Lecture 3Notes courtesy of: Prof. Yoav Peles

ENGR-1100 Introduction toEngineering Analysis

Lecture outline

• Forces as vectors• Resultant of two concurrent forces

• Resultant of three or more concurrentforces

• Resolution of force into components

Characteristics of a force

• A force is characterized by the following:

(1) Magnitude

(2) Direction

(3) Point of application

FB

FA

F

B

x

y

F

θ

Force representation

x

y

F6m

5m

x

y

z

θx

θy

θz

F

x

y

z

F

7 m

5 m

4 m

Two dimensions

Three dimensions

Principle of Transmissibility

• The external effect of a force on a rigidbody is the same for all points ofapplication of the force along its line ofaction (force is a sliding vector!)

Push PullLine of action

Forces classification

• Contact or surface forces• Example: push or pull

• Body forces• Example: gravitational forces, magnetic forces

Concurrent forces

• A force system is said to be concurrent ifthe action lines of all forces intersect at acommon point

Example(a) Determine the x and y scalar components of the force

shown in the figure.

(b) Express the force in Cartesian vector form.

F=275 lb

x

y

570

Solution

Fx = 275 * cos(570) = 149.8 lb

Fy = 275 * sin(570) = 230.6 lb

x

F=275 lby

570

Fy

Fx

F= Fxi+ Fyj=(149.8 i + 230.6 j) (lb)

3-D rectangular components of a force

x

y

z

F

Fx=Fxi

Fz=Fzk

Fy=Fyjθx θy

θz

F = Fx + Fy + Fz= Fx i + Fy j + Fz k

= F cos θx i + F cos θy j + F cos θz k = FlF

WherelF= cos θx i + cos θy j + cos θz k is a unitvector along the line of action of the force.

The scalar components of a forceFx = F cos θx ; Fy = F cos θy ; Fz = F cos θz ;

θx=cos-1(Fx/F); θy=cos-1(Fy/F); θz=cos-1(Fz/F);

F= Fx2 + Fy

2 + Fz2

x

y

z

F

Fx

Fyθx θy

θz

Fz cos2 θx +cos2 θy +cos2 θz=10< θ <1800

Azimuth angle

x

y

z

F

θ φ

θ- azimuth angleφ- elevation angle

θ

x

y

z

Fxy

Fz

Fxy

Fxy = F cos φ ;

Fz

Fz = F sin φ

Fx

Fx= Fxy cos θ =F cos φ cos θ

Fy

Fy= Fxy sin θ =F cos φ sin θ

Finding the direction of a force bytwo points along its line of action

x

y

z

FΒΑ

xB

xA

yB

yA

zBzA

cosθx=xB-xA

(xB-xA)2+ (yB-yA)2+ (zB-zA)2

cosθy=yB-yA

(xB-xA)2+ (yB-yA)2+ (zB-zA)2

zB-zA

(xB-xA)2+ (yB-yA)2+ (zB-zA)2cosθz=

Α (xA, yA, zA) ; Β (xB, yB, zB)

Example

For the force showna) Determine the x, y, and z scalar components of the force.b) Express the force in Rectangular form.

x

y

zF=475 N

370

300

Solution

x

y

zF=475 N

370

300

Fz

Fxy

Fz = F sin φ = F sin(600) =411.4 N

Fxy = F cos φ = F cos(600) =237.5 N

φ

F = (-142.9 i – 189.7 j + 411.4 k) Nb)

Fx

Or if we follow the obtained formula:Fx= Fxy cos θ =237.5*cos(2330 )= -142.9 NFy= Fxy sin θ =237.5*sin(2330) = -189.7 N

x

y

z

Fz

370

Fxy=237.5 N

θ

Fx= Fxy cos θ =−237.5*sin(370 )= -142.9 N

Fy= Fxy sin θ =−237.5*cos(370) = -189.7 N

Fy

Resultant by rectangularcomponents

A = Ax i + Ay j + Az k

B = Bx i + By j + Bz k

The sum of the two forces are:

R=A+B=(Ax i + Ay j + Az k) + (Bx i + By j + Bz k) =

Rx= (Ax +Bx )i; Ry= (Ay +By )j; Rz= (Az +Bz )k

z

x

y

A

B

(Ax +Bx )i+ (Ay +By ) j + (Az +Bz) k

The magnitude: R= Rx2 + Ry

2 + Rz2

The direction:

x=cos-1(Rx/R); y=cos-1(Ry/R); z=cos-1(Rz/R);

Rx= (Ax +Bx )i; Ry= (Ay +By )j; Rz= (Az +Bz )k

ExampleDetermine the magnitude and direction of theresultant of the following three forces.

Solution:

F1=350 i

F2=500*cos(2100) i +500*sin(2100)j

F3= 600*cos(1200) i + 600*sin(1200)j

F1=350 i

F2= -433 i -250 j

F3= -300 i + 519.6 j

R=F1+F2+F3=-383 i + 269.6 j

R=F1+F2+F3=-383 i + 269.6 j

The force magnitude:

x=cos-1(Rx/R)

R= Fx2 + Fy

2 =

R= 468.4 N

x=cos-1(Rx/R)= cos-1(-383/468.4)=144.80

x=144.80

R

The direction:

x3832 + 269.62

Example

x

y

z

F1 =300 lb

600

1.5 ft6 ft2 ft

4.5 ft

F2 =240 lb

Determine:

a) The magnitude and direction (x, y, z)

of the resultant force.

b) The magnitude of the rectangular

component of the force F1 along

the line of action of force F2.

c) The angle between force F1 and F2.

Solution

• F1= F1l1

l1 = 1.5/(1.52+62+4.52)1/2 i + 6/(1.52+62+4.52)1/2 j+4.5/(1.52+62+4.52)1/2 k

l1 = 0.196 i + 0.784 j+ 0.588 k

F1 = (58.8 i + 235.3 j+ 176.5 k) lb

x

y

z

F1 =300 lb

60

1.5 ft6 ft2 ft

4.5 ft

F2 =240 lb

L1=(22+1.52)1/2=2.5 ft

x

y

z

F1 =300 lb

600

1.5 ft6 ft2 ft

4.5 ft

F2 =240 lb

L1

L2

L2=2.5 tan(600)=4.33 ft

• F2= F2 e2

e2 = 1.5/(1.52+(-2)2+4.332)1/2 i -2/(1.52+(-2)2+4.332)1/2 j+4.33 /(1.52+(-2)2+4.332)1/2 k

e2 = 0.3 i - 0.4 j+ 0.866 k

F2= (72 i - 96 j+ 207.8 k) lb

R= F1 + F2 = 130.8 i + 139.35 j+ 384.3 k lb

R= Rx2 + Ry

2 + Rz2 = 130.82 + 139.352+384.32 = 429 lb

x=cos-1(Rx/R); y=cos-1(Ry/R);z=cos-1(Rz/R);

R= 429 lb

x=cos-1(130.8/429)

y=cos-1(139.35/429)

z=cos-1(384.3/429)

x=72.20

y=71.10

x=26.40