1. Chapter 13 : Kinetics of A Particle – Force and acceleration

2. Newton’s Law of Motion
First Law: A particle originally at rest, or moving in a straight line with a constant velocity, will remain in this state provided the particle is not subjected to an unbalanced force.
Second Law: A particle acted upon by an unbalanced force F experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force.

3. Newton’s Law of Motion
Third Law: The mutual forces of action and reaction between two particles are equal, opposite and collinear.
Equation of motion: F = ma
Newton’s Law of Gravitational Attraction. A law governing the mutual attractive gravitational force acting between them.

4. Newton’s Law of Motion
Mass and Weight. Mass is a property of matter by which we can compare the response of one body with that of another. It is an absolute quantity since the measurement can be made at any location. Weight of a body is not absolute since it is measured in a gravitational field, hence its magnitude depends on the location measured.

5. Newton’s Law of Motion
SI System of Units. The mass of the body is specified in kilograms and the weight must be calculated using the equation of motion, F = ma
W = mg (N)
(g = 9.81 m/s2)

6. The Equation of Motion The equation of motion may be written as
Consider particle P of mass m and subjected to the action of two forces, F1 and F2.

7. The Equation of Motion
From the free body diagram, the resultant of these forces produces the vector ma, it magnitude and direction can be represented graphically on the kinetic diagram.

8. The Equation of Motion
Note that if FR = ΣF = 0, then acceleration is zero, so that the particle will either remain at rest or moves along a straight line with a constant velocity.
Such a condition is called static equilibrium, Newton’s First Law of Motion

9. Equation of Motion for a System of Particles
Consider a system of n particles isolated within an enclosed region in space.
Arbitrary ith particle having a mass of mi is subjected to a system of internal forces which resultant force is represented by fi and a resultant external force Fi.

10. Equation of Motion for a System of Particles
The free body diagram for the ith particle are shown. Applying equation of motion yields
ΣF = ma; Fi + fi = miai
If equation of motion is applied to each of the other particles, these equations can be added together vectorially,
ΣFi + Σ fi = Σmiai

11. Equation of Motion for a System of Particles
Since internal forces between particles all occur in equal but opposite collinear pairs, the summation of these internal forces will equal zero.
ΣFi = Σmiai
If rG is a position vector which locates the center of mass G of the particles, then mrG = Σmiri where m = Σmi is the total mass of all the particles
Differentiating twice w.r.t time yields
maG = Σmiai

12. Equation of Motion for a System of Particles
Therefore,
ΣF = maG
The sum of the external forces acting on the system of particles is equal to the total mass of the particles times the acceleration of its center of mass G.

13. Equation of Motion: Rectangular Coordinates
When a particle is moving relative to an inertial x, y, z frame of reference, the forces acting on the particle, and its acceleration may be expressed in term of their i, j, k components
ΣF = ma
ΣFxi + ΣFyj + ΣFzk = m(axi + ayj + azk)

14. Equation of Motion: Rectangular Coordinates
We may write the following three scalar equations:

15. Example 13.1
The 50-kg crate rests on a horizontal plane for which the coefficient of kinetic friction is μk = 0.3. If the crate is subjected to a 400-N towing force, determine the velocity of the crate in 3 s starting from rest.

16. Example 13.1
Free-Body Diagram. The weight of the crate is W = mg = 50 (9.81) = N. The frictional force has a magnitude F = μkNC and acts to the left, since it opposes the motion of the crate. The acceleration a is assumed to act horizontally, in the positive x direction. There are 2 unknowns, namely NC and a.

17. Example 13.1
Equations of Motion.
Solving for the two equations yields

18. Example 13.1
Kinematics. Acceleration is constant, since the applied force P is constant. Initial velocity is zero, the velocity of the crate in 3 s is

19. Equations of Motion: Normal and Tangential Coordinates
When a particle moves over a curved path which is known, the equation of motion for the particle may be written in the tangential, normal and binormal directions. We have
ΣF = ma
ΣFtut + ΣFnun + ΣFbub = mat +man
Here ΣFt, ΣFn, ΣFb represent the sums of all the force components acting on the particle in the tangential, normal and binormal directions.

20. Equations of Motion: Normal and Tangential Coordinates
Since the particle is constrained to move along the path, there is no motion in the binormal direction

21. Equations of Motion: Normal and Tangential Coordinates
at (=dv/dt) represents the time rate of change in the magnitude of velocity
Therefore if ΣFt acts in the direction of motion, the particle’s speed will increase. If it acts in the opposite direction, the particle will slow down.
an (=v2/ρ) represents the time rate of change in the velocity’s direction.

22. Equations of Motion: Normal and Tangential Coordinates
Since this vector always acts in the positive n direction, i.e. toward the path’s center of curvature, then ΣFn, which causes an, also act in this direction.

23. Equations of Motion: Cylindrical Coordinates
When all forces acting on a particle are resolved into cylindrical components, i.e. along the unit-vector directions ur, uθ, uz, the equation of motion may be expressed as
ΣF = ma
ΣFrur + ΣFθuθ + ΣFzuz = marur +maθuθ+mazuz

24. Equations of Motion: Cylindrical Coordinates
We may write the following three scalar equations of motion:

25. Equations of Motion: Cylindrical Coordinates
Tangential and Normal Forces.
Determination of the resultant force components ΣFr, ΣFθ, ΣFz causing a particle to move with a known acceleration.
If acceleration is not specified at given instant, directions or magnitudes of the forces acting on the particle must be known or computed to solve.
Consider the force P that causes the particle to move along a path r = f(θ)

26. Equations of Motion: Cylindrical Coordinates
The normal force N which the path exerts on the particle is always perpendicular to the tangent of the path.
Frictional force F always acts along the tangent in the opposite direction of motion.

27. Equations of Motion: Cylindrical Coordinates
The directions of N and F can be specified relative to the radial coordinate by using the angle ψ, which is defined between the extended radial line and the tangent to the curve.

28. Equations of Motion: Cylindrical Coordinates
If ψ is positive, it is measured from the extended radial line to the tangent in a CCW sense or in the positive direction θ
If it is negative, it is measured in the opposite direction to positive θ

29. PROCEDURE FOR ANALYSIS
Free-Body Diagram
Establish the r, θ, z inertial coordinate system and draw the particle’s free body diagram.
Assume that ar, aθ, az act in the positive directions of r, θ, z if they are unknown.
Identify all the unknowns in the problem.

30. PROCEDURE FOR ANALYSIS
Equations of Motion
Apply the equations of motion

31. PROCEDURE FOR ANALYSIS
Kinematics
Determine r and the time derivatives and then evaluate the acceleration components
If any of the acceleration components is computed as a negative quantity, it indicates that is acts in it negative coordinate direction.

32. Example 13.10
The 2-kg block moves on a smooth horizontal track such that its path is specfied in polar coordinates by the parametric equations r = (3t2) m and θ = (0.5t) rad where t is in seconds. Determine the magnitude of the tangential force F causing the motion at the instant t = 1 s.

33. Example 13.10
Free-Body Diagram. The normal force N, and the tangential force F are located at an angle from the r and θ axes. By expressing r = f(θ), we yield r = 12θ2. When t = 1 s, θ = 0.5 rad.

34. Example 13.10
Because is a positive quantity, it is measured counterclockwise from the r axis to the tangent (same direction as θ). There are four unknowns: F, N, ar and aθ

35. Example 13.10
Equations of Motion.
Kinematics.

36. Example 13.10
Substituting into the two equations of motion and solving,
F = N
N = 9.22 N