5
APPLICATIONS CHAPTER OF NEWTONS LAW
Friction
Drag Forces
Motion Along Curved Path
The Center of Mass
1
FRICTION
Friction is physical phenomenon that appear at contact of two objects,
when you move or attempt to move one object relatively another.
Polished steel surface
Source of friction:
Interaction between surfaces on
on molecular level
Nickel
atoms
Gold atoms
2
Static Friction
Static friction formula f s max = s Fn
Where constant s is called coefficient static friction.
Object begin to slide if magnitude applied force F s Fn
Friction forces
Free body diagram for the block
Fn
Direction of friction force
always opposite direction of
motion
In case sliding block on
horizontal surface the
F
fs
magnitude of normal force
is equal to weight
Fg
Measuring coefficient static friction
using incline plane
3
Kinetic Friction
Kinetic friction formula is applied to the mowing object.
f k = k Fn
Constant k is the coefficient of kinetic friction.
Usually for the same couple surfaces s k
From your experience, it is difficult to start slide object but when it start
it is easy to continue push.
The figure demonstrate how friction force
depend from applying force.
When object moving kinetic friction
force remain constant
Balanced forces
1st Newtown law
at rest
In motion
4
5
The frictional force depends only on the type of surfaces and how
hard the surfaces are pressed together.
Normal force
Friction force
We can reduce friction by
lubricating the surfaces. This
means that the surfaces no longer
rub directly on each other, but
slide past on a layer of liquid.
Fgn
Fg
Acting force
6
Four identical blocks of same mass are moving on a surface for
which the coefficient of kinetic friction between each block and
the surface is k. The velocity of each block is indicated by the
vector on the block. For which block is the force of friction
between the surface and the block greatest?
A.
B
C
D
E
1
2
3
4
The force of friction is the same for all blocks.
7
Rolling Friction
Rolling friction formula f r = r Fn
Constant r is coefficient of rolling friction.
r Depends on nature of contacting
surfaces
v
8
Conceptual problem:
Blocks made of the same material and similar shape but differing in
mass lie on the bed of a truck that is moving along a straight horizontal
road. When track accelerate all of the blocks will slide. What is
relation between minimum accelerations to start sliding of each block
with different mass ?
Solution:
Of these three forces, the only one that
acts in the direction of the acceleration
is the static friction force.
Apply Newtons 2nd law
f s = s Fg = s mg = ma x
ax = s g
Because ax is independent of m and Fg, the critical
accelerations are the same.
9
Conceptual problem:
Attach a rubber band to the block and pull gently and steadily on
the rubber band in the horizontal direction. At some point, the
block will start moving, but it will not move smoothly. Instead, it
will start moving, stop again, start moving again, stop again, and so
on. Explain why the block moves this way.
Answer:
As the rubber band (spring) is extended, the force exerted by the
rubber band on the block increases. Once that force is greater than
the maximum value of the force of static friction on the block, the
block will begin to move. However, as it accelerates, it will shorten
the length of the band, decreasing the force that the rubber band
exerts on the block. As this happens, the force of kinetic friction
can then slow the block to a stop, which starts the cycle over again.
10
A block of mass m is at rest on an inclined plane that makes an
angle of 30 with the horizontal, as shown in the figure. Which
of the following statements about the force of static friction is
true?
A
B
C
D
E
fs > mg
fs > mg cos 30
fs = mg cos 30
fs = mg sin 30
None of the statements is true.
11
Problem:
On an icy roadway, the coefficient of friction between the tires of car
and roadway is reduced to one-quarter of its value on a dry road. As a
result, the maximum speed vmax at which the car can safely turn on
curve of radius R is reduced. What is a new value for this speed.
Solution:
Fx = f s,max
f s,max = s mg
2
vmax
=m
R
vmax = s gR
F y = Fn mg = 0
v'max = 's gR
's gR
v'max
' s
=
=
v max
s gR
s
v'max =
' s
v max v' =
max
s
v
1
4 max
= 0.5vmax = 50%vmax
12
A block of mass m rests on a plane that is inclined at an angle with
the horizontal. What is maxim angle when block start slide if the
coefficient of static friction is s
Apply Fx = ma x
f s Fg sin = 0
or, because Fg = mg, f s mg sin = 0
Apply F y = ma y
(1)
Fn mg cos = 0 (2)
Divide equation (1) by equation (2) to obtain:
fs
tan =
Fn
Substitute for fs = sFn and simplify to obtain:
tan =
s Fn
= s
Fn
= tan 1 s
13
Two blocks of masses m1 and m2 are sliding down an incline as
shown in figure. They are connected by a massless rod. The
coefficients of kinetic friction between the block and the surface are
1 for block 1 and 2 for block 2. Determine the acceleration of the
two blocks.
Apply F = m a to block 1
Fx = T1 + m1 g sin f k,1 = m1a x (1)
F y = Fn,1 m1 g cos = 0
f k,1 = 1Fn,1
Apply F = m a to block 2
Fx = m2 g sin T2 f k,2 = m2a x (2)
F y = Fn,2 m 2 g cos = 0
f k,2 = 2 Fn,2
m 2 g sin
14
Let T1 = T2 = T,
Use the definition of the kinetic friction force to eliminate fk and Fn
from equations (1) and (2)
m1a x = m1 g sin + T 1m1 g cos (3)
m 2a x = m 2 g sin T 2 m 2 g cos (4)
Add equations (3) and (4) to eliminate T and solve for ax:
1m1 + 2 m 2
a x = g sin
cos
m1 + m 2
15
In figure m1 = 4.0 kg and the coefficient of static friction between
the block and the incline is 0.40. Find the range of possible values
for m2 for which the system will be in static equilibrium.
Noting that Fg,1 = m1 g , apply
Fx = T f s,max m1 g sin 30 = 0 (1)
F = m a to the block whose mass is m1:
F y = Fn,1 m1 g cos 30 = 0 (2)
Using f s, max = s Fn , substitute equation (2) in equation (1) to obtain:
T m g cos 30 m g sin 30 = 0 (3)
s
1
1
Noting that Fg,2 = m 2 g , m 2 g T = 0 (4)
Add equations (3) and (4) to eliminate T
m 2 = m1 ( s cos 30 + sin 30)
6
m 2 = ( 4.0 kg ) [ ( 0.40 ) cos 30 + sin 30] m 2, + = 3.4 kg1and m 2,- = 0.61 kg
Summary
Horizontal surface
the maximum static friction force
f s,max = s mg
f s, max = s Fn
if Fapp f s ,max object is rest and f s , frict = Fapp
if object have const . velocity Fapp = f k , frict
if object move with acceleration Fapp f k , frict = ma x
Incline plane
same condition we can apply to incline
only Fn = mg cos and Fapp = mg sin
Fapp
if Fapp f s ,max object is rest and f s , frict = Fapp
If additional force F applied to same
direction as Fapp then: Fapp ,total = F + Fapp
object move with acceleration Fapp ,total f k , frict = ma x
17
If force is applied under some angle
Fapp
the maximum static friction force
f s, max = s Fn
Fn = Fg Fapp sin
x
f s ,max = s ( Fg Fapp sin )
if Fapp , x
Fg = mg
Fapp , x = Fapp cos
f s ,max object is rest and f s , frict = Fapp , x
if object have const . velocity Fapp , x = f k , frict
if object move with acceleration Fapp , x f k , frict = ma x
18
Resistive Force
Resistive force act on object that move in media always is in opposite
direction of velocity.
General formula for this force is:
(
)
Fres = b1v + b2 v 2 v
Resistive force on spherical object with radius r, mass m, and speed v is
(
)
Fres = C1rv + C 2 r 2v 2 v
kg
m3
kg
ms
Firs term associate with viscosity and second with pressure
Fres
At condition when
acceleration is 0 and
2
Fg = mg = (C1rvterm + C 2 r 2v term ) Velocity of object is terminal velocity.
2
C1rvcrit = C2r 2vcrit
Fg
Fres
=
Fg
vcrit = C1 C2r
C
v
Fres = C1rv 1 + 2 rv = C1rv 1 +
v
C1
crit
Regime 1 when v < <vcrit (viscose liquids) For air C1=3.1*10^-4
19
Regime 2 when v >> v crit (gases)
and C2=0.85 for liquids
Drag Force
2
Drag force formula Fd = b v
The parameter b is function of area
of object exposed to the flow.
If object falling under influence of
gravity then acceleration will define
by dragging force,
mg bv 2 = ma y
bv 2
a y= g
m
Terminal speed define when a y approach zero
(drag force balancing gravitational force).
1/ 2
mg
vT =
b
bv 2
Terminal speed mean that object move with
ma y
constant velocity (1st Newtons Law in action) mg
v2
a y = g 1 2
20
vT
As an object falls faster and faster,
the air resistance will get more and
more. Eventually, the force of air
resistance will be equal to the
object's weight, so the object will
stay at that speed, and stop
21
accelerating.
A skydiver of mass 60.0 kg can slow himself to a constant speed of
200 km/h by orienting body horizontally in arc position. In this
position he presents the maximum cross-sectional area and thus
maximize the air-drag (a) force. What is the magnitude of the drag
force on the skydiver? (b) If the drag force is given by bv^2, what is
the value of b?
(a) Apply F y = ma y
(1) Fd mg = ma y or, because ay = 0, F = mg
d
(
)
Fd = ( 60.0 kg ) 9.81 m/s 2 = 588.6 N = 589 N
(b) Substitute Fd = bv t2 in equation (1)
mg Fd
bv t2 = mg
b= 2 = 2
vt
vt
b=
588.6 N
km
1h
200
h 3600 s
2
0.42 kg/m
22
(1)
MOTION ALONG A CURVED PATH
Satellite moving in circular orbit around Earth with
constant tangential velocity.
( vt ) 2 + r 2 = ( r + h ) 2
( vt ) 2 + r 2 = r 2 + 2rh + h 2
h( 2r + h ) = ( vt ) 2
For short int erval time we can neglect h compared with 2r
( v 2 r ) t 2
h
2
a t2
this formula is similar to x =
2
The satellite moving with constant speed and appear
falling every time distance h on circular orbit with
v2
acceleration
r
23
vt
MOTION ALONG A CURVED PATH
Conical pendulum
Newtowns Second Law
v2
Fx = ma x T sin = m = Fnet
r
F y = ma y T cos mg = 0
The net force is centripetal force.
v2
m
Fnet
r
T=
=
sin
sin
Fnet v 2
ax =
=
m
r
v = axr
Period rotation =
2r
v
24
A block of mass m1 is attached to a cord of length L1, which is fixed at
one end. The block moves in a horizontal circle on a frictionless
tabletop. A second block of mass m2 is attached to the first by a cord
of length L2 and also moves in a circle on the same frictionless
tabletop, as shown in figure. If the period of the motion is P, find the
tension in each cord.
Free body diagram
25
Apply
Fx = max
2
v1
T1 T2 = m1
L1
Apply
to the block whose mass is m1:
(1)
Fx = max
2
v2
T2 = m2
L1 + L2
to the block whose mass is m2:
(2)
Relate the speeds of each block to their common period P
v1 =
2L1
2 ( L1 + L2 )
and v2 =
P
P
In the second an first force equations,
substitute for v2 and v1 , and simplify to obtain:
T2 = [ m2 ( L1 + L2 ) ]
2
P
2
2
T1 = [ m2 ( L1 + L2 ) + m1L1 ]
P
2
26
Problem:
You swing a bucket containing mass m of water in a vertical
circle of radius r. If the speed is vtop at the top of the circle find (a)
the force FPW exerted by the bucket on the water at the top of the
circle, and (b) the minimum value of vtop for the water to remain in
the bucket (c) What is the force exerted by the bucket on the
water at the bottom of the circle, where the speed of the bucket is
vbot.
27
Solution:
For position of bucket on the top.
2
m v top
FPW + mg =
r
2
v top
FPW = m
g
r
For minimum speed on the top FPW = 0
mg =
2
m v top min
v top min = rg
r
For position of bucket on the bottom.
FPW
2
2
v bot
m vbot
mg =
FPW = m
+ g
r
r
28
An airplane is flying in a horizontal circle at a speed of 480 km/h. The
plane is banked for this turn, its wings tilted at an angle of 40 from the
horizontal. Assume that a lift force acting perpendicular to the wings
acts on the aircraft as it moves through the air. What is the radius of
the circle in which the plane is flying?
29
v2
Apply F = m a to the plane: F = F sin = m
x
lift
r
and F y = Flift cos mg = 0
Eliminate Flift between these equations to obtain by dividing:
v2
v2
r=
tan =
g tan
rg
2
km
1h
480
h 3600 s
= 2.2 km
r=
9.81 m/s 2 tan40
(
)
30
A curve of radius 150 m is banked at an angle of 10. An 800-kg car
negotiates the curve at maximum 85 km/h without skidding. Neglect the
effects of air drag and rolling friction. Find (a) the normal force exerted
by the pavement on the tires, (b) the frictional force exerted by the
pavement on the tires, (c) the coefficient of static friction between the
pavement and the tires.
v2
Fx = Fn sin + f s cos = m
r
F y = Fn cos f s sin mg = 0
Multiply the x equation by sin and y equation by cos
v2
2
f s sin cos + Fn sin = m sin
r
2
Fn cos f s sin cos mg cos = 0
v2
Add these equations to eliminate fs: Fn mg cos = m sin
r
2
2
v
v
Fn = mg cos + m sin = m g cos + sin
r
r
fs
s =
Fn cos mg
fs =
31 Fn
sin
You ride a bicycle in a 20-m-radius circle on a horizontal surface. The
resultant force exerted by the surface on the bicycle (normal force plus
frictional force) makes an angle of 15 with the vertical. What is your
speed?
Apply F = m a to the bicycle:
mv 2
and F y = Fn mg = 0
Fx = f s =
r
Relate Fn and fs to :
mv 2
fs
v2
tan =
=r=
Fn mg rg
Solving for v yields:
v = rg tan = 7.3 m / s
32
Summary
v2
Centripetal acceleration: a c =
r
v2
Centripetal force:Fc = ma c = m
r
d vt
Tangential acceleration: a t =
dt
The magnitude of total acceleration at curve path is: a =
a = ac + at
At constant speed (uniform circular motion) a t = 0
2
ac + at
2
When you make a turn on road with friction at constant maxim speed
without skidding define by:
mv 2
f s = s Fn = s mg =
r
33
THE CENTER OF MASS
The center of mass call:
The point in a system of bodies at which the mass of the system
may be considered to be concentrated and at which external forces
may be considered to be applied.
Parabola
Center mass define by;
Mxcm = m1 x1 + m 2 x2
where M = m1 + m 2
Choose origin of coordinate for location of
first mass and distance between mass is d.
baton thrown into the air
Mxcm = m1 0 + m 2 d
xcm
m2d
=
m1 + m 2
34
35
Center mass in three dimension for N particles.
Mxcm = m1 x1 + m 2 x2 + m3 x3 + .... + m N x N
or for x coordinate
Mxcmr = m i x i
i
Where M = m i total mass
i
for y coordinate
Mycm = m i yi
i
for z coordinate
Mzcm = m i z i
i
in vector notation Mrcm = m i ri
Where ri = xi i + yi + z i k
j
j
and rcm = xcm i + ycm + zcm k
r d m = r d m Center mass in three
rcm d m = r d m rcm =
dimension for solid body.
M
36
dm
Three balls A, B, and C, with masses of 3.0 kg, 1.0 kg, and 1.0 kg,
respectively, are connected by massless rods, as shown in figure.
What are the coordinates of the center of mass of this system?
The x coordinate of the center of mass is given by:
x cm
m A x A + m B xB + mC xC
=
= 2m
m A + mB + mC
The y coordinate of the center of mass
is given by:
y cm
1.4
c.m.
m A y A + m B y B + m C yC
=
= 1.4m
mA + mB + mC
37
Alternate Solution Using Vectors
The vector expression for the center
of mass is:
m i ri
M rcm = m i ri
rcm = i
i
M
i
j
where rcm = xcm + ycm + zcm k
1.4
c.m.
The position vectors for the objects located at A, B, and C are:
i
j
rA = 2.0 + 2.0 + 0k m rB = ( + + 0k )m rC = ( 3.0 + 0 + 0k ) m
i
j
ij
m A rA + m B rB + mC rC
rcm =
m A + m B + mC
(
[
)
(
)
(
)
(
)]
( 3.0 kg ) 2.0 + 2.0j + 0k m + (1.0 kg ) + j + 0k m + (1.0 kg ) 3.0 + 0j + 0k m
i
i
i
rcm =
( 3.0 kg + 1.0 kg + 1.0 kg )
= ( 2.0 m ) + (1.4 m ) + 0k
i
j
38
Two identical uniform rods each of length L are glued together so that
the angle at the joint is 90. Determine the location of the center of
mass (in terms of L) of this configuration relative to the origin taken
to be at the joint.
The x coordinate of the center of mass is given by:
( 0) m1 + ( 1 L) m2
x1,cm m1 + x2,cm m 2
2
xcm =
xcm =
m1 + m 2
m1 + m 2
or, because m1 = m2 = m,
xcm =
ycm =
( 0) m + ( 1 L) m
2
m+m
( 1 L) m + ( 0) m
2
m+m
=1L
4
=1L
4
The center of mass of this system is located at
( 1 L, 1 L)
4
4
39
Problem:
Find the center of mass of the uniform sheet of plywood.
Solution:
Divide sheet of plywood on two rectangle 1 and 2.
m1 xcm1 + m 2 xcm 2
xcm =
m1 + m 2 = M
m1 + m 2
m1 ycm1 + m 2 ycm 2
M
Area of sheet1 = A1 sheet 2 = A2 m i M = Ai A
ycm =
A1
A2
A1
A2
xcm =
xcm1 +
xcm 2 ycm =
ycm1 +
ycm 2
A
A
A
A
A1 = 0.32m 2 A2 = 0.04m 2 A = A1 + A2
xcm1 = 0.4m
ycm1 = 0.2m
xcm 2 = 0.7 m
ycm 2 = 0.5m
xcm = 0.43m
ycm = 0.23m
40
Find the location of the center of mass of a nonuniform rod 0.40 m in
length if its linear mass density ( x ) varies linearly from 1.00 g/cm
at one end to 5.00 g/cm at the other end. Specify the center-of-mass
location relative to the less-massive end of the rod.
The x coordinate of the center of mass
xdm
xcm =
dm = ( x ) dx , xcm = x ( x ) dx ( x ) = b + ax
dm
( x ) dx
( 0 ) = b = 1.00 g / cm ( 40 ) = 1.0 g / cm + a 40 cm a = 0.1 g / cm 2
xcm
( x ) = 1.00 g/cm + ( 0.10 g/cm 2 ) x
40 cm
2
x[1.00 g/cm + ( 0.10 g/cm ) x ]dx
= 40 cm
= 24 cm
0
[1.00 g/cm + (0.10 g/cm 2 ) x ]dx
41
0
Find the center of mass of a uniform semicircular disk of radius R.
y dA
ycm =
y = r sin
(1)
M
Express dA in terms of r and :
dA = r d dr
M = Ahalf disk = 1 R 2
2
Substitute in equation (1) and evaluate ycm:
R
y cm =
r 2 sin d dr
00
M
2 R 2
2 3 4
=
r dr = 3M R = 3 R
M0
42
Motion of the Center Mass
drcm
dri
dr1
dr2
M
= m1
+ m2
+ ... = m i
dt
dt
dt
dt
i
Mv cm = m1v1 + m 2 v 2 + ... = m i v i
i
Macm = m1a1 + m 2a2 + ... = m i ai
i
mi ai = Fi
i
i
Fi = Fint i + Fexti
i
i
i
Fint i = 0 Third Newtowns law
i
for stable system of particles
Fnet = Fexti = Macm
i
The center mass of system move like single mass M = m i
under the influence of external force.
43