Chapter 2 Kinetics of a Particle:

Force and Acceleration

Dr. Md Tasyrif bin Abdul Rahman

Applied Mechanics Division

School of Mechatronic Engineering

Universiti Malaysia Perlis (UniMAP)

tasyrif@unimap.edu.my

v = vo + act

s = so + vot + act2

v2 = vo2 + 2ac(s-so)

2

1

2v

an

22

tn aaa

vat vdvdsat

Normal & Tangential

Components

Rectilinear motion

constant acceleration

Curvilinear motion

KINEMATICS

Summary for previous chapter

KINETICS

Newtons Method Work and

Energy Method

Impulse and

momentum

CHAPTER 13

2.1 Newtons Laws of Motion

Todays Objectives:

Students will be able to:

a) Write the equation of motion

for an accelerating body.

b) Draw the free-body and

kinetic diagrams for an

accelerating body.

In-Class Activities:

Applications

Newtons laws of motion

Newtons law of

gravitational attraction

Equation of motion for a

particle or system of particles

A particle originally at rest, will remain in rest

Newtons first law

SF=0

Newtons second law

A particle acted upon by an unbalanced force,

experiences acceleration that has the same

direction as the force and a magnitude that is directly

proportional to the force

SF=ma

STATICS

DYNAMICS

Mutual forces of action and reaction between two particles are

equal, opposite, and collinear. STATICS

Newtons third law

NEWTONS LAWS OF MOTION

APPLICATIONS

The motion of an object depends on the

forces acting on it.

Knowing the drag force, how can we

determine the acceleration or velocity of

the parachutist at any point in time?

A parachutist relies on the atmospheric

drag resistance force to limit his velocity.

APPLICATIONS (continued)

A freight elevator is lifted using a

motor attached to a cable and pulley

system as shown.

How can we determine the tension

force in the cable required to lift the

elevator at a given acceleration?

Is the tension force in the cable greater than the weight

of the elevator and its load?

NEWTONS LAW OF GRAVITATIONAL ATTRACTION

F = G(m1m2/r2)

where F = force of attraction between the two bodies,

G = universal constant of gravitation, 66.73(10-12) m3/kgs2

m1, m2 = mass of each body, and

r = distance between centers of the two bodies.

When near the surface of the earth, the only gravitational

force having any sizable magnitude is that between the earth

and the body. This force is called the weight of the body.

Any two particles or bodies have a mutually attractive

gravitational force acting between them. Newton postulated

the law governing this gravitational force as

MASS AND WEIGHT

It is important to understand the difference between the

mass and weight of a body!

Mass is an absolute property of a body. It is independent of

the gravitational field in which it is measured. The mass

provides a measure of the resistance of a body to a change

in velocity, as defined by Newtons second law of motion

(m = F/a).

The weight of a body is not absolute, since it depends on the

gravitational field in which it is measured. Weight is defined

as

W = mg

where g is the acceleration due to gravity.

UNITS: SI SYSTEM VS. FPS SYSTEM

SI system: In the SI system of units, mass is a base unit and

weight is a derived unit. Typically, mass is specified in

kilograms (kg), and weight is calculated from W = mg. If the

gravitational acceleration (g) is specified in units of m/s2, then

the weight is expressed in newtons (N). On the earths surface,

g can be taken as g = 9.81 m/s2.

W (N) = m (kg) g (m/s2) => N = kgm/s2

FPS System: In the FPS system of units, weight is a base unit

and mass is a derived unit. Weight is typically specified in

pounds (lb), and mass is calculated from m = W/g. If g is

specified in units of ft/s2, then the mass is expressed in slugs.

On the earths surface, g is approximately 32.2 ft/s2.

m (slugs) = W (lb)/g (ft/s2) => slug = lbs2/ft

2.2 The Equation of Motion

The motion of a particle is governed by Newtons second law, relating

the unbalanced forces on a particle to its acceleration. If more than one

force acts on the particle, the equation of motion can be written

F = FR = ma

where FR is the resultant force, which is a vector summation of all the

forces.

To illustrate the equation, consider a

particle acted on by two forces.

First, draw the particles free-

body diagram, showing all

forces acting on the particle.

Next, draw the kinetic diagram,

showing the inertial force ma

acting in the same direction as

the resultant force FR.

KEY POINTS

1) Newtons second law is a Law of Nature--experimentally

proven and not the result of an analytical proof.

2) Mass (property of an object) is a measure of the resistance

to a change in velocity of the object.

3) Weight (a force) depends on the local gravitational field.

Calculating the weight of an object is an application of

F = ma, i.e., W = m g.

4) Unbalanced forces cause the acceleration of objects. This

condition is fundamental to all dynamics problems.

PROCEDURE FOR THE APPLICATION OF THE

EQUATION OF MOTION

1) Select a convenient inertial coordinate system. Rectangular,

normal/tangential, or cylindrical coordinates may be used.

2) Draw a free-body diagram showing all external forces

applied to the particle. Resolve forces into their

appropriate components.

3) Draw the kinetic diagram, showing the particles inertial

force, ma. Resolve this vector into its appropriate

components.

4) Apply the equations of motion in their scalar component

form and solve these equations for the unknowns.

2.4 Equations of Motion: Rectangular Coordinates

Todays Objectives:

Students will be able to apply

Newtons second law to

determine forces and

accelerations for particles in

rectilinear motion.

In-Class Activities:

Applications

Equations of motion using

rectangular (Cartesian)

Coordinates

EQUATION OF MOTION

The equation of motion, F = m a, is best used when the problem

requires finding forces (especially forces perpendicular to the

path), accelerations, velocities or mass. Remember, unbalanced

forces cause acceleration!

Three scalar equations can be written from this vector equation.

The equation of motion, being a vector equation, may be

expressed in terms of its three components in the Cartesian

(rectangular) coordinate system as

F = ma or Fx i + Fy j + Fz k = m(ax i + ay j + az k)

or, as scalar equations, Fx = max , Fy = may , and Fz = maz .

PROCEDURE FOR ANALYSIS

Free Body Diagram

Establish your coordinate system and draw the particles

free body diagram showing only external forces. These

external forces usually include the weight, normal forces,

friction forces, and applied forces. Show the ma vector

(sometimes called the inertial force) on a separate diagram.

Make sure any friction forces act opposite to the direction

of motion! If the particle is connected to an elastic spring,

a spring force equal to ks should be included on the FBD.

PROCEDURE FOR ANALYSIS (continued)

Equations of Motion

If the forces can be resolved directly from the free-body

diagram (often the case in 2-D problems), use the scalar

form of the equation of motion. In more complex cases

(usually 3-D), a Cartesian vector is written for every force

and a vector analysis is often best.

A Cartesian vector formulation of the second law is

F = ma or

Fx i + Fy j + Fz k = m(ax i + ay j + az k)

Three scalar equations can be written from this vector equation.

You may only need two equations if the motion is in 2-D.

PROCEDURE FOR ANALYSIS (continued)

The second law only provides solutions for forces and

accelerations. If velocity or position have to be found,

kinematics equations are used once the acceleration is

found from the equation of motion.

Kinematics

Any of the tools learned in Chapter 1 may be needed to

solve a problem. Make sure you use consistent positive

coordinate directions as used in the equation of motion

part of the problem!

EXAMPLE 1

Given: A crate of mass m is pulled by a cable attached to a truck.

The coefficient of kinetic friction between the crate and

road is mk.

Find: Draw the free-body and kinetic diagrams of the crate.

Plan: 1) Define an inertial coordinate system.

2) Draw the crates free-body diagram, showing all

external forces applied to the crate in the proper

directions.

3) Draw the crates kinetic diagram, showing the inertial

force vector ma in the proper direction.

EXAMPLE (continued)

1) An inertial x-y frame can be defined as fixed to the ground.

Solution:

3) Draw the kinetic diagram of the crate:

The crate will be pulled to the right. The

acceleration vector can be directed to the

right if the truck is speeding up or to the

left if it is slowing down.

2) Draw the free-body diagram of the crate:

The weight force (W) acts through the

crates center of mass. T is the tension

force in the cable. The normal force (N)

is perpendicular to the surface. The

friction force (F = uKN) acts in a direction

opposite to the motion of the crate.

y

x

W = mg T

30

N F = uKN

ma

Free body diagram Kinetics diagram

Example of Free body diagram Kinetics diagram

CONCEPT QUIZ

1. The block (mass = m) is moving upward with a speed v. Draw the FBD if the kinetic friction coefficient is mk.

A) B)

C) D) None of the above. mg

mkmg

N

mg

N

mkN

N

mg

mkN

v

CONCEPT QUIZ (continued)

2. Packaging for oranges is tested using a machine that exerts ay = 20 m/s

2 and ax = 3 m/s2, simultaneously.

Select the correct FBD and kinetic diagram for this condition.

A) B)

C) D)

=

may

max

W

Ry

Rx

= max

W

Ry

Rx

=

may

max

W

Ry

=

may

Ry

y

x

GROUP PROBLEM SOLVING

Given: The block and cylinder have a mass of m. The coefficient of kinetic friction at all surfaces of contact is . Block A is moving to the right.

Find: Draw the free-body and kinetic diagrams of each block.

Plan: 1) Define an inertial coordinate system. 2) Draw the free-body diagrams for each block, showing all

external forces. 3) Draw the kinetic diagrams for each block, showing the

inertial forces.

GROUP PROBLEM SOLVING (continued) Solution:

The friction force opposes the motion of block A relative to the surfaces on which it slides.

1) An inertial x-y frame can be defined as fixed to the ground.

2) Draw the free-body diagram of each block:

3) Draw the kinetic diagram of each block:

y

x

Block A:

T

NA FfA = NA

WA = mg y

x

2T

WB = mg

Block B:

Block A: maA

Block B:

maB

EXAMPLE 2

The crate has a mass of 50 kg. If the crate is subjected to a 400[N] towing

force as shown, determine the velocity of the crate in 3[s] starting from rest.

ms= 0.5, mk= 0.3,

Equations of Motion :

Kinematics : The acceleration is constant, P is constant

The crate has a mass of 80 kg. If the magnitude of P is

increased until the crate begins to slide, determine the

crates initial acceleration. ms= 0.5, mk= 0.3,

EXAMPLE 3

20o

T

80(9.81) = 784.8[N]

N

Ff = 0.5N

ms= 0.5 mk= 0.3

[solution]

: verge of slipping

: impending motion Equations of equilibrium :

SFx=0 ;

SFy=0 ;

Tcos20o 0.5N = 0 ..(i)

N + Psin20o 784.8 = 0 ...(ii)

T = 353.29 [N] , N = 663.79 [N]

20o

353.29 [N]

784.8 [N]

N

Ff = 0.3N

a

[solution]

Equations of Motion :

SFx=max ;

SFy=may ; N 784.8 + 353.29sin20o = 80(0)

N = 663.97 [N] 353.29cos20o 0.3(663.97) = 80a

a = 1.66 [m/s2]

EXAMPLE 4

Given: The motor winds in the cable with a constant acceleration such that the 20-kg crate moves a distance s = 6 m in 3 s, starting from rest. mk = 0.3.

1) Draw the free-body and kinetic diagrams of the crate.

2) Using a kinematic equation, determine the acceleration of the crate.

3) Apply the equation of motion to determine the cable tension.

Find: The tension developed in the cable.

Plan:

EXAMPLE 4 (continued)

1) Draw the free-body and kinetic diagrams of the crate:

Solution:

Since the motion is up the incline, rotate the x-y axes so the x-axis aligns with the incline. Then, motion occurs only in the x-direction.

There is a friction force acting between the surface and the crate. Why is it in the direction shown on the FBD?

= 30

x

y 20 a W = 20 g T

N

Fk= 0.3 N

EXAMPLE (continued) 2) Using kinematic equation s = v0 t + a t

2

6 = (0) 3 + a (32) a = 1.333 m/s2

3) Apply the equations of motion

+ Fy = 0 -20 g (cos30) + N = 0

=> N = 169.9 N

+ Fx = m a T 20g(sin30) 0.3 N = 20 a

T = 20 (981) (sin30) + 0.3(169.9) + 20 (1.333)

T = 176 N

= 30

x

y 20 a W = 20 g T

N

Fk= 0.3 N

Since both forces and velocity are involved, this problem requires both kinematics and the equation of motion.

1) Draw the free-body and kinetic diagrams of the car.

2) Apply the equation of motion to determine the acceleration.

3) Using a kinematic equation, determine velocity & distance.

GROUP PROBLEM SOLVING

Given: Ore cars mass is 400 kg, vo = 2 m/s.

The force in the cable is F = (3200 t2) N, where t is in seconds.

Plan:

Find: v and s when t = 2 s.

GROUP PROBLEM SOLVING (continued)

Solution:

1) Free-body and kinetic diagrams of the mine car:

W = 400 g T

N

x y

8

15

17

400 a

Note that the mine car is moving along the x-axis.

2) Apply the equation of motion

+ Fx = m a T 400g (8/17) = 400 a 3200 t2 400 (9.81) (8/17) = 400 a a = (8 t2 4.616) m/s2

=

GROUP PROBLEM SOLVING (continued)

t a v v c o

2 c o o t (1/2)a t v s s

) s - (s 2a ) (v v o c 2

o 2

The two blocks A and B having mass of 10 kg and 30 kg respectively,

that shown in Figure 5 are originally at rest. Neglect the masses of the

pulleys and the effect of friction in the pulleys. When the blocks are

released, determine:

(a) Acceleration of each block.

(b) Tension of the cable.

Quiz 3

2.5 Equations of Motion: Normal and Tangential Coordinates

Todays Objectives:

Students will be able to apply

the equation of motion using

normal and tangential

coordinates. In-Class Activities:

Applications

Equation of motion in n-t

coordinates

APPLICATIONS

Race tracks are often banked in the

turns to reduce the frictional forces

required to keep the cars from sliding

at high speeds.

If the cars maximum velocity and a

minimum coefficient of friction

between the tires and track are

specified, how can we determine the

minimum banking angle (q) required

to prevent the car from sliding?

APPLICATIONS (continued)

Satellites are held in orbit around

the earth by using the earths

gravitational pull as the centripetal

force the force acting to change

the direction of the satellites

velocity.

Knowing the radius of orbit of

the satellite, how can we

determine the required speed of

the satellite to maintain this orbit?

NORMAL & TANGENTIAL COORDINATES

When a particle moves along a

curved path, it may be more

convenient to write the equation of

motion in terms of normal and

tangential coordinates.

The normal direction (n) always points toward the paths

center of curvature. In a circle, the center of curvature is the

center of the circle.

The tangential direction (t) is tangent to the path, usually set

as positive in the direction of motion of the particle.

EQUATIONS OF MOTION

This vector equation will be satisfied provided the individual

components on each side of the equation are equal, resulting in

the two scalar equations: Ft = mat and Fn = man .

Here Ft & Fn are the sums of the force components acting in

the t & n directions, respectively.

Since the equation of motion is a

vector equation , F = ma,

it may be written in terms of the

n & t coordinates as

Ftut + Fnun = mat + man

Since there is no motion in the binormal (b) direction, we can also

write Fb = 0.

NORMAL AND TANGENTIAL ACCERLERATIONS

The tangential acceleration, at = dv/dt, represents the time rate of

change in the magnitude of the velocity. Depending on the direction

of Ft, the particles speed will either be increasing or decreasing.

The normal acceleration, an = v2/, represents the time rate of change

in the direction of the velocity vector.

Remember,

an always acts toward the paths center of curvature. Thus, Fn will

always be directed toward the center of the path.

Recall, if the path of motion is defined

as y = f(x), the radius of curvature at

any point can be obtained from

= [1 + ( )2]3/2

dy

dx

d2y

dx2

SOLVING PROBLEMS WITH n-t COORDINATES

Use n-t coordinates when a particle is moving along a

known, curved path.

Establish the n-t coordinate system on the particle.

Draw free-body and kinetic diagrams of the particle. The

normal acceleration (an) always acts inward (the positive n-

direction). The tangential acceleration (at) may act in either the

positive or negative t direction.

Apply the equations of motion in scalar form and solve.

It may be necessary to employ the kinematic

relations:

at = dv/dt = v dv/ds an = v2/

At the instant q = 60o, the boys center of mass

is momentarily at rest.

Determine the speed and the tension in each of the

two supporting cords of the swing when q = 90o.

The boy has a weight of 300 N ( 30 kg).

Neglect his size and the mass of the seat and cords

Example

W

n

t

2T q

Free-body diagram

mat

n

t

man

Kinetic diagram

At q = 60o, v = 0 At q = 90o, v = ?

T = ?

Equations of Motion :

SFn=man ;

SFt=mat ;

2T Wsinq = man 2T 300sinq = 30(v2/3) (i)

an = v2/

= v2/3

Wcosq = mat 300cosq = 30at 10cosq = at (ii)

n

mat

t

man

W t

2T q

n

2T 300sinq = 30(v2/3) (i)

10cosq = at (ii)

Kinematics : (to relate at and v)

vdv = at ds

vdv = 10cosq dq v = 2.68 [m/s]

s = q ds = dq

v

0

90

60

d)cos10(vdv qq

Solving, we get T = 186 [N]

The speed of the boy at q = 90o

Homework during semester break

1. Download tutorial 1 from portal and try to solve it. Submit as

assignment 5. (14 April 2015) #Lecturer Notes : School of Mechatronic Engineering : Semester 2 Sidang

Akademik 2014/2015 : ENT142 Engineering Dynamics : Ex-Dip

:Tutorial

2. Solve below problems and submit as assignment 6

(14 April 2015);

Text book: Dynamics (R.C. Hibbeler-13th edition)

Chapter 2)

1. 13-15

2. 13-21

3. 13-22

4. 13-37

5. 13-72

3. Mid Term Examination

Date: 16th April 2015 (thursday)

Time: 8.00pm (1hour 15mnts)

Venue: to be decided