vector mechanics for engineers: dynamics - chapter 12 notes

7/23/2019 Vector Mechanics for Engineers: Dynamics - Chapter 12 notes

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Chapter

12

Vector M echanics for Engineers: Dynamics

F. Beer & Al

RVCC

Kinetics of Particles:

Newton’s Second Law

ENGR 133-51 - Engineering Mechanics II : Dynamics



12- 2

Contents

• Newton’s Second Law of

Motion• Linear Momentum of a

Particle

• System of Units

• Equations of Motion

• Dynamic Equilibrium

• Angular Momentum of a Particle

• Equation of Motion in Radial & TransverseComponents

• Conservation of Angular Momentum

• Newton’s Law of Gravitation



12- 3

Introduction

Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or continue to move in a straight line

(Statics Ch. 2)

Newton’s Second Law: If the resultant force acting on a particle is

not zero, then the particle will have an acceleration such that

am F

(Dynamics)

Recalling Newton’s Laws:

Newton’s Third Law: The forces of action and reaction between two

particles have the same magnitude and line of action with oppositesense (Statics Ch. 6)

M m

F -F



12- 4

Newton’s Second Law of Motion

• Newton’s Second Law: If the resul tant force acting on a

particle is not zero, the particle will have an acceleration

proportional to the magni tude of resul tant and in the

direction of the resul tant .

• Consider a particle subjected to constant forces,

m

a

F

a

F

a

F mass,constant

3

3

2

2

1

1

• When a particle of mass m is acted upon by a force

the acceleration of the particle must satisfy

, F

am F

• Acceleration must be evaluated with respect to a Newtonian frame of reference, i.e., one that is not

accelerating or rotating.

• If force acting on particle is zero, particle will not

accelerate (a =0), i.e., it will remain stationary or continue

on a straight line at constant velocity.



12- 5

Problem 1



12- 6

Systems of Units

• Of the units for the four primary dimensions (force,

mass, length, and time), three may be chosen arbitrarily.

The fourth must be compatible with Newton’s 2nd Law.

• International System of Units (SI Units): base units are

the units of length (m), mass (kg), and time (second).

The unit of force is derived,

22 s

mkg1

s

m1kg1 N1

• U.S. Customary Units: base units are the units of force(lb), length (m), and time (second). The unit of mass is

derived,

ft

slb1

sft1

lb1slug1

sft32.2

lb1lbm1

2

22


http://slidepdf.com/reader/full/vector-mechanics-for-engineers-dynamics-chapter-12-notes 7/2312- 7

Systems of Units

kg0.4536mass- pound1

kg14.59ft

slb1

slug1 Mass

N4.448lb1 Force

m0.3048ft1 Length

2

US CS → SI



Equations of Motion

• Newton’s second law provides

am F

• Solution for particle motion is facilitated by resolving

vector equation into scalar component equations, e.g.,

for rectangular components,

z m F ym F xm F

ma F ma F ma F

k a jaiamk F j F i F

z y x

z z y y x x

z y x z y x

FBD

(Forces)

Kinetic Diagram

(motion)



Sample Problem 12.1

A 200-lb block rests on a horizontal

plane. Find the magnitude of the force

P required to give the block an accelera-

tion or 10 ft/s2

to the right. The coef-ficient of kinetic friction between the

block and plane is m k 0.25.



Problem 3.a

12- 10

A 180-lb person stands in an elevator.

1) What is the apparent weight of the person while the elevator accelerate

upward with a constant value of 8 ft/s2? (FBD is the person)

2) What is the apparent weight if the elevator is accelerating downward with thesame value?

3) What is the tension in the rope of the elevator when it is moving downward?

(FBD is the person and the elevator)

4) For what value of acceleration of the elevator the person would appear

weightless? (FBD is the person)



Problem 3.b

Hints:

1) Deceleration is not constant!



Problem 4

Hints:

1) determine if A slips studying

static equilibrium on A (is the

force of friction necessary for

equilibrium exceeding themaximum frictional force?).

Only then you can consider if

Newton’s second law applies.

2) FBD for A with its own

coordinate system3) If finding that A slips, draw FBD

for B with its own coordinates

and apply also to B the dynamic

equilibrium equation

The two crates are released from rest. Their masses are mA = 40 kg and mB 30 kg, and the

coefficient of friction between crate A and the inclined surface are μs = 0.2 and μk = 0.15.

Are the two crates in static equilibrium? If not, what is the acceleration of the crates?



Sample Problem 12.4

The 12-lb block B starts from rest and

slides on the 30-lb wedge A, which is

supported by a horizontal surface.

Neglecting friction, determine (a) the

acceleration of the wedge, and (b) the

acceleration of the block relative to the

wedge.



Dynamic Equilibrium

• Alternate expression of Newton’s second law,

ector inertial vamam F

0

• With the inclusion of the inertial vector, the system

of forces acting on the particle is equivalent to

zero. The particle is in dynamic equil ibr ium .

• Methods developed for particles in static

equilibrium may be applied, e.g., coplanar forces

may be represented with a closed vector polygon.

• Inertia vectors are often called inertial forces as

they measure the resistance that particles offer to

changes in motion, i.e., changes in speed or

direction.



@Hibbeler

Acceleration



@Hibbeler

Deceleration



Equations of Motion

• Newton’s second law provides

am F

• For tangential and normal components,

2v

m F dt

dvm F

ma F ma F

nt

nnt t

Recalling that tangential component of

acceleration reflects change of speed and

normal component reflects change of direction.



Sample Problem 12.5

The bob of a 2-m pendulum describes

an arc of a circle in a vertical plane. If

the tension in the cord is 2.5 times the

weight of the bob for the positionshown, find the velocity and accel-

eration of the bob in that position.

A i l i ACB f 2 h h i C h i h d



Problem 7

Hints:- What is the equation you are going to use to find v…?

- To get there you will need to find ρ as a function of L

and angles (L = total length of rope), then

- Draw the FBDs. To optimize your equations, is

convenient to choose x , y with x coinciding with n

- Apply Newton’s Law

A single wire ACB of 2 m passes through a ring at C that is attached to

a sphere. The sphere revolves at a constant speed v in the horizontal

circle shown. Knowing that ϴ1=60o and ϴ2=30o and that the tension is

the same in both portions of the wire, determine the speed v .

A i il i ’ li i d i f f ff i



Problem 8A civil engineer’s preliminary design for a freeway off -ramp is

circular with radius R= 60m. If she assumes that the coefficient of

static friction between tires and road is at least μs=0.4, what is the

maximum speed at which vehicles can enter the ramp without loosing

traction?

Hints:

• The reason why we can turn following a curved path, is because of friction force

between the road and the tires (we cannot turn on a sheet of ice).

• This necessary force of friction is related to the normal acceleration of a circular motion

• Evaluating the maximum frictional force related to the static coefficient allows to

calculate the maximum velocity without slipping.

• Draw FBDs considering front view of the car



Sample Problem 12.6

Determine the rated speed of a

highway curve of radius = 400 ft

banked through an angle q = 18o.

[The rated speed of a banked highway

curve is the speed at which a car

should travel if no lateral friction force

is to be exerted at its wheels]



Problem 9

Hints:

1. The rated speed of a banked highway

curve is the speed at which a car should

travel if no lateral friction force is to be

exerted at its wheels.2. Consider that in a banked curve, the

normal acceleration is always horizontal.

3. Draw the FBD and analyze dynamic

equilibrium in a general case

(considering a frictional force that in (a)

is zero) with x // to the road.4. Specify each of the cases.

FBDF FBDK



Homework:

Solving Problems on Your Ownpage ~704

vector mechanics for engineers: dynamics - chapter 12 notes

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