1. ENGR-1100 Introduction to Engineering Analysis
Pawel Keblinski
Materials Science and Engineering MRC115
Office hours: Tuesday 1-3
phone: (518)

2. ENGR-1100 Introduction to Engineering Analysis
TA: Igor Bolotnov, Mechanical, Aerospace and Nuclear Engineering JEC5204
Office hours: Tuesday 3-5
phone: (518)

3. ENGR-1100 Introduction to Engineering Analysis
SI (Supplemental Instruction)
(Begins Wed. Sept. 8)
Sun. 8pm-10pm - DCC 330
Wed 8pm-10pm - Sage 5510
Drop-in Tutoring - DCC 345
(Begins Tue. Sept. 7th)
Sundays - 3-5pm
M-Th - 7-9pm

4. ENGR-1100 Introduction to Engineering Analysis
Lecture 3

5. Lecture Outline Rectangular components of a force
Resultant force by rectangular components

6. Rectangular Components of a Force
A force F can be resolved into a rectangular component Fx along the x-axis and a rectangular component Fy along the y-axis . The forces Fx and Fy are the vector components of the force F. Fy Fx x y F

7. F Fy Fx F = Fx + Fy= Fx i + Fy j q
The force F and its two dimensional vector components Fx and Fy can be written in Cartesian vector form by using unit vectors i and j.
F = Fx + Fy= Fx i + Fy j
F=| F |
Fx = F cos q F= Fx2+ Fy2
Fy = F sin q q=tan-1(Fy/Fx)

8. Example
Determine the x and y scalar component of the force shown in figure 2-47.
Express the force in Cartesian vector form.
F=275 lb x y
Figure 2-47
570

9. Solution F=275 lb Fy 570 Fx Fx=cos(570) *275=149.8 lb
Fy=sin(570) *275=230.6lb
F= Fxi+ Fyj=149.8 i j

10. Class Assignment: Exercise set 2-49
please submit to TA at the end of the lecture
Determine the x and y scalar components of the force shown in figure 2-49.
Express the force in Cartesian vector form. y
Solution
a) Fx=-440lb
Fy=-177.9lb
b) F=-440 i j lb x
220
F=475 lb
Figure 2-49

11. Three Dimension Rectangular Components of a Force
F = = Fx i + Fy j + Fz k
= F cos qx i + F cos qy j + F cos qz k = FeF
Where eF= cos qx i + cos qy j + cos qz k is a unit vector along the line of action of the force. x y z F
Fx=Fxi
Fz=Fzk
Fy=Fyj qx qy qz

12. The Scalar Components of a Force
Fx = F cos qx ; Fy = F cos qy ; Fz = F cos qz ;
qx=cos-1(Fx/F); qy=cos-1(Fy/F); qz=cos-1(Fz/F);
F= Fx2 + Fy2 + Fz2
cos2 qx +cos2 qy +cos2 qz=1
0< q <1800 x y z F Fz qx qy qz Fy Fx

13. Azimuth Angle x y z F x y z Fz = F sin f ; Fy= Fxy sin q=F cos f sin q
Fxy = F cos f; Fx
Fx= Fxy cos q=F cos f cos q
q- azimuth angle
f- elevation angle

14. Finding the direction of a force by two points along its line of actionx y z F B A xB xA yB yA zB zA
A(xA, yA, zA) ; B(xB, yB, zB)
cosqx=
xB-xA
(xB-xA)2+ (yB-yA)2+ (zB-zA)2
yB-yA
(xB-xA)2+ (yB-yA)2+ (zB-zA)2
cosqy=
zB-zA
(xB-xA)2+ (yB-yA)2+ (zB-zA)2
cosqz=

15. Example For the force shown in the figure
Determine the x, y, and z scalar components of the force.
Express the force in Cartesian form. x y z
F=745 N
370
300

16. Solution z F=745 N Fz Fxy y x Fz = F sin f = 745 sin(600)=411.4 N
370
300 Fz
Fz = F sin f = 745 sin(600)=411.4 N
Fxy = F cos f = 745 cos(600)=237.5 N f
Fxy

17. x y z Fz Fxy=237.5 N Fx Fy F=-142.9 i – 189.7 j + 411.4 k b)
370
Fxy=237.5 N q Fx Fy
Fx= Fxy cos q=-237.5*sin(370)= N
Fy= Fxy sin q=-237.5*cos(370)= N
Or if we follow the obtained formula:
Fx= Fxy cos q=237.5*cos(2330)= N
Fy= Fxy sin q=237.5*sin(2330)= N
F= i – j k b)

18. Class Assignment: Exercise set 2-55
please submit to TA at the end of the lecture
For the force shown in Fig. P2-58
Determine the x,y, and z scalar components of the force.
Express the force in Cartesian form. z
Solution
a) Fx=-583 lb
Fy=694lb
Fz=423 lb
b) F=-583 i +694 j k lb
F=1000 lb
250 y
1300 x

19. Resultant Force by Rectangular Componentsz x y A B
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) =
(Ax +Bx )i
+ (Ay +By ) j
+ (Az +Bz) k
Rx= (Ax +Bx );
Ry= (Ay +By );
Rz= (Az +Bz )

20. Rx= (Ax +Bx );
Ry= (Ay +By );
Rz= (Az +Bz )
R= Rx2 + Ry2 + Rz2
The magnitude:
The direction:
qx=cos-1(Rx/R); qy=cos-1(Ry/R); qz=cos-1(Rz/R);

21. Example
Determine the magnitude and direction of the resultant force of the following three forces.
F3= 600*cos(1200) i + 600*sin(1200)j+0 k
Solution:
F2=500*cos(2100) i +500*sin(2100)j+0 k
F1=350 i+ 0 j + 0 k
F1=350 i
F2= -433 i -250 j
F3= -300 i j
R=F1+F2+F3=-383 i j

22. R=F1+F2+F3=-383 i j
The force magnitude: R
The direction: qx
R= Fx2 + Fy2 + Fz2 =
R= N
qx=cos-1(Rx/R)
qx=cos-1(Rx/R)= cos-1(-383/468.4)=144.80
qx=144.80

23. Class Assignment: Exercise set 2-71
please submit to TA at the end of the lecture
Determine the magnitude R of the resultant of the forces and the angle qx between the line of action of the resultant and the x-axis, using the rectangular component method y
F1=600 lb
Solution:
R=1696 lb
F2=700 lb
450
150 x
300
F3=800 lb

24. The scalar (or dot) product
The scalar product of two intersecting vectors is defined as the product of the magnitudes of the vectors and the cosine of the angle between them q A B
A•B=B•A=AB cos(q)
0< q <1800
Finding the rectangular scalar component of vector A along the x-axis
Ax=A•i=A cos(qx)
Along any direction
An=A•en=A cos(qn) A n y q en et
At=A-An

25. The scalar product of the two vectors written in Cartesian form are:
A•B = (Ax i + Ay j + Az k) • (Bx i + By j + Bz k) =
Ax Bx (i•i) + Ax By (i•j) + Ax Bz (i•k)+
Ay Bx (j•i) + Ay By (j•j) + Ay Bz (j•k)+
Az Bx (k•i) + Az By (k•j) + Az Bz (k•k)
Since i, j, k are orthogonal:
i•j= j•k= k•j=(1)*(1)*cos(900)=0
i•i= j•j= k•k=(1)*(1)*cos(00)=1
Therefore: A•B = Ax Bx + Ay By + Az Bz

26. Determine the angle q between the following vectors:
Example
Determine the angle q between the following vectors:
A=3i +0j +4k and B=2i -2j +5k
A•B=AB cos(q)
cos(q)= A•B/ AB
A = = 5
B = = 5.74
AB= 28.7
A•B=3*2+0*(-2)+4*5=26
cos(q)= 26/28.7
q= 25.10

27. Orthogonal (perpendicular) vectors
(w1 ,w2 ,w3) z w
(v1 ,v2 ,v3) v y x
w and v are orthogonal if and only if w·v=0

28. Properties of the dot product
If u , v, and w are vectors in 2- or 3- space and k is a scalar, then:
u·v= v·u
u·(v+w)= u ·v + u·w
k(u·v)= (ku) ·v = u·(kv)
v·v> 0 if v=0, and v·v= 0 if v=0

29. Using dot product for a force systemx y z
F1 =300 lb
600
1.5 ft
6 ft
2 ft
4.5 ft
F2 =240 lb
Determine:
The magnitude and direction (qx, qy, qz)
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 a between force F1 and F2.

30. x y z
F1 =300 lb
600
1.5 ft
6 ft
2 ft
4.5 ft
F2 =240 lb
Solution
F1= F1 e1
e1 = 1.5/( )1/2 i + 6/( )1/2 j+
4.5/( )1/2 k
e1 = i j k
F1 = 58.8 i j k lb

31. F2= F2 e2 z L1=(22+1.52)1/2=2.5 ft L2 L2=2.5 tan(600)=4.33 ft y L1 x
F1 =300 lb
600
1.5 ft
6 ft
2 ft
4.5 ft
F2 =240 lb
L1=( )1/2=2.5 ft L2
L2=2.5 tan(600)=4.33 ft L1
F2= F2 e2
e2 = 1.5/(1.52+(-2) )1/2 i -2/(1.52+(-2) )1/2 j+
4.33 /(1.52+(-2) )1/2 k
e2 = 0.3 i j k
F2= 72 i - 96 j k lb

32. R= F1 + F2 = i j k lb
R= Rx2 + Ry2 + Rz2 = = 429 lb
R= 429 lb
qx=cos-1(Rx/R); qy=cos-1(Ry/R); qz=cos-1(Rz/R);
qx=cos-1(130.8/429)
qx=72.20
qy=cos-1(139.35/429)
qy=71.10
qz=cos-1(384.3/429)
qz=26.40

33. b) The magnitude of the rectangular component of the force F1
b) The magnitude of the rectangular component of the force F1 along the line of action of force F2.
F1•e2=(58.8 i j k)•(0.3 i j k)=
58.8* *(-0.4)+176.5*0.866=76 lb

34. c) The angle a between force F1 and F2.
F1•F2= F1 F2 cos(a)
cos(a)= F1•F2 / F1 F2
F1•F2= 58.8 i j k)•(72 i - 96 j k )=
58.8* *(-96)+176.5*207.8=18321 lb
F1 F2=72000
cos(a)=18321/72000
a =

35. Class assignment: Exercise set 2-63
Two forces are applied to an eye bolt as shown in Fig. P2-63.
Determine the x,y, and z scalar components of vector F1.
Express vector F1 in Cartesian vector form.
Determine the angle a between vectors F1 and F2. x y z
3 ft
4 ft
6 ft
F2=700 lb
F1=900 lb
Fig. P2-63

36. Solution +72 d1 = d1 = x12+ y12+ z12 (-6)2 +32 =9.7 ft
F2=700 lb
F1=900 lb
+72
d1 =
d1 = x12+ y12+ z12
(-6)2
+32
=9.7 ft
F1x = F1 cos(qx) =900 *{(–6)/9.7}=-557 lb
F1y = F1 cos(qy) =900 *{3/9.7}=278.5 lb
F1z = F1 cos(qz) =900 *{7/9.7}=649.8 lb
b) Express vector F1 in Cartesian vector form.
F1 = -557 i j k lb

37. c) Determine the angle a between vectors F1 and F2.
cos(a)= F1•e2 / F1 e2 x y z
3 ft
4 ft
6 ft
F2=700 lb
F1=900 lb
F1 = -557 i j k lb
d2 = x22+ y22+ z22
d2 =
(-6)2
+62
+32
=9 ft
F1•e2 = (-557)*(-2/3)+278.5*2/ *0.33=771.4lb
e2 =
-6/9 i
+ 6/9j
+ 3/9 k
= i j k
cos(a)=771.4/900
a=310

38. Class Assignment: Exercise set 2-78
please submit to TA at the end of the lecture
Determine the magnitude R of the resultant of the forces and the angles qx, qy, qz between the line of action of the resultant and the x-, y-, and z-coordinate axes, using the rectangular component method.
Solution:
R=28.6 kN
qx= 82.20
qy= 69.60
qz= 22.00