1. 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.

2. S F = ma . S F = L 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 .
S F = 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.

3. S Fx = max S Fy = may S Fz = maz
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 ax az
S Fx = max S Fy = may S Fz = maz x z
Using tangential and normal components, y an dv dt
S Ft = mat = m at P v2 r
S Fn = man = m O x
Using radial and transverse components, aq . .. ar
S Fr = mar= m(r - rq2) r P .. . q .
S Fq = maq = m(rq + 2rq) x O

4. 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 ), and the direction is always towards or away from the source .
Using these assumptions, we can derive an expression for the rotation frequency:
The result implies that under the action of a central force, the rotation frequency depends only on the radial distance away from the attractor.

5. Using this result, we can arrive at an expression for radial force purely in terms of the radial distance
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.
One example that we have discussed previously is the flat earth approximation. When is it valid?
baseball
where h Re
when
earth
where

6. Example Problem

7. Free Body Diagram for Block B

8. Free Body Diagram for Block A
Force on Block A due to substrate

9. Newton’s 1st law provides us with an expression that the force balance equals zero. (i.e.)
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
These 4 equations have 4 unknowns listed below, and the solution can be solved using linear algebra

10. Thus, the critical slipping threshold can be expressed as:
Since we’re only interested in Block B, let’s try a change in coordinates x y y’ x’
Thus, the critical slipping threshold can be expressed as:

11. Case 1. Assume that the acceleration applied to the block is zero
Therefore, the Block will slide under the action of gravity if the angle exceeds θ.
Case 2. Neglect Gravity
Note: This is the result derived in the book.

12. Case 3. Account for everything
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.

14. Given:
Constraints: Motion of Slider A is confined to the X-axis
Motion of Slider B is confined to the Y-Axis
Distance between Sliders A and B is constant
Free Body Diagrams:

15. Additional Constraints
In these two equations, the only thing we don’t know are the accelerations of sliders A and B, however they are interdependent
Thus, we really have 2 equations with 2 unknowns… ,

16. Note: The answer shown in the book assumes gravity is zero
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?

17. Problem 1 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 ms = 0.15 and mk = 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. P q A B

18. Problem 1
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 ms = 0.15 and mk = 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. P q A B
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.

19. Problem 1
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 ms = 0.15 and mk = 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. P q A B
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 = mk N.
The friction force on a surface when motion is impending
is F = ms N.

20. Problem 1
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 ms = 0.15 and mk = 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. P q A B
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.

21. aA = aB Problem 1 Solution Kinematics.
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. P q A B q A B aA aB
aA = aB

22. = = WA = mA g WA = (30 kg)(9.81 m/s2) WA = 294.3 N WB = mB g
Problem 1 Solution P q A B
Kinetics; draw a free body diagram.
Block A :
mA a = 30 a N T
250 N
WA= N
Fk = mk N =
WA = mA g
WA = (30 kg)(9.81 m/s2)
WA = N
WB = mB g
WB = (15 kg)(9.81 m/s2)
WB = N
Block B :
WB= N T N N’
Fk = mk N
F’k = mk N’
mB a = 15 a =

23. Problem 1 Solution
Block A :
WA= N T
mA a = 30 a
30o =
250 N N
Fk = mk N
Apply Newton’s second law.
+ SFy = 0: N - (294.3) cos 30o = N = N
Fk = mk N = 0.10 (254.9) = N
SFx = ma: (294.3) sin 30o T = 30 a
T = 30 a (1)

24. = + SFy = 0: N’ - N - (147.15) cos 30o = 0 N’ = 382.31 N
Problem 1 Solution
Block B :
WB= N T N N’
Fk = mk N
F’k = mk N’ =
30o
mB a = 15 a
+ SFy = 0: N’ - N - (147.15) cos 30o = N’ = N
F’k = mk N’ = 0.10 (382.31) = N
SFx = ma: T - Fk - F’k - (147.15) sin 30o = 15 a
T = 15 a (2)
Solving equations (1) and (2) gives:
T = 215 N a = 5.21 m/s2

25. Problem 1 Solution P q A B
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
Find P for impending motion.
For impending motion both blocks are in equilibrium:
Block A
Block B N T P
WA= N
Fm = ms N
30o
WB= N N’
F’m = ms N’

26. N = 254.87 N and N’ = 382.31 N, F’m = ms N’ = 0.15 (382.31) = 57.35 N
Problem 1 Solution N T P
WA= N
Fm = ms N
30o
WB= N N’
F’m = ms N’ A B
From + SFy = 0 find again
N = N and N’ = N,
and thus
Fm = ms N = 0.15 (254.87) = N
F’m = ms N’ = 0.15 (382.31) = N

27. + SFx = 0: P + (294.3) sin 30o - 38.23 - T = 0 (3)
Problem 1 Solution N T P
WA= N
Fm = ms N
30o
WB= N N’
F’m = ms N’ A B
For block A:
SFx = 0: P + (294.3) sin 30o T = (3)
For block B:
SFx = 0: T (147.15) sin 30o = (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.

28. Problem 2 12 lb 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. B
30 lb A
30o

29. Problem 2 A B
12 lb
30 lb
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.
30o
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.

30. Problem 2 A B
12 lb
30 lb
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.
30o
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.

31. Problem 2 Solution A B
12 lb
30 lb
30o
Kinematics.
aB/A B aA
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 A
30o aA

32. = = Problem 2 Solution 12 lb Kinetics; draw a free body diagram. B
Block B : B
12 lb N1
mB aB/A = aB /A 12
32.2
mA aA = aA = A
30 lb N1 N2
mA aA = aA 30
32.2
Block A : =

33. = + SFx = ma: (N1 + 30) sin 30o = aA (1) Problem 2 Solution Block A :
Apply Newton’s second law.
30o N1 A A = x N2 x 30
32.2
30 lb
mA aA = aA
SFx = ma: (N1 + 30) sin 30o = aA (1) 30
32.2

34. = + SFy = may : 12 - N1 = aA sin 30o (2)
Block B :
Problem 2 Solution 12
32.2
mB aB/A = aB /A
12 lb = B
30o B 12
32.2
mA aA = aA N1
+ SFy = may : N1 = aA sin 30o (2)
SFx = max: = aB /A aA cos 30o (3) 12
32.2
Solving equations (1), (2), and (3) gives:
aA = ft/s o aB /A = ft/s2