Kinetics of Particles
So far, we have only studied the kinematics of particle motion, meaning
we have studied the relationships between position, velocity, and
acceleration, with only cursory reference to the forces which produced the
motion.
In some sense, the kinematics is the study of dynamics from a purely
experimental standpoint. That is, if we are able to map the positions,
velocities, and accelerations of multiple particles, then we can ascertain
the force which produced the motion. Note: By the word “particle”, it is
implied that the object is contained entirely within a point.
In the next part of this course, we will be interested in the forward model.
That is, if we know the forces on the particle, what should the resulting
motion be? The reason I call this a forward model is because the
classification of different forces is purely theoretical and can only be
backed up by experimental measurements.
8-1
NEWTON’S SECOND LAW
Denoting by m the mass of a particle, by S F the sum, or
resultant, of the forces acting on the particle, and by a the
acceleration of the particle relative to a newtonian frame of
reference, we write
S F = ma
Introducing the linear momentum of a particle, L = mv,
Newton’s second law can also be written as
.
SF=L
which expresses that the resultant of the forces acting on a
particle is equal to the rate of change of the linear momentum
of the particle.
8-2
ay
y
To solve a problem involving the motion of a
particle, S F = ma should be replaced by
equations containing scalar quantities. Using
rectangular components of F and a, we have
P
az
ax
x
z
S Fx = max
S Fy = may
S Fz = maz
Using tangential and normal components,
y
an
at
P
x
O
O
r
Using radial and transverse components,
aq
r
dv
S Ft = mat = m
dt
v2
S Fn = man = m
q
P
ar
x
.. . 2
S Fr = mar= m(r - rq )
..
..
S Fq = maq = m(rq + 2rq)
8-3
Central Forces
In many forces, such as gravitational, electrical forces, etc., the magnitude of
the force only depends on the distance away from the source (i.e. r  2 ), and
the direction is always towards or away from the source rˆ .





2
F  r , q   m a  F r rˆ  Fq qˆ  m r  r q rˆ  m r q  2 rq qˆ

 Fq  0  m r q  2 rq

Using these assumptions, we can derive an expression for the rotation
frequency:
1 d
r 2q   r q  2 rq  0
r dt
The result implies that under the
2
r q  const .  h
action of a central force, the
h
rotation frequency depends only

q  2
on the radial distance away from
r
the attractor.
8-4
Using this result, we can arrive at an expression for radial force
purely in terms of the radial distance
2


h


2


q  2
F r  m r  r q  m  r  r  2  


r
r




The above represents a nonlinear differential equation, to which the solution
will give us r as a function of time. We will return to this later in the course.

h

One example that we have discussed previously is the flat earth
approximation. When is it valid?
baseball
F r  Gm B m e
h
Re
earth
1
r
2

Fr  m B
1
R e
 h
Gm e
R
2
e
2
1
r
2
1


h 
2

R e  1 
R e 

 mBg
r  Re  h
where
2
1

2
Re
where
h  R e
when
g 
Gm e
2
Re
8-5
Example Problem
8-6
Free Body Diagram for Block B

f  N
q

N BA
q

mg

F B  F Bx xˆ  F By yˆ
F Bx   f cos q   N BA sin q   m B a x
F By  f sin q   N BA cos q   m B g  m B a y
8-7
Free Body Diagram for Block A

N As
Force on Block A due
to substrate

F Applied

N AB

F A  F Ax xˆ  F Ay yˆ
q

mAg

f   N AB
F Ax  F Applied  f cos q   N AB sin q   m A a x
F Ay   f sin q   N AB cos q   N As  m A g  m A a y  0
8-8
Newton’s 1st law provides us with an expression that the force
balance equals zero. (i.e.)
f  N
AB
  N BA
What we would like to do is solve for the forces if Block B is on the verge of
slipping, but has not yet done so. This allows us to develop a relationship
between the accelerations of Blocks A and B
F Bx   f cos q   N BA sin q   m B a x
F By  f sin q   N BA cos q   m B g  m B a y  0
F Ax  F Applied  f cos q   N AB sin q   m A a x
F Ay   f sin q   N AB cos q   N As  m A g  m A a y  0
These 4 equations have 4 unknowns listed below, and the solution can
be solved using linear algebra
F applied , N AB , a x , N
8-9
Since we’re only interested in Block B, let’s
try a change in coordinates

y

f   N BA
y’
N BA
x
q
x’

mBg
F Bx    N BA  m B g sin q   m B a cos q 
F By  N BA  m B g cos q   m B a sin q 
Thus, the critical slipping threshold can be expressed as:
N BA  m B a sin q   m B g cos q
   m B a sin q   m B g cos q
 
a cos q   g sin q

  m B a cos q   m B g sin q 


a sin q   g cos q 
a
g
a
g
cos q   sin q
sin q   cos q


tan q   a

a
g
g
tan q   1
8 - 10
Case 1. Assume that the acceleration applied to the
block is zero
tan q   a
 
a
g
g
tan q   1
 tan q

q  tan
1

Therefore, the Block will slide under the action of gravity if the angle exceeds θ.
Case 2. Neglect Gravity
tan q   a
 
a
g
g
tan q   1

1
tan q 
q  tan
1
1 
 

Note: This is the result derived in the book.
8 - 11
Case 3. Account for everything
tan q   a
 
a
g
g
tan q   1
 a 

g
1
q  tan 
1  a g






So what is the effect of applying acceleration to the block? You
can think of it in two ways. One is that the acceleration delays the
action of gravity. Depending on the friction coefficient, the block
may slide downwards when the applied force is removed.
Alternatively, one can say that the acceleration keeps the block
from slipping at higher angles.
Note: We still haven’t solved for the acceleration explicitly.
Once the block starts to slip, we have to take into account the
relative motion of Blocks A and B.
8 - 12
8 - 13
Given:
x A  0 . 9
m
s
x A  0 .4 m
l  0 .5 m
Constraints: Motion of Slider A is confined to the X-axis
Motion of Slider B is confined to the Y-Axis

F B  F Bx xˆ  F By yˆ  ma y yˆ

F A  F Ax xˆ  F Ay yˆ  ma x xˆ
Distance between Sliders
A and B is constant
x A
 xB   y A  yB   l
2
2
2
Free Body Diagrams:
F By   m B g  T BA sin q   m B a y  m B y B
F Ax   T AB cos q   P  m A a x  m A x A
8 - 14
Additional Constraints
d
dt
 x
d
2
dt
2
A
 x

2
 yB
2

A
 x A  

 yB
1
xA
 x
2

d
dt
2  
2
A
l
2
 0  2 x A x A  2 y B y B  0  y B   x A
xA
yB
2
2
0  x A x A  x A  y B  y B y B
2
 y B  y B y B

In these two equations, the only thing we don’t know are the
accelerations of sliders A and B, however they are interdependent
 m B g  T BA sin q   m B y B
 T AB cos q   P  m A x A  
mA
xA
x
2
A
2
 y B  y B y B
Thus, we really have 2 equations with 2 unknowns…

y B , T BA .
8 - 15
y B  
 T AB cos q   P  m A x A  
m y
T AB  A B sin q   cos q
 mB xA

  

mA  2
x
 x A  y B2  gy B  P A
x A 
mA
T AB
T AB
2 kg 

 0 . 9
0 . 4 m 
m
s


  0 .9

mB
sin q   g
 T

mA  2
 x A  y B2  y B   BA sin q   g  
 m

x A 
B


m  2
x
2
 A  x A  y B  gy B  P A
xA 
mA
2
T BA
m
s




 m A y B







sin
q

cos
q
 m x

 B A

1
2
0 .4 m 
m 
0 . 4 m   3 2 kg  0 . 3 m  4 


   10 2  0 . 3 m   40 N


  5
5
0 .3 m 
2 kg  
3 kg  0 . 4 m 
s 


1
T AB  46 . 6 N
 aB  
T BA
 aA  
1
mB
xA
sin q   g  
 x
2
A
2
 y B
46 . 6 N 3
3 kg
  9 . 32
5
m
s

 0 . 9


 yB yB   
0 . 4 m 
1
m
s
2

2

  0 .9

m
s
2
0 .4 m 
m 
m

  0 . 3 m   9 . 32 2    1 . 365 2
0 .3 m 
s  
s

Note: The answer shown in the book assumes gravity is zero. What are the
correct tensions and accelerations if gravity is taken into account? What also
must we do if friction between the sliders and wall are taken into account?
8 - 16
Problem 1
P
A
B
q
Block A has a mass of 30 kg and block
B has a mass of 15 kg. The coefficients
of friction between all plane surfaces of
contact are s = 0.15 and k = 0.10.
Knowing that q = 30o and that the
magnitude of the force P applied to
block A is 250 N, determine (a) the
acceleration of block A , (b) the tension
in the cord.
8 - 17
Problem 1
P
A
B
q
Block A has a mass of 30 kg and block
B has a mass of 15 kg. The coefficients
of friction between all plane surfaces of
contact are s = 0.15 and k = 0.10.
Knowing that q = 30o and that the
magnitude of the force P applied to
block A is 250 N, determine (a) the
acceleration of block A , (b) the tension
in the cord.
1. Kinematics: Examine the acceleration of the particles.
2. Kinetics: Draw a free body diagram showing the applied
forces and an equivalent force diagram showing the vector
ma or its components.
8 - 18
Problem 1
P
A
B
q
Block A has a mass of 30 kg and block
B has a mass of 15 kg. The coefficients
of friction between all plane surfaces of
contact are s = 0.15 and k = 0.10.
Knowing that q = 30o and that the
magnitude of the force P applied to
block A is 250 N, determine (a) the
acceleration of block A , (b) the tension
in the cord.
3. When a problem involves dry friction: It is necessary first to
assume a possible motion and then to check the validity of the
assumption. The friction force on a moving surface is F = k N.
The friction force on a surface when motion is impending
is F = s N.
8 - 19
Problem 1
P
A
B
q
Block A has a mass of 30 kg and block
B has a mass of 15 kg. The coefficients
of friction between all plane surfaces of
contact are s = 0.15 and k = 0.10.
Knowing that q = 30o and that the
magnitude of the force P applied to
block A is 250 N, determine (a) the
acceleration of block A , (b) the tension
in the cord.
4. Apply Newton’s second law: The relationship between the
forces acting on the particle, its mass and acceleration is given
by S F = m a . The vectors F and a can be expressed in terms of
either their rectangular components or their tangential and normal
components. Absolute acceleration (measured with respect to
a newtonian frame of reference) should be used.
8 - 20
Problem 1 Solution
Kinematics.
P
A
B
Assume motion with block A moving
down.
If block A moves and accelerates down
the slope, block B moves up the slope
with the same acceleration.
q
aA = aB
A
aA
aB
B
q
8 - 21
Problem 1 Solution
P
A
Kinetics; draw a free body diagram.
Block A :
B
q
T
WA= 294.3 N
WA = mA g
WA = (30 kg)(9.81 m/s2)
WA = 294.3 N
WB = m B g
WB = (15 kg)(9.81 m/s2)
WB = 147.15 N
mA a = 30 a
=
250 N
Fk = k N
N
Block B :
N
WB= 147.15 N
T
Fk = k N
F’k = k N’
mB a = 15 a
=
N’
8 - 22
Problem 1 Solution
Block A :
WA= 294.3 N
30o
T
mA a = 30 a
=
250 N
Fk = k N
+ SFy = 0:
N
Apply Newton’s second law.
N - (294.3) cos 30o = 0
N = 254.87 N
Fk = k N = 0.10 (254.9) = 25.49 N
+
SFx = ma: 250 + (294.3) sin 30o - 25.49 - T = 30 a
371.66 - T = 30 a
(1)
8 - 23
Problem 1 Solution
Block B :
30o
N
WB= 147.15 N
T
Fk = k N
mB a = 15 a
=
N’
F’k = k N’
+ SFy = 0:
N’ - N - (147.15) cos 30o = 0
N’ = 382.31 N
F’k = k N’ = 0.10 (382.31) = 38.23 N
+
SFx = ma: T - Fk - F’k - (147.15) sin 30o = 15 a
T - 137.29 = 15 a
(2)
Solving equations (1) and (2) gives:
T = 215 N
a = 5.21 m/s2
8 - 24
P
A
Problem 1 Solution
Verify assumption of motion.
Check: We should verify that blocks
actually move by determining the value of
the force P for which motion is impending
B
q
Find P for impending motion.
For impending motion both blocks are in equilibrium:
Block A
Block B
WA= 294.3 N
30o
P
Fm = s N
T
30o
N
WB= 147.15 N
T
Fm = s N
N
F’m = s N’
N’
8 - 25
Problem 1 Solution
WA= 294.3 N
30o
T
A
P
Fm = s N
30o
N
Fm = s N
N
WB= 147.15 N
T
B
F’m = s N’
N’
From + SFy = 0 find again
N = 254.87 N and N’ = 382.31 N,
and thus
Fm = s N = 0.15 (254.87) = 38.23 N
F’m = s N’ = 0.15 (382.31) = 57.35 N
8 - 26
Problem 1 Solution
WA= 294.3 N
30o
T
A
P
Fm = s N
30o
N
WB= 147.15 N
Fm = s N
N
T
B
F’m = s N’
N’
For block A:
+
SFx = 0: P + (294.3) sin 30o - 38.23 - T = 0
(3)
For block B:
+
SFx = 0: T - 38.23 - 57.35 - (147.15) sin 30o = 0
(4)
Solving equations (3) and (4) gives P = 60.2 N.
Since the actual value of P (250 N) is larger than the value for
impending motion (60.2 N), motion takes place as assumed. 8 - 27
Problem 2
12 lb
B
30 lb A
30o
A 12-lb block B rests as shown on
the upper surface of a 30-lb wedge
A. Neglecting friction, determine
immediately after the system is
released from rest (a) the acceleration
of B relative to A, (b) the acceleration
of B relative to A.
8 - 28
Problem 2
12 lb
B
30 lb A
30o
A 12-lb block B rests as shown on
the upper surface of a 30-lb wedge
A. Neglecting friction, determine
immediately after the system is
released from rest (a) the acceleration
of B relative to A, (b) the acceleration
of B relative to A.
1. Kinematics: Examine the acceleration of the particles.
2. Kinetics: Draw a free body diagram showing the applied
forces and an equivalent force diagram showing the vector
ma or its components.
8 - 29
Problem 2
12 lb
B
30 lb A
30o
A 12-lb block B rests as shown on
the upper surface of a 30-lb wedge
A. Neglecting friction, determine
immediately after the system is
released from rest (a) the acceleration
of B relative to A, (b) the acceleration
of B relative to A.
3. Apply Newton’s second law: The relationship between the
forces acting on the particle, its mass and acceleration is given
by S F = m a . The vectors F and a can be expressed in terms of
either their rectangular components or their tangential and normal
components. Absolute acceleration (measured with respect to
a newtonian frame of reference) should be used.
8 - 30
Problem 2 Solution
12 lb
B
Kinematics.
30 lb A
30o
Since the wedge is constrained
to move on the inclined surface,
its acceleration aA is in the
direction of the slope.
The acceleration aB of the block
can be expressed as the sum of
the acceleration of A and the
acceleration of B relative to A.
aB = aA + aB/A
B
aB/A
aA
A
30o
aA
8 - 31
Problem 2 Solution
12 lb
B
Kinetics; draw a free body diagram.
Block B :
30 lb A
mB aB/A = 12 aB /A
32.2
12 lb
o
30
=
B
B
mA aA = 12 aA
32.2
N1
Block A :
N1
A
30 lb
=
N2
A
mA aA = 30 aA
32.28 - 32
Problem 2 Solution
Block A :
30o
N1
A
x
=
30 lb
+
Apply Newton’s second law.
N2
x
A
mA aA = 30 aA
32.2
SFx = ma: (N1 + 30) sin 30o = 30 aA
32.2
(1)
8 - 33
Block B :
12 lb
= 30
B
o
N1
mB aB/A = 12 aB /A
32.2
B
mA aA = 12 aA
32.2
+ SFy = may :
+
Problem 2 Solution
SFx = max:
12 - N1 = 12 aA sin 30o
32.2
12
0 = 32.2
aB /A - 12 aA cos 30o
32.2
(2)
(3)
Solving equations (1), (2), and (3) gives:
aA = 20.5 ft/s2
30o
aB /A = 17.75 ft/s2
8 - 34