Chapt02 Lecture Updated

8/10/2019 Chapt02 Lecture Updated

http://slidepdf.com/reader/full/chapt02-lecture-updated 1/29

Chapter-2: Statics of Particles

(Algebraic) Operations on Vectors (Vector Algebra)

NegationAddition (Done in Chapter ‐1)SubtractionMultiplication of a vector by a scalar



2 - 2



Introduction

2 - 3

• The objective for the current chapter is to investigate the effects of forceson particles:

- replacing multiple forces acting on a particle with a single

equivalent or resultant force,- relations between forces acting on a particle that is in a

state of equilibrium .

• The focus on particles does not imply a restriction to miniscule bodies.Rather, the study is restricted to analyses in which the size and shape ofthe bodies is not significant so that all forces may be assumed to be

applied at a single point.



Resultant of Two Forces

2 - 4

• force: action of one body on another;characterized by its point of application ,magnitude , line of action , and sense .

• Experimental evidence shows that thecombined effect of two forces may berepresented by a single resultant force.

• The resultant is equivalent to the diagonal ofa parallelogram which contains the twoforces in adjacent legs.

• Force is a vector quantity.



Vectors

2 - 5

• Vector : parameter possessing magnitude and directionwhich add according to the parallelogram law. Examples:displacements, velocities, accelerations.

• Vector classifications:- Fixed or bound vectors have well defined points of

application that cannot be changed without affectingan analysis.

- Free vectors may be freely moved in space withoutchanging their effect on an analysis.

- Sliding vectors may be applied anywhere along theirline of action without affecting an analysis.

• Equal vectors have the same magnitude and direction.

• Negative vector of a given vector has the same magnitudeand the opposite direction.

• Scalar : parameter possessing magnitude but notdirection. Examples: mass, volume, temperature



Addition of Vectors

2 - 6

• Trapezoid rule for vector addition

• Triangle rule for vector addition

B

B

C

C

QP R BPQQP R

+=−+= cos2222

• Law of cosines,

• Law of sines,

AC

R B

Q A sinsinsin ==

• Vector addition is commutative,

PQQP +=+

• Vector subtraction



Addition of Vectors

2 - 7

• Addition of three or more vectors throughrepeated application of the triangle rule

• The polygon rule for the addition of three ormore vectors.

• Vector addition is associative,

S QPS QPS QP ++=++=++

• Multiplication of a vector by a scalar



Resultant of Several Concurrent Forces

2 - 8

• Concurrent forces : set of forces which all pass through the same point.

A set of concurrent forces applied to a particle may be replaced by a singleresultant force which is the vector sum of theapplied forces.

• Vector force components : two or more forcevectors which, together, have the same effectas a single force vector.



Sample Problem 1

2 - 9

The two forces act on a bolt at A. Determine their resultant.

SOLUTION:

• Graphical solution - construct a parallelogram with sides in the samedirection as P and Q and lengths in

proportion. Graphically evaluate theresultant which is equivalent in directionand proportional in magnitude to the thediagonal.

• Trigonometric solution - use the trianglerule for vector addition in conjunction

with the law of cosines and law of sinesto find the resultant.



Sample Problem 1

2 - 10

• Graphical solution - A parallelogram with sidesequal to P and Q is drawn to scale. Themagnitude and direction of the resultant or of

the diagonal to the parallelogram are measured,

°== 35 N98 α R

• Graphical solution - A triangle is drawn with Pand Q head-to-tail and to scale. The magnitudeand direction of the resultant or of the third side

of the triangle are measured,°== 35 N98 α R



Sample Problem 1

2 - 11

• Trigonometric solution - Apply the triangle rule.From the Law of Cosines,

( ) ( ) ( )( ) °−+=

−+=

155cos N60 N402 N60 N40

cos222

222 BPQQP R

A

A

R

Q B A

R B

Q A

+°=°=

°=

=

=

20

04.15 N73.97

N60155sin

sinsin

sinsin

α

N73.97= R

From the Law of Sines,

°= 04.35α



Sample Problem 2

2 - 12

a) the tension in each of the ropesfor α = 45 o,

b) the value ofα

for which thetension in rope 2 is a minimum.

A barge is pulled by two tugboats.

If the resultant of the forcesexerted by the tugboats is 5000 lbfdirected along the axis of the

barge, determine

SOLUTION:• Find a graphical solution by applying the

Parallelogram Rule for vector addition. The parallelogram has sides in the directions ofthe two ropes and a diagonal in the directionof the barge axis and length proportional to5000 lbf.

• The angle for minimum tension in rope 2 isdetermined by applying the Triangle Ruleand observing the effect of variations in α .

• Find a trigonometric solution by applyingthe Triangle Rule for vector addition. Withthe magnitude and direction of the resultantknown and the directions of the other two

sides parallel to the ropes, apply the Law ofSines to find the rope tensions.



Sample Problem 2

2 - 13

• Graphical solution - Parallelogram Rulewith known resultant direction andmagnitude, known directions for sides.

lbf 2600lbf 3700 21 ==

T T

• Trigonometric solution - Triangle Rulewith Law of Sines

°=

°=

° 105sinlbf 5000

30sin45sin21 T T

lbf 2590lbf 3660 21 == T T



Sample Problem 2

2 - 14

• The angle for minimum tension in rope 2 isdetermined by applying the Triangle Ruleand observing the effect of variations in α .

• The minimum tension in rope 2 occurs whenT 1 and T 2 are perpendicular.

( ) °=30sinlbf 50002T lbf 25002

=T

( ) °= 30coslbf 50001T lbf 43301 =T

°−°= 3090α °= 60α



Rectangular Components of a Force: Unit Vectors

2 - 15

• Vector components may be expressed as products ofthe unit vectors with the scalar magnitudes of thevector components.

F x and F y are referred to as the scalar components of

jF iF F y x +=

F

• May resolve a force vector into perpendicularcomponents so that the resulting parallelogram is arectangle. are referred to as rectangularvector components and

y x F F F +=

y x F F and

• Define perpendicular unit vectors which are parallel to the x and y axes.

ji and



Addition of Forces by Summing Components

2 - 16

S QP R ++=

• Wish to find the resultant of 3 or moreconcurrent forces,

( ) ( ) jS QPiS QP

jS iS jQiQ jPiP j Ri R

y y y x x x

y x y x y x y x+++++=

+++++=+

• Resolve each force into rectangular components

∑= ++= x

x x x xF S QP R

• The scalar components of the resultant are equalto the sum of the corresponding scalarcomponents of the given forces.

∑=++=

y y y y y

F S QP R

x y y x R

R R R R 122 tan −=+= θ

• To find the resultant magnitude and direction,



Sample Problem 3

2 - 17

Four forces act on bolt A as shown.Determine the resultant of the forceon the bolt.

SOLUTION:

• Resolve each force into rectangularcomponents.

• Calculate the magnitude and directionof the resultant.

• Determine the components of theresultant by adding the correspondingforce components.



Sample Problem 3

2 - 18

SOLUTION:• Resolve each force into rectangular components.

9.256.96100

0.1100110

2.754.27800.759.129150

4

3

21

−+

−

+−++

−−

F

F

F F

comp ycomp xmag force

22 3.141.199 += R N6.199= R

• Calculate the magnitude and direction.

N1.199 N3.14tan =α °= 1.4α

• Determine the components of the resultant by

adding the corresponding force components.

1.199+= x R 3.14+= y R



Equilibrium of a Particle

2 - 19

• When the resultant of all forces acting on a particle is zero, the particle isin equilibrium .

• Particle acted upon bytwo forces:- equal magnitude- same line of action- opposite sense

• Particle acted upon by three or more forces:- graphical solution yields a closed polygon- algebraic solution

00

0

==

==

∑∑∑

y x F F

F R

• Newton’s First Law : If the resultant force on a particle is zero, the particle willremain at rest or will continue at constant speed in a straight line.



Free ‐Body Diagrams

2 - 20

Space Diagram : A sketch showingthe physical conditions of the

problem.

Free-Body Diagram : A sketch showingonly the forces on the selected particle.



Sample Problem 4

2 - 21

In a ship-unloading operation, a3500-lb automobile is supported bya cable. A rope is tied to the cableand pulled to center the automobileover its intended position. What isthe tension in the rope?

SOLUTION:

• Construct a free-body diagram for the particle at the junction of the rope and

cable.

• Apply the conditions for equilibrium bycreating a closed polygon from the

forces applied to the particle.• Apply trigonometric relations to

determine the unknown forcemagnitudes.



Sample Problem 4

2 - 22

SOLUTION:

• Construct a free-body diagram for the particle at A.

• Apply the conditions for equilibrium.

• Solve for the unknown force magnitudes.

°=

°=

° 58sinlb3500

2sin120sin AC AB T T

lb3570= ABT

lb144= AC T



Sample Problem 5

2 - 23

It is desired to determine the drag forceat a given speed on a prototype sailboathull. A model is placed in a testchannel and three cables are used toalign its bow on the channel centerline.For a given speed, the tension is 40 lbin cable AB and 60 lb in cable AE .

Determine the drag force exerted on the

hull and the tension in cable AC .

SOLUTION:

• Choosing the hull as the free body,draw a free-body diagram.

• Express the condition for equilibriumfor the hull by writing that the sum ofall forces must be zero.

• Resolve the vector equilibriumequation into two componentequations. Solve for the two unknowncable tensions.



Sample Problem 5

2 - 24

SOLUTION:

• Choosing the hull as the free body, draw afree-body diagram.

°=

==

25.60

75.1ft4ft7

tan

α

α

°=

==

56.20

375.0ft4ft1.5

tan

β

β

• Express the condition for equilibriumfor the hull by writing that the sum of

all forces must be zero.0=+++= D AE AC AB F T T T R



Sample Problem 5

2 - 25

• Resolve the vector equilibrium equation intotwo component equations. Solve for the twounknown cable tensions.

( ) ( )( ) ( )

( )

( )

( ) jT

iF T

R

iF F

iT jT iT

jT iT T

ji jiT

AC

D AC

D D

AC AC

AC AC AC

AB

609363.084.19

3512.073.34

0

lb069363.03512.0

56.20cos56.20sin

lb84.19lb73.3426.60coslb4026.60sinlb40

−++

++−=

=

=

−=+=

°+°=

+−=°+°−=



Sample Problem5

2 - 26

( )( ) jT

iF T

R

AC

D AC

609363.084.19

3512.073.34

0

−++

++−=

=

This equation is satisfied only if each componentof the resultant is equal to zero

( )( ) 609363.084.1900

3512.073.3400−+==

++−==∑∑

AC y

D AC x

T F F T F

lb66.19lb9.42+=

+=

D AC

F

T



Rectangular Components in Space

2 - 27

• The vector iscontained in the

plane OBAC .

F • Resolve intohorizontal and vertical

components.

F

y y F F θ cos=

yh F F θ sin=

• Resolve intorectangular components

hF

φ θ

φ

φ θ φ

sinsin

sin

cossincos

y

h y

y

h x

F

F F

F F F

=

=

=

=




2 - 28

• With the angles between and the axes,F

( )

k jiF

k jiF

k F jF iF F

F F F F F F

z y x

z y x

z y x

z z y y x x

θ θ θ λ λ

θ θ θ

θ θ θ

coscoscos

coscoscos

coscoscos

++==

++=

++=

===

• is a unit vector along the line of action of

and are the directioncosines for

F

F

λ

z y x θ θ θ cosand ,cos,cos




2 - 29

Direction of the force is defined bythe location of two points,

( ) ( )222111 ,,and ,, z y x N z y x M

( )

d Fd F

d Fd F

d Fd F

k d jd id d

F F

z zd y yd x xd

k d jd id N M d

z z

y y

x x

z y x

z y x

z y x

===

++=

=

−=−=−=

++==

1

and joiningvector

121212

λ

λ

Documents

Chapt02 Lecture Updated