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

2. Mechanics -> Static
Mechanics of
rigid bodies
Mechanics of
deformable bodies
Mechanics of
fluids
Static
Kinematics
Kinetics

3. Lecture outline Newton’s law. Law of gravitation.
Basic concept of physical problem solving.
Fundamentals of vectors

4. Newton’s Law of Motion Law #1: Body is in equilibrium SFi=0
Body will remain in rest or continue to move with same speed and direction unless unbalanced force is acted on the body or particle.

5. Newton’s Law of Motion F=ma Law #2: Body not in equilibrium SFi=0
Change of motion of a body is proportional to the net force imposed on the body in the direction of the net force
Where: F is the external force acting on the body.
m is the mass of the body.
a is the acceleration of the body in the direction of the force.
F=ma

6. Newton’s Law of Motion
Law #3: The forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense.

7. Law of gravitation M.m F=G F
Two bodies of mass M and m are mutually attracted to each other with equal and opposite forces F and –F of magnitude F given by the formula:
M.m r2
F=G
Where r is the distance between the center of mass of the two bodies. And G is the universal gravitational constant
G=3.439(10-8)ft3/(slug*s2) in the U.S customary system of units.
G=6.673(10-11)m3/(kg*s2) in SI system of units M m r F

8. Mass and weight Me.m F=G =mg Me
The mass m of a body is an absolute quantity.
The weight W of a body is the gravitational attraction exerted on the body by the earth or by another massive body such as another planet.
At the surface of the earth:
Me.m
re2
F=G
=mg
Where: Me is the mass of the earth.
re is the mean radius of the earth Me
re2
g=G
At sea level and latitude 450
g=32.17 ft/s2 = m/s2

9. Example 1
Determine the gravitational force, in kilonewtons, exerted by the earth on the moon. M m r
Me.mm r2
F=G
G= 6.673* m3/(kg*s2)
Me=5.976*1024 kg
mm=7.35*1022 kg
r=3.844*108 m
F=1.98*1017 kN

10. Example 2
At what distance from the surface of the earth, in kilometers, is the weight of the body equal to one-half of its weight on the earth’s surface? W2
Me.m
r12
W1=G
On earth
Me.m
r22
W2=G
Above earth W1 D r1
earth
The requirement: W2=1/2W1
W1/W2= r22 /r12=2
r2=√2 r1
D=r2-r1= (√2-1) r1= (√2-1)*6370 km = 2638km

11. Class assignment: Exercise set 1-1, 1-2, 1-17.
please submit to TA at the end of the lecture
1-1) Calculate the mass m of a body that weighs 600 lb at the surface of the earth.
1-2) Calculate the weight W of a body at the surface of the earth if it has a mass m of 675 kg.
1-17) If a man weight 210 lb at sea level,
determine the weight W of the man
a) At the top of Mt. Everest (29,028 ft
above sea level)
b) In a satellite at an altitude of 200 mi.
Solution:
1-1) M=18.65 slug
1-2) W=6.62 kN
1-17) a=209 lb
b=190.3 lb

12. Units of measurement
The U.S customary system of units (the British gravitational system)
Base units are foot (ft) for length, pound (lb) for force, and second (s) for time.
Pound is defined as the weight at sea level and altitude of 450 of a platinum standard.
The international system of units (SI)
Three class of units
(1) base units
(2) supplementary units.
(3)derived units.

13. Base units Supplementary units Derived units Quantity unit Symbol
Length
meter m
Mass
kilogram kg
time
second s
Supplementary units
Quantity
unit
Symbol
Plane angle
radian
rad
Solid angle
steradian sr
Derived units
Quantity
unit
Symbol
Area
Square meter m2
Volume
Cubic meter m3
Linear velocity
Meter per second
m/s

14. SI/U.S. customary units conversion
Quantity
U.S. customary to SI
SI to U.S. customary
Length
1 ft = m
1 m = ft
Velocity
1 ft/s = m/s
1 m/s = ft/s
Mass
1 slug = kg
1 kg = slug

15. Dimensional homogeneity
Equation does not depend on the units of measurement.
F=mg is valid whether the force is measured in newtons or lb and the mass in slugs, kilograms or grams, provided g is measured in the same units of length, time and mass as m and F.
The F=9.8m is not dimensional homogeneous since the equation applies only if the length is measured in meters and the time in second.

16. Method of problem solving
Statement of the problem
Identification of physical principles
Sketch and tabulation
Mathematical modeling
Solution
Checking

17. Class assignment: Exercise set 1-25
please submit to TA at the end of the lecture
Determine the weight W, in U.S. Customary units, of an 85-kg steel bar under standard conditions (sea level at a latitude of 45 degrees).
Solution:
W=187.4 lb

18. Scalar and vectors
A scalar quantity is completely described with only a magnitude (a number).
- Examples: mass, density, length, speed, time, temperature.
A vector quantity has both magnitude and direction and obeys the parallelogram law of addition.
- Examples: force, moment, velocity, acceleration.

19. Vector B A Terminal point Initial point Direction of arrow
direction of vector
Length of arrow
magnitude of vector

20. The sum of two vectors – geometrical representation
Two vectors can be vectorially added using the parallelogram law. R F1 F2
Position vector F1 so that its initial point coincides with the terminal point of F2. The vector F1+F2 is represented by the vector R.

21. Vectors in rectangular coordinate systems- two dimensionalx y
(v1,v2) v
(v1,v2) are the terminal point of vector v
V = Vx i + Vy j

22. The sum of two vectors – analytic representation (two dimensional )x y
(v1+w1,v2+w2) v2 w1
(w1,w2) w2 w v
(v1,v2) v1
v+w=(v1+w1,v2+w2)
v +w = (v1 + w1 )i + (v2 + w2 ) j

23. The sum of two vectors – rectangular components (Three dimensional )z x y
(a1,a2,a3)
(b1,b2,b3) a b
a+b=(a1 +b1,a2+b2, a3 +b3)
a +b = (a1 + b1 )i + (a2 + b2 ) j+ (a3 + b3 ) k

24. Vectors with initial point not on the origin
P1(x1 ,y1 ,z1) y z v w x
P2(x2 ,y2 ,z2)

25. Example
Find the components of the vector having initial point P1 and terminal point P2
P1(-1,0,2), P2(0,-1,0)
Solution:
v= (0+1,-1-0,0-2)=(1,-1,-2)

26. Class assignment: Exercise set 3.1-3a-3c,3e,6a

27. Vector arithmetic
If u,v,w are vectors in 2- or 3-space and k and l are scalar, then the following relationship holds:
u+v=v+u
u+0=0+u=u
k(lu)=(kl)u
(k+l)u=ku+lu
(u+v)+w=u+(v+w)
u+(-u)=0
k(u+v)= ku+ kv
1u=u

28. The norm of a vector (the length of a vector)x y
(v1,v2) v v2 v1
From the Theorem of Pythagoras it follows that the length of a vector is :
v = v12 + v22

29. Three dimensional spacez
(a1,a2,a3) a a3 y a1 a2 x
a = a12 + a22 + a32

30. Example
Let u=(2,-2,3), v=(1,-3,4), w=(3,6,-4). Evaluate the following expression:
3u-5v+w
Solution
3u-5v+w =3(2,-2,3)-5(1,-3,4)+(3,6,-4)
=(6,-6,9)-(5,-15,20)+(3,6,-4)=
3u-5v+w =(4,15,-25)
3u-5v+w = (-15)2 =19.9

31. Class assignment: Exercise set 3.2-3e

32. Unit vector
Any vector can be written as a product of its magnitude and a unit vector in the direction of the given vector.
A vector F in the positive n direction can be written as follows:
F=|F|en=Fen
Where en is the unit vector and can be written as follows:
en= = i j k
F Fx Fy Fz
F F F F

33. Class assignment:
Find the magnitude of any unit vector

34. Hanchen Stopped here on Aug 28, 2004

35. Right-hand system
The Cartesian coordinates axes are arranged as a right-hand system.

36. Multiplication of Cartesian vectors
The scalar (or dot) product.
The vector (or cross) product.

37. 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

38. 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

39. 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

40. 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

41. 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

42. 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

43. 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

44. 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

45. Characteristics of a Force
A force is characterizes by the following:
(1) Magnitude,
(2) Direction,
(3) Point of application. F B F B F A

46. Force Representation Two dimensions Three dimensions y y F F x x x y zq x 5m x
Three dimensions x y z qx qy qz F x y z F
7 m
5 m
4 m

47. Principle of Transmissibility
The external effect of a force on a rigid body is the same for all points of application of the force along its line of action.
Push
Pull
Line of action

48. Forces Classification
Contacting or surface forces
For example: push or pull
Body forces
For example: gravitational forces, magnetic forces.

49. Concurrent forces
A force system is said to be concurrent if the action lines of all forces intersect at a common point

50. Two Concurrent Forces
Any two concurrent forces F1 and F2 acting on a body can be replaced by a single force, called the resultant R.
F1+F2=R
The two forces can be vectorially added using the parallelogram law. F1 R F2

51. F2 R F1 a b c The law of sines: Sin a sin b sin g = =
= =
b = sin-1(F2 sin f/R)
The law of cosines:
c2 = a2 + b2 - 2ab cos g
R2 = F12 + F22 - 2F1F2 cos g
R2 = F12 + F22 + 2F1F2 cos f

52. Example
300
450
A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboat is a 5000 lb force directed along the axis of the barge, determine the tension in the ropes.

53. Solution T1 T2 T2 T1 T1 T2 5000 sin 450 sin 300 sin 1050 = =
5000 lb
450
300
1050
450
300
5000 lb T2 T1
T T
sin 450 sin sin 1050
= =
T1=3660 lb T2=2590 lb

54. Example – 2-16
Determine the magnitude of the resultant R and the angle q between the x-axis and the line of action on the resultant for the following problem:

55. f tan(l)=65/120 l=28.440 tan(d)=70/120 d=30.260 f=180-l-d=121.30 Using
We get
R2 = F12 + F22 + 2F1F2 cos f
R2 = *825*740 cos(121.30)
R= 771 N

56. b = sin-1(F2 sin f/R) Using We get b = sin-1(740sin(121.3)/771)=55.140l q
b = sin-1(F2 sin f/R)
Using
We get
b = sin-1(740sin(121.3)/771)=55.140
q=b+l= =83.580

57. Class Assignment: Exercise set 2-13
please submit to TA at the end of the lecture
Use the law of sines and the law of cosines to determine the magnitude of the resultant R and the angle q between the x-axis and the line of action of the resultant.
You could use :
b = sin-1(F2 sin f/R)
R2 = F12 + F22 + 2F1F2 cos f
Solution
R= lb
q=46.230

58. Solution
600 lb
500 lb f l g b R
Using the cosine equation: R2 = F12 + F22 + 2F1F2 cos f
tan(l)=2/5
tan(g)=4/1
l=21.80
g=75.960
f=g-l=54.16
R2 = *600*500 cos(54.160)
R= lb

59. b q R f l g b R b = sin-1(F2 sin f/R) Using We get
500 lb b
600 lb
980.5 lb q R
600 lb
500 lb f l g b R
b = sin-1(F2 sin f/R)
Using
We get
b = sin-1(500sin(54.16)/980.5)
b=24.420
q=l+b= =46.230

60. Three or More Concurrent Forces
The previous two concurrent forces can be extended to cover three or more forces F1 +F2 +…+Fn =R F1 F2 F3 F4
F34
F12
F12 F3 F4 R
F12
F34

61. Example
Use the laws of sines and cosines to determine the magnitude of the resultant R and the angle q between the x-axis and the line of action of the resultant.

62. Solution
250 lb
consider first two forces
(the 250 lb and 350 lb forces ) b
350 lb
f12 a g
Using the cosine equation: R122 = F12 + F22 + 2F1F2 cos f12
tan(a)=1/3
tan(g)=4/3
a=18.40
g=53.130
f12=g-a=34.730
R122 = *350*250 cos(34.730)
R12= lb

63. b b f a q12 g b = sin-1(F2 sin f12/R) Using We get
250 lb
250 lb lb b
350 lb b f a
q12
350 lb g
b = sin-1(F2 sin f12/R)
Using
We get
b = sin-1(250sin(34.73)/573.4)
b=14.380
q12=a+b= =32.780

64. Using the cosine equation: R2 = R122 + F32 + 2R12F3 cos f12,3lb
500 lb
32.780
450
f12,3
Using the cosine equation: R2 = R122 + F32 + 2R12F3 cos f12,3
f12,3= =77.780
R2 = *573.4 *500cos(77.780)
R= 837 lb

65. Solution: q123 f12,3 b3 b3 = sin-1(F3 sin f12,3/R) Using We getlb
q123
f12,3
450 b3
500 lb
b3 = sin-1(F3 sin f12,3/R)
Using
We get
b3 = sin-1(573.4sin(77.78)/837)
b3 =42.030
q123= =2.970
Solution:
R= 837 lb

66. Class Assignment: Exercise set 2-22
please submit to TA at the end of the lecture
Use the law of sines and the law of cosines to determine the magnitude of the resultant R and the angle q between the x-axis and the line of action of the resultant.
Solution:
R=468; q=35.1