (a) (taper pendulum), a rope with length l has one
of his ends fixed on the ceiling and the other end hung with a ball of mass m ,when
the ball is pushed, it circulates with a constant angular speed  around the vertical
axis going through the center of the circle O .What is the angle
and the vertical axis ? (ignore the friction of air)
Solution:
 between the rope
Fig. 3-8 Example 5
The ball is subject to gravity P and the pulling force of the rope FT , the equation of
3.
motion is
Force
FT  P  m a
(1)
Where a is the acceleration of the small ball.
Since the small ball carries out a circular motion on the horizontal plane with a
constant linear speed v  r ,take an arbitrary point A on the circle and select a
natural coordinate system, the unit vectors along the coordinate axes are en and et ,the
magnitude of the normal acceleration is an 
v2
and the tangential acceleration
R
is at  0 , and the direction of the velocity v at any point on the circle is
perpendicular to the plane formed by P and FT . Choose coordinate system as per
Fig. 3-8(b), the component from of Eq.(1) becomes
FT cos   P  0
From the figure we know that r  l sin  ,and the above two equations give
FT  m 2l
and
cos  
mg
g
 2
2
m l  l
and
  arccos
g
 2l
(Answer)
We see that the large the  , the large the angle  between the rope and the vertical
axis.
What is worth mentioning is that in the early stage of the steam engine ,Watt basing
on the principle that the angle  changes with the angular speed  created the
timing device of the steam engine. Fig. 3-8 (c) is an illustration of the timing device,
when the speed of rotation goes beyond a certain limit, the angle between the rotating
arm and the vertical axis reaches a threshold to shut the valve down, so that the steam
entering the cylinders gets reduced; when the rotation speed is too low, the angle is so
small that the valve gets opened up, so that the amount of steam gets increased in the
3.
Force
cylinders, therefore the speed of rotation can be adjusted accordingly. Even today
many machines still have this kind of timing device equipped on them.
Example 6 Our 10.0kg mystery box rests on a horizontal floor. The coefficient of
s  0.40 and the coefficient of kinetic friction is k  0.30 .
Determine the force of friction, F fr , acting on the box if a horizontal external applied
static friction is
force Fs is exerted on it of magnitude: (a)0 N (b)10 N (c)20 N ( d )38 N and
(e ) 4 N
0 .
Solution: The free-body diagram of the box is shown in Fig.3-9 a. Examine it
carefully. In the vertical direction there is no motion, so
F
y
 may  0 yields FN  mg  0 .
Hence the normal force for all cases is
Fig. 3-9 a Example 6
Fig. 3-9 b Example 6
FN  mg  (10.0kg )(9.8m / s 2 )  98 N
(a) Since no force is applied in this first case, the box doesn’t move, and Ffr  0 .
(b) The force of static friction will oppose any applied force up to a maximum of
s FN  (0.40)(98.0 N )  39 N
The applied force is FA  10 N . Thus the box will not move; Since
F
x
 FA  Ffr  0
3.
Force
Ffr  10 N .
then
(Answer)
(c) An applied force of 20N is also not sufficient to move the box. Thus Ffr  20 N
to balance the applied force.
(Answer)
(d) The applied force of 38N is still not quite large enough to move the box; so the
friction force has now increased to 38N to keep the box at rest.
(Answer)
(e) A force of 40N will start the box moving since it exceeds the maximum force of
static friction, s FN  (0.40)(98.0 N )  39 N .Instead of static friction, we now have
kinetic friction, and its magnitude is
Ffr  k FN  (0.30)(98.0 N )  29 N
There is now a net (horizontal) force on the box of magnitude
F  40N  29N  11N
so the box will accelerate at a rate
ax   F / m  11N /10kg  1.1m / s 2
(Answer)
as long as the applied force is 40N . Figure.3-5b shows a graph that summarizes this
Example.
Problem Solving
1 (1) Two horizontal forces F1 and F2 act on a
4.0kg disk that slides over frictionless ice, on which an
xy coordinate systems is laid out. Forces F1 is in the
positive direction of the x axis and has a magnitude of
7.0N . Force F2 has a magnitude of 9.0N . Figure
3-32 gives the x component vx of the velocity of
the disk as a function of time t during the sliding .What is
the angle between the constant direction of forces F1 and F2 ?
Fig. 3-6 Problem 1
Solution: According to Fig.3-6, the acceleration of the disk along the x axis can be
yielded as
ax  vx 5m/ s-(-4)m/ s 3.0m/ s2
t
3s-0s
3.
Force
We next consider the forces acting on the disk. The force F1 is along the x axis
and has the magnitude 7.0N . The horizontal component of the force F2 is
F2 x  F2 cos  , in which 
is the angle between the forces F1 and F2 . Thus we can write Newton’s second law
for components along the x axis as
F1  F2 cos   max
which here means
7.0 N  (9.0 N ) cos   (4.0kg )(3.0 m s 2 )  12.0 N
5
9
 , yields   c o s1  5 6 . 3
solving for
(Answer)
2 (5) In Figure. 3-7, elevator cabs A and B are connected by a short cable and can be
pulled upward or lowered by the cable above of A. Cab A has mass 1700kg ; cab B
has mass 1300kg . A 12.0kg box of catnip lies on the floor of cab A. The tension in
the cable connecting the cabs is 1.9110 N . What is the magnitude of the normal
4
force on the box from the floor?
Solution: As shown in cabs’ free-body diagram of Fig.3-7a, there are two forces
acting on them, the tension T in the cabs and the gravitational force Fg . We can
relate them to the cab’s acceleration via Newton’s second law, which we write as
T  Fg  Ma
Because the two forces on the cabs and the cabs’ acceleration are all in the vertical
direction along the y axis, we can rewrite the equation above as
T  (mA  mB  mc ) g  (mA  mB  mc )a
from which
a
[1.91104 N  (1700kg  1300kg  12kg )(9.8m / s 2 )]
 3.5m / s 2
(1700kg  1300kg  12kg )
3.
Force
“minus sign” means a is in the negative direction of the y axis.
Now, we can relate the net force on the box of catnip to the box’s acceleration with
Newton’s second law as
Fc  N  m c a
mc g  N  mc a
where
Solving for N and substituting known values, we find
T
y
G
Fig. 3-7 Problem 2
Fig. 3-7a Problelm 2
N  (12.0kg )(9.8m / s 2 )  (12.0kg )(3.5m / s 2 )  77 N
(Answer)
So the magnitude of the normal force on the box from the floor is 77N .
3 (8) Figure 3-8 shows four penguins that are being playfully pulled along very
slippery (frictionless) ice by a curator. The masses of three penguins and the tension in
m  12kg , m3  15kg , m4  20kg , T2  111N and
T4  222 N . Find the penguin mass m2 that is not given.
two
of
the
cords
are
Solution: Penguin 3 and penguin 4 can be seen as a system. Relating the net force on
the system to its acceleration with Newton’s second law, we can write the law along
the x axis as
Fig. 3-8 Problem 3
Fnet  Ma
3.
here,
from which
Force
T4  T2  (m3  m4 )a
a  (222 N  111N ) (15kg  20kg ) 
111
m / s2
35
Since penguin 2 and penguin 1 can also be seen as a system, we can use Newton’s
second law again
  M a
Fnet
here,
T2  (m1  m2 )a
Since the forces T2 and T2 are a third-law force pair, T2 has the same magnitude as
the force T2 . Substituting known values, we can obtain
m2  23kg
(Answer)
4 (9) A 10kg monkey climbs up a massless rope that runs over a frictionless tree
limb and back down to a 15kg package on the ground (Fig.3-9). (a) What is the
magnitude of the least acceleration the monkey must have if it is to lift the package off
the ground? If, after the package has been lifted , the monkey stops its climb and holds
onto the rope, what are the (b) magnitude and (c) direction of the monkey’s
acceleration and (d) the tension in the rope?
T
y
T
mg
Mg
Fig. 3-9 Problem 4
Fig. 3-9a Problem 4
Solution: (a) As shown in Fig. 3-9a, we can relate the net force to the monkey’s
acceleration via Newton’s second law. Thus,
T  Fg ,mon  ma
3.
Force
where m is the mass of the monkey.
If the monkey wants to lift the package off the ground, the tension in the rope must
have the same magnitude as the package’s gravitational force at least, which means
Tmin  Fg , pac  Mg
where M is the mass of the package.
T  mg  ma
here,
so , substituting known values, we find
Tmin  mg Mg  mg

m
m
2
(15kg )(9.8m / s )  (10kg )(9.8m / s 2 )

 4.9m / s 2
10kg
amin 
(b) After the package has been lifted, the monkey stops its climb and holds onto the
rope. In this case, the monkey and the package both can be seen as a particle. And we
can relate the net forces on them to their accelerations with Newton’s second law
separately. As shown in Fig. 3-39b, we find
T  Mg   Ma
T  mg  ma
Then solving for a yields
a
(M  m) g (15kg  10kg )(9.8m / s 2 )

 2.0m / s 2 (Answer)
M m
15kg  10kg
(c) The monkey will have a acceleration in the positive direction of the y axis. So
the direction of the acceleration is upward.
(d) Solving for T yields
T  120 N
So the tension in the rope is 120N .
(Answer)
5 (10) Figure 3-10 shows a man sitting in a bosun’s chair that dangles from a
massless rope, which runs over a massless, frictionless pulley and back down to the
man’s hand .The combined mass of man and chair is 95.0kg . With what force
magnitude must the man pull on the rope if he is to rise (a) with a constant velocity
2
and (b) with an upward acceleration of 1.30m / s ? (Hint: A free-body diagram can
3.
Force
really help.)
If the rope on the right extends to the ground and is pulled by a co-work, with what
force magnitude must the co-work pull for the man to rise ©
with a constant velocity and (d) with an upward acceleration of
1.30m / s 2 ? What is the magnitude of the force on the ceiling
from the pulley system in (e) part a (f) part b, (g) part c, (h) part
d?
Solution: (a) There are two forces acting on the system of the
man and chair, tension T in the rope and gravitational force
Fg . And the key idea is that in this situation, there are two
lengths of rope connecting the system, which shows that the
Fig. 3-10 Problem 5
force acting on the system is twice the tension in one rope. If the man is to rise with a
constant velocity, his acceleration is equal zero. So relating the net forces to the
acceleration via Newton’s second law, yields
2Ta  Fg  Ma  0
2Ta  Mg  0
here,
Substituting the known values, we find
Ta 
1
1
Mg  (95.0kg )(9.8m / s 2 )  465.5 N
2
2
(Answer)
Since the tension in the rope and the force the man pulling on the rope are force and
reaction force,
the force which the man must pull on the rope has the same
magnitude as the tension Ta .
2
(b) If the man is to rise with an upward acceleration 1.30m / s , we can utilize the
Newton’s second law again to solve the problem. Thus,
2Tb  Fg  Ma
in the scalar form, we write
2T  Mg  Ma
from which
Tb 
Ma  Mg 95.0kg (1.30m / s 2  9.8m / s 2 )

 527 N
2
2
(Answer)
3.
Force
(c) If the rope on the right extends to the ground and is pulled by a co-worker,
the tension acting directly on the system has the same magnitude as the force the
co-worker pulls on the rope. We can relate the net force to the acceleration with
Newton’s second law as
Tc  Mg  Ma  0
from which
Tc  Mg  (95.0kg )(9.8m / s 2 )  931N
(Answer)
Td  Mg  Ma
(d)
substituting known values and solving for T yields
Td  M (a  g )  (95.0kg )(1.30m / s 2  9.8m / s 2 )  1054.5 N
(Answer)
(e) Because the rope wraps halfway around the pulley, the net force on the ceiling
from the pulley has the magnitude 2T . Thus
(f)
F  2Ta  931N
(Answer)
F  2Tb  1054.5 N
(Answer)
(g)
F  2Tc  1862 N
(Answer)
(h)
F  2Td  2109 N
(Answer)
6 (11) A hot-air balloon of mass M is descending vertically with downward
acceleration of magnitude a . How much mass (ballast) must be thrown out to give
the balloon upward acceleration of magnitude a ? Assume that the upward force from
the air (the lift ) dose not change because of the decrease in mass.
Solution: During the descending process, the net forces acting on the hot-air balloon
include the lift force f from the air and the gravitational force Fg . Relating the net
forces to the hot-air balloon’s acceleration a via Newton’s second law, we can find
f  Fg  Ma
from which
f  Mg  Ma  M ( g  a ) (upward is the positive direction)
3.
Force
Now, in order to give the balloon an upward acceleration of magnitude a , we
assume the mass of the ballast which must be thrown out has the magnitude of m .
Since the upward force from the air remains constant, we can write Newton’s second
law as
f  Fg  (M  m)a
where
 f  ( M  m) g  ( M  m ) a
m
Solving for m , yields
2Ma
ga
(Answer)
7 (12) Figure 3-11 show a box of mass m1  1.0kg on a frictionless plane inclined
at angle   30 . It connected by a cord of negligible mass to a box of mass
m2  3.0kg on a horizontal frictionless surface. The pulley is frictionless and
massless. (a) If the magnitude of the horizontal force F1 is 2.3N , what is the
tension in the connecting cord ?
(b) What is the largest value the magnitude of
F may have without the cord becoming slack?
Solution: (a) The free-body diagram of two boxes is shown in Fig. 3-41a. The forces
in x axis acting on box A are horizontal force F and the tension T in the
A m1
Fig. 3-11 Problem 7
T
F
T
B m2
Fg sin 
Fig. 3-11a Problem 7
connecting cord. Relating the net force to box A s acceleration via Newton’s
second law, we find
T  F  m1a
in the scalar form, we write
T  F  m1a
(1)
As for box B, since it is connected to box A by a cord , it has the same acceleration
as that of box A. There are three forces, the gravitational force Fg , the tension T and
3.
Force
the normal force N , acting on box B. We can write the Newton’s second law
along x axis
m2 g sin   T  m2 a
(2) combining equation (1) and (2) and solving for T and a , yields
a  1.8m / s 2
T  3.1N
(Answer)
(b) With the horizontal force F growing bigger, the acceleration a of the system
will increase, which will make the magnitude of the tension T become smaller and be
zero at last. In this case, the rope will become slack. So when the magnitude of tension
T decrease to zero, the horizontal force
F
will have the largest magnitude.
m g sin   m2 a
{ 2
Fmax  m2 a
solving for F , yields
Fmax  14.7 N
(Answer)
8 (15) Figure 3-12 is an overhead view of a 12kg tire that is to be pulled by three
horizontal ropes. One rope’s force ( F1  50 N ) is indicated. The forces form the other
ropes are to be oriented such that the tire’s acceleration magnitude a is least. What is
a if (a) F2  30 N , F3  20 N (b)
F2  30 N , F3  10 N ; and (c) F2  F3  30 N .
that least
Solution: The key idea here is to find the minimum
value of tire’s acceleration. According to Newton’s
second law, we know that if the net force acting on
the tire is minimum, the acceleration of the tire to be
minimum.
Fig. 3-12 Problem 8
(a) If F2  30 N , F3  20 N , we can assume that these two forces are oriented in the
opposite direction of F1 , which means that the net force of these three forces is zero,
so from Newton’s second law we know that the least acceleration a equals zero too.
3.
Force
amin  0
Thus,
(Answer)
(b) When F2  30 N , F3  10 N , we can easily find out that these three forces can’t
be balanced, that is to say the acceleration can’t be zero. If F2 and F3 are also be
oriented in the opposite direction of the force F1 to make the tire have the least
acceleration, the net force will be
f net  F1  F2  F3  50 N  30 N  10 N  10 N
so the least acceleration is
amin 
f net 10 N

 0.83m / s 2
m 12kg
(Answer)
(c) With F2  F3  30 N , since these three forces are all between 0N and 80N ,
they can be balanced, which shows that with these forces acting on the tire, it will have
the least acceleration zero. So we find
amin  0
(Answer)
9 (20) Figure 3-13a shows a mobile hanging from a ceiling; it consists of two metal
pieces ( m1  3.5kg , m2  4.5kg ) strung together by cords of negligible mass. What
is the tension in (a) the bottom cord and (b) the top cord? Figure 3-13b shows a mobile
consisting of three metal pieces. Two of the masses are m3  4.8 N , m5  5.5 N . The
tension in the top cord is 199N . What is the tension in (c) the lowest cord and (d) the
middle cord?
Solution: (a) As shown in Fig. 3-13a, there are two forces acting on the metal 2, the
tension T in the bottom cord and the gravitational force Fg 2 .Choose the upward as
the positive direction. Since the metal pieces are at rest, the acceleration of the metal
pieces are equal zero. Relating the net force to the acceleration via Newton’s second
law, we will yield
T  Fg 2  m2 a  0
from which
T  m2 g  (4.5kg )(9.8m / s 2 )  44 N
(Answer)
3.
Force
(b) Newton’s second law can be used again to solve the tension in the top cord.
We can relate the net force to the acceleration of metal 1 as
T  ( Fg1  Fg 2 )  (m1  m2 )a  0
Solving for T we get
T  (3.5kg  4.5kg )(9.8m / s 2 )  78N
(Answer)
(a) As illustrated in Fig. 3-13b, the tension in the
lowest cord T and the gravitational force Fg 5
acting on the metal 5 makes the acceleration of
metal 5 zero. So we can relate the net force and the
acceleration by writing the Newton’s second law
as
T  Fg 5  m5 a  0
solving for the tension in the lowest cord, yields
Fig. 3-13 Problem 9
T  (4.5kg )(9.8m / s 2 )  54 N
(Answer)
(b) Since three metal pieces can be seen as a system, we can relate the net force of
the system to the acceleration
T  (m3  m4  m5 ) g  0
solving for m4 we will yield
m5 
199 N  (4.8kg  5.5kg )(9.8m / s 2 )
 10kg
9.8m / s 2
Again, we relate the net force to the acceleration of the system containing metal 4 and
metal 5 via Newton’s second law as
T  (m4  m5 ) g  0
substituting known values and solving for the tension in the middle cord, yields
T  (10kg  5.5kg )(9.8m / s 2 )  152 N
(Answer)
3.
Force
10 (22) A 0.20kg hockey puck has a velocity of 2.0m / s toward the east as it
slides over the surface of a frozen lake. What are the magnitude and direction of the
average force that must act on the puck during a 0.5s interval to change its velocity
to (a) 5.0m / s , west , and (b) 5.0m / s , south?
S
Solution: We choose the positive direction towards east.
(a) As shown in the right figure, the average
v  5.0m / s
v  2.0m / s
acceleration of the puck is
a
vt  v0 5.0m / s  2.0m / s

 14.0m / s 2
t
0.5s
W
O
From Newton’s second law, we get the average force that
acting on the puck
F  ma  (0.2kg )(14.0m / s 2 )  2.8 N
So the magnitude of the average force is 2.8N ,
and the direction is toward the west.
N
Fig.3-14 Problem10
(b) During the 0.5s interval, the puck’s velocity has been changed to
5.0m/ s toward the south. The acceleration component a1 in the east direction can
be written as
0  2.0m / s
 4.0m / s 2
0.50 s

In the south direction, the acceleration component a 2 is
a1 
a2 
5.0m / s
 10.0m / s 2
0.50 s
Therefore, the magnitude of the average acceleration of the pack is
a  a12  a 22  (4.0m / s 2 ) 2  (10.0m / s 2 ) 2  11m / s 2
According to Newton’s second law, we get the average force as
F  ma  0.20kg(11.m / s 2 )  2.2 N
E
3.
and
  tan 1
Force
a2
10m / s 2
 tan 1
 tan 1 (2.5)
2
a1
4.0m / s
(Answer)
 22
11 (23) In about 1915 , Henry Sincosky of Philadelphia suspended himself from a
rafter by gripping the rafter with the thumb of each hand on one side and the fingers
on the other side ( Fig. 3-14 ). Sincosky’s mass was 79kg . If the coefficient of static
friction between hand and rafter was 0.70 , what was the least
magnitude of the normal force on the rafter from each thumb or
opposite fingers? (After suspending himself, Sincosky chinned
himself on the rafter and then moved hand-overhand along the
rafter. If you do not think Sincosky’s grip was remarkable, try
to repeat his stunt.)
Solution: As shown in Fig. 3-14, if Henry wants to suspend
himself from a rafter by gripping the rafter with the thumb of
each hand on one side and the fingers on the other side, his
finger must give the rafter a very big normal force to keep
himself not falling down and moving hand-overhand along
Fig. 3-14 Problem 11
the rafter. Relating the net force to the acceleration by writing Newton’s second law as
f  Fg  ma
which here means
2 N  mg  ma
so if the acceleration of the man in vertical direction becomes zero, the normal force
the man acting on the rafter will become the least magnitude. Thus,
Nmin 
mg (79kg )(9.8m / s 2 )

 553N
2
2  0.70
(Answer)
12 (26) You testify as expert witness in a case involving an accident in which car A
slid into the rear of car B, which was stopped at a red light along a road headed down a
hill (Fig.3-15). You find that the slope of the hill is   12.0 , that the cars were
separated by distance d  24.0m when the driver of car A put the car into a slide(it
3.
Force
lacked any automatic antibrake-lock system), and that the speed of car A at the
onset of braking was v0  18.0m / s. With what speed did car A hit car B if the
coefficient of kinetic friction was (a) 0.60 (dry road surface ) and (b) 0.10 (road
surface covered with wet leaves )?
Solution: (a) As shown in the free-body diagram of the car A in Fig. 3-15a, the forces
FN
fk
A

Fig. 3-15 Problem 12
Fg
Fig. 3-15a Problem 12


acting on it are the frictional force f k , the normal force F N and the gravitational

force Fg , whose magnitude is equal to mg . Then, from Newton’s second law, we
have




f k  Fg  FN  ma
Rewriting the equation above for components along the x and y axes in Fig. 3-15a, we
have
FN  mg cos   ma y  0
FN  mg cos 
so
Similarly, for the x axis we have
f k  mg sin   max
in which
f k   k FN
solving for a x yields
ax  k g cos   g sin 
 0.60  9.8 m s 2 cos12.0  9.8 m s 2 sin12.0  3.7m / s 2
3.
Force
Assuming the speed was v t when car A hit car B , along the x axis we get
vt2  v02  2ax d
Substituting known values and solving for v t yields
vt  (18.0 m s ) 2  2  3.7 m s 2 (24.0m)  12.1m / s
(Answer)
(b) If road surface was covered with wet leaves, which makes the coefficient of kinetic
friction smaller, the acceleration of car B along x axis would be
ax  k g cos   g sin 
(Answer)
 0.10  9.8 m s 2 cos12.0  9.8 m s 2 sin12.0  1.1m / s 2
(c) So the speed vt with which car A hit car B was
vt  (18.0 m s) 2  2  (1.1 m s 2 )(24.0m)  19.4m / s
(Answer)
13 (27) A loaded penguin sled weighing 80N rests on a plane inclined at angle
  20 to the horizontal (Fig. 3-16). Between the sled and the plane, the coefficient
of static friction, is 0.25 , and the coefficient of kinetic friction is 0.15 . (a) What is
the least magnitude of the force F , parallel to the plane, that will prevent the sled
from slipping down the plane ? (b)What is the minimum magnitude F that will start
the sled moving up the plane ? (c) What value of F is required to move the sled up the
N
fs F
A

Fg
Fig. 3-16 Problem 13
Fig. 3-16a Problem 13
plane at constant velocity ?
Solution: (a) As shown in Fig. 3-16a, there are four forces acting on the sled, the



normal force N from the plane, the tension F , the gravitational force Fg and the
3.
Force

static frictional force f s . Relating the net force to the sled’s acceleration with
Newton’s second law, we write
   

N  F  f s  Fg  ma
Rewriting the components versions of the equation above yields
N  mg cos   0
F  f s  mg sin   ma x
When the acceleration along x axis tends to be zero, and the static frictional force
along the inclined plane is upward along the plane, the least magnitude of

F preventing the sled from slipping down the plane will be
F min  mg sin    s mg cos   80 N (0.342  0.25  0.940)  8.6 N (Answer)

(b) With the tension F growing bigger, the static frictional force will become kinetic
frictional force. Again, the components of Newton’s second law can be written as
N  mg cos   0
F  f s  mg sin   ma x


If the net force of F and Fg along the x axis gets larger than the static force, the
sled will begin to move up, and a x  0 at this time.
Fmin  mg sin    s mg cos   80 N (0.342  0.25  0.940)  48N (Answer)
(c) When the sled is required to move up with a constant velocity, the forces acting on
the sled will be balanced, which means that the components of acceleration a x and
a y are all zero.
N  mg cos   0
F  f k  mg sin   0

Substituting the known values and solving for F , we get
Fmin  mg sin    k mg cos   80 N (0.342  0.15  0.940)  38.6 N (Answer)
3.
Force
14 (28) In Fig. 3-17, a box of Cheerios (mass mc  1.0kg ) and a box of
mw  3.0kg ) are accelerated across a horizontal surface by a
horizontal force F applied to the Cheerios box . The magnitude of the frictional
force on the Cheerios box is 2.0N , and the magnitude of the frictional force on the
Wheaties box is 4.0N . If the magnitude of F is 12N , what is the magnitude of
Wheaties (mass
the force on the Wheaties box from the Cheerios box ?
Solution: Two boxes can be seen as a system, the free-body diagram of which is
shown in Fig. 3-54a. According to Newton’s second law, we write
fw
fc
mc  mw
F
Fig. 3-17 Problem 14

 

F  f  Ma
Fig. 3-17a Problem 14
where f is the total frictional force on the system and M is the total mass of the
system.
F  ( f C  f w )  (mC  mW )a
here,

Solving of a , yields
a
12 N  2 N  4 N
 1.5m / s 2
1.0kg  3.kg

Now, we consider the Wheaties’s motion. The force from the Cheerios FCW and the

frictional force f W acting on the Wheaties, make it have the same magnitude of
acceleration as the system we studied above. So the Newton’s second law can be
written as
FW  fW  mw a
From which ,
FW  4.0 N  3.0kg(1.5m / s 2 )  8.5 N
(Answer)
So the magnitude of the force on the Wheaties box from the Cheerios box is 8.5N .