1. Engineering Mechanics
د. عبد الرزاق طوقان

2. Introduction
Mechanics is a physical science concerned with the behavior of bodies subjected to forces. To study mechanics we need:
Input: decompose problem into components
know degree of components
draw a mathematical model
Processing: define laws governing mechanics
provide details needed to solve problem
design methodology to solve problem (manual, computer)
Output: deliver solution in a nice way
develop capability to solve problems (intuition)
enhance intuition to values (dean)

3. Decomposition Bodies are Forces are Rigid Flexible Fluids Static
Dynamics
Kinematics
Kinetics

4. Degree of components (hierarchy of creation)

5. Define laws governing mechanics (versus theories)
constitutive relationships
equilibrium equations
compatibility equations
Theories:
Based on:
Assumptions based on:
Available knowledge is constrained with:
True dynamics is

6. Details needed: definitions
Space: is the geometric region occupied by bodies
A particle: is a body of negligible dimensions
Rigid body: a body whose relative movement between its parts is negligible
Mass:
Quantity of matter in a body (or measure of the inertia of a body (resistance to change of motion)).
Force: the action of one body on another
Time: is the measure of the succession of events

7. Details needed to solve problem: System of Units
Base units are units of mass, length and time
QUANTITY
SI UNITS
U.S (ENGLISH UNITS)
Mass (M)
Kilogram (kg)
Slug (slug)
Length (L)
Meter (m)
Foot (ft)
Time (T)
Second (s)
Force (F)
Newton* (N)
Pound (Ib)
(1ft = m, 1lb = N. 1slug = 1lb s²/ft = kg)
*Derived unit N = (1 kg)(1 m/ s²)
1 Newton is the force required to give a mass of 1kg an acceleration of 1m/ s²
The weight of 1 kg Mass is: W = mg = (1kg)(9.81m/ s²) = 9.81 N

8. Details needed to solve problem: System of Units: SI Prefixes
When a numerical quantity is either very large or very small, the units used may be modified by using prefix
EXPONENTIAL FORM
PREFIX
SI SYMBOL
Multiple
109
giga G
106
mega M
103
kilo k
Sub-multiple
10-3
milli m
10-6
micro μ
10-9
nano n

9. Design methodology for solving problems
Input
Problem to be solved
Physics of problem
Mathematical model
Processing
Propose theory
Formulate equations
Solve equations
Output
1. Verify laws
2. Build engineering sense
3. Start a new cycle

10. develop capability to solve problems
Only by working many problems can you truly understand the basic principles and how to apply them (muscles cannot be built by reading hundreds of books without practice!!!)

11. End of introduction
Let Learning Continue

12. RECTILINEAR KINEMATICS: 1D MOTION
CHAPTER 12
RECTILINEAR KINEMATICS: 1D MOTION
Objectives:
-find the kinematic quantities of a particle traveling along a straight path.

13. APPLICATIONS
The motion of large objects, such as rockets …, can often be analyzed as if they were particles.
Why?

14. Mechanism definition Motion of rigid bodies are
Translational:
Rotational
General
Definition of a particle

15. POSITION AND DISPLACEMENT
Define rectilinear motion
The position relative to the origin, O, is defined by r, or s, units?
The displacement is change in position.
Vector :  r = r’ - r
Scalar :  s = s’ - s
The total distance, sT, is …

16. VELOCITY
Velocity is... It is a vector quantity, its magnitude is called speed, units m/s
The average velocity is
vavg = r/t
The instantaneous velocity is
v = dr/dt
Speed is … v = ds/dt
Average speed is (vsp)avg = sT/  t

17. ACCELERATION Acceleration is …, its units are … Vector form: a = dv/dt
Scalar form: a = dv/dt = d2s/dt2
We can also express: a ds = v dv

18. ò ò Thus • Differentiate position to get velocity and acceleration.
v = ds/dt ; a = dv/dt or a = v dv/ds
• Integrate acceleration for velocity and position.
Velocity: ò = t o v dt a dv s ds or ò = t o s dt v ds
Position:
• so and vo are the initial position and velocity at t = 0.

19. CONSTANT ACCELERATION
(particular case)
Constant acceleration (e.g. gravity ac =g= m/s2 ), then t a v c o + =
yields ò dt dv 2 s
(1/2)a ds ) - (s 2a (v

20. EXAMPLE
Given: A motorcyclist travels along a straight road at a speed of 27 m/s. When the brakes are applied, the motorcycle decelerates at a rate of -6t m/s2.
Find: The distance the motorcycle travels before it stops.

21. ò ò EXAMPLE (solution) 1) Determine the velocity.
a = dv / dt => dv = a dt
=> v – vo = -3t2 => v = -3t2 + vo ò - = t o v dt dv ) 6 (
2) Find time to stop (v = 0). Use vo = 27 m/s.
0 = -3t => t = 3 s
3) Calculate distance using so = 0:
v = ds / dt => ds = v dt =>
=> s – so = -t3 + vot
=> s – 0 = -(3)3 + (27)(3) => s = 54 m ò + - = t o s dt v ds ) 3 ( 2

22. GROUP PROBLEM SOLVING
Given: Ball A is released from rest at a height of 12m at the same time that ball B is thrown upward, 1.5m from the ground. The balls pass one another at a height of 6m
Find: The speed at which ball B was thrown upward.

23. GROUP PROBLEM SOLVING (solution)
1) Time required for ball A to drop to 6m.
t = 1.1s
2) Ball B: sBo = 1.5m, a=…, s=…
(vB)o = 9.5m/s

24. End of
Let Learning Continue

25. ERRATIC MOTION Objectives:
Determine position, velocity, and acceleration using graphs.

26. APPLICATION
Having a v-s graph, can we determine a at s = 300m?
How?

27. GRAPHING
-Better to handle complex motions difficult to describe with formulas.
-Graphs provide a visual description of motion.
-Graphs are true meaning of dynamics

28. S-T GRAPH
V-t: Find slope of s-t (v = ds/dt) at various points.

29. V-T GRAPH
a-t: Find slope of v-t (a = dv/dt) at various points.

30. A-T GRAPH v-t is the area under the a-t curve.
We need initial velocity of the particle.

31. A-S GRAPH
Area under a-s curve= one half difference in the squares of the speed (recall ∫a ds = ∫ v dv ).
½ (v1² – vo²) = ò s2 s1
a ds
We need initial velocity

32. V-S GRAPH
Knowing velocity v and the slope (dv/ds) at a point, acceleration is.
a = v (dv/ds)

33. EXAMPLE Given: v-t graph for a train moving between two stations
Find: a-t graph and s-t graph

34. EXAMPLE (Solution:) a0-30 = dv/dt = 40/30 = 4/3 m/s2 a30-90 = 0
a = -4/3 m/s2 4 -4 3
a(m/s2)
t(s)

35. EXAMPLE (continued)

36. GROUP PROBLEM SOLVING Given: The v-t graph shown
Find: The a-t graph, average speed,
and distance traveled for the 30 s interval

37. GROUP PROBLEM SOLVING
a(m/s²)
t(s)

38. GROUP PROBLEM SOLVING (continued)
Ds0-10 = ò v dt = = 400/3 m
Ds10-30 = ò v dt = =
s0-30 = = m
vavg(0-30) = total distance / time

39. End of 12.3
Let Learning Continue

40. CURVILINEAR MOTION: RECTANGULAR COMPONENTS:
2D MOTION
Objectives:
a) Describe the motion of a particle traveling along a curved path.
b) Relate kinematic quantities in terms of the rectangular components of the vectors.

41. Methodology Understand previous 1D
Perform analogical solutions between 1D and 2D models
Build up experience with 2D models
2D Models:
Rectangular: 3 distances
Cylindrical: 2 distances + 1 angle
Spherical: distance + 2 angles

42. APPLICATIONS
Motion of a plane can be tracked with radar relative to a point recorded as a function of time. How to determine v and a?
A car travels down a fixed, helical path at a constant speed. How to determine v and a? To design the track, is it important to predict a?

43. POSITION AND DISPLACEMENT
Define a curvilinear motion?
A particle moves along a curve defined by the path function, s.
The position is designated by r = r(t).
the displacement is Δr = r’ - r

44. VELOCITY Average velocity is: vavg = Δ r/ Δ t .
Instantaneous velocity is: v = dr/dt .
v is always tangent to the path
What is speed v? Δ s → Δ r as t→0, then v = ds/dt.

45. ACCELERATION Average acceleration is: aavg = Dv/Dt = (v’ - v)/Dt
Instantaneous acceleration is: a = dv/dt = d2r/dt2
A hodograph (also named velocity diagram) is a plot of the locus of points defining the arrowhead of velocity vectors. It is used in physics, astronomy and fluid mechanics.
Acceleration is tangent to the hodograph.
Is it tangent to the path function?

46. RECTANGULAR COMPONENTS: POSITION
The position can be defined as
r = x i + y j + z k
where
x = x(t), y = y(t), and z = z(t) .
Magnitude is: r = (x2 + y2 + z2)0.5
Direction is defined by the unit vector: ur = (1/r)r

47. RECTANGULAR COMPONENTS: VELOCITY
Velocity vector is :
v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt
Since i, j, k are constants then?
Magnitude is
v = [(vx)2 + (vy)2 + (vz)2]0.5
Direction is tangent to the path.

48. RECTANG. COMPONENTS: ACCELERATION
Acceleration vector is:
a = dv/dt = d2r/dt2 = axi + ayj + azk
where ax = = = dvx /dt, ay = = = dvy /dt,
az = = = dvz /dt vx x vy y vz z • ••
Magnitude is a = [(ax)2 + (ay)2 + (az)2 ]0.5
What about direction of a?

49. EXAMPLE Given: The motion of two particles (A and B) is
rA = [3t i + 9t(2 – t) j] m
rB = [3(t2 –2t +2) i + 3(t – 2) j] m
Find: The point at which the particles collide and their speeds just before the collision.

50. EXAMPLE (Solution:) - Collision: rA = rB, so xA = xB and yA = yB .
- x-components: 3t = 3(t2 – 2t + 2)
t2 – 3t + 2 = 0 => t = 2 or 1 s
- y-components: 9t(2 – t) = 3(t – 2)
3t2 – 5t – 2 = => t = 2 or – 1/3 s
- So, t = 2s. Substituting yields: xA = xB = 6 m, yA = yB = 0

51. EXAMPLE (continued) Velocity vectors: . .
vA = drA/dt = = [3i + (18 – 18t)j] m/s
At t = 2 s: vA = [3i – 18j] m/s . xA i + yA j • •
vB = drB/dt =xBi + yBj = [(6t – 6)i + 3j] m/s
At t = 2 s: vB = [6i + 3j] m/s
Speed is vA = ( ) = 18.2 m/s
vB = ( ) = 6.71 m/s

52. GROUP PROBLEM SOLVING
Given: A particle travels along a path y = 0.5x2. The x-component of velocity is vx = (5t) m/s.
When t = 0, x = y = 0.
Find: The particle’s distance from the origin and the magnitude of its acceleration when t = 1 s.

53. ò GROUP PROBLEM SOLVING (Solution) 1) x-components:
Velocity: vx = x = dx/dt =
Position: =
Acceleration: ax = x = vx = •• ò x dx t •
2) y-components:
Position: y =
Velocity: vy = dy/dt =
Acceleration: ay = vy = •

54. GROUP PROBLEM SOLVING (continued)
3) The displacement from the origin is:
r = x i + y j =
At t = 1 s, r =
Distance: d = r =
The magnitude of the acceleration is:
Acceleration vector: a =
Magnitude: a = = 37.8 m/s2

55. End of
Let Learning Continue

56. MOTION OF A PROJECTILE (Section 12.6)
Objectives:
Analyze the free-flight motion of a projectile.

57. APPLICATIONS
At what θ and vo, the ball must be kicked to make a field goal? And to get the maximum distance?
What is the maximum h a fireman can project water to, and at what θ ?

58. CONCEPT OF PROJECTILE MOTION
Projectile motion is a two rectilinear motions: horizontal with zero acceleration and vertical with gravity acceleration.
Consider two balls: the red falls from rest, the yellow is given a horizontal velocity. Each picture is taken after the same time interval. Why both balls remain at the same elevation at any instant? What about the horizontal distance between successive photos of the yellow ball? Explain!

59. KINEMATIC EQUATIONS: HORIZONTAL MOTION
No air resistance is assumed!
Since ax = 0, vox remains constant and the position in the x direction can be determined by:
x = xo + (vox)(t)

60. KINEMATIC EQUATIONS:
VERTICAL MOTION
Since the positive y-axis is directed upward, ay = -g. Application of the constant acceleration equations yields:
vy = voy – g(t)
y = yo + (voy)(t) – ½g(t)2
vy2 = voy2 – 2g(y – yo)
For any given problem, only two of these three equations can be used. Why?

61. ( ) ( ) ( ) Example 1 Given: vo and θ
Find: The equation that defines y as a function of x.
Solution: Using vx = vo cos θ and vy = vo sin θ
We can write: x = (vo cos θ)t or
y = (vo sin θ)t – ½ g(t)2
t = x
vo cos θ
y = (vo sin θ)
x g x
vo cos θ 2 vo cos θ – 2
( ) ( ) ( )
By substituting for t:

62. Example 1 (continued):
Simplifying the last equation, we get:
y = (x tanθ) –
g x2
2vo2
(1 + tan2θ)
( )
The above equation describes the path of a particle in projectile motion. The equation shows that the path is parabolic.
Remark: a drop of a particle from 9m height is equivalent to a car moving with 47km/hr velocity if it hits a wall!

63. Example 2: Given: Snowmobile is going 15 m/s at point A.
Find: The distance it travels (R) and the time in the air.

64. GROUP PROBLEM SOLVING
Given: Skier leaves the ramp at θA = 25o and hits the slope at B.
Find: The skier’s initial speed vA.

65. GROUP PROBLEM SOLVING (continued)
Solution:
Motion in x-direction:
Motion in y-direction:
vA = m/s

66. End of 12.6
Let Learning Continue

67. NORMAL AND TANGENTIAL COMPONENTS (Section 12.7)
CURVILINEAR MOTION:
NORMAL AND TANGENTIAL COMPONENTS (Section 12.7)
Objectives:
Determine the normal and tangential components of velocity and acceleration of a particle traveling along a curved path.

68. APPLICATIONS
Cars experience an acceleration due to change in velocity (magnitude and\or direction)
Why would you care about the total acceleration of the car?
If the motorcycle starts from rest and increases its speed at a constant rate, how can we determine its velocity and acceleration at the top of the hill?

69. NORMAL AND TANGENTIAL COMPONENTS
Normal (n) and tangential (t) coordinates are used when a particle moves along a curved path and the path of motion is known
The origin is located on the particle
t-axis is tangent to path and +ve in direction of motion, n-axis is perpendicular to the t-axis and +ve toward the center of curvature of the curve.

70. NORMAL AND TANGENTIAL COMPONENTS
(continued)
n and t directions are defined by the unit vectors un and ut, respectively.
Radius of curvature, ρ, is the perpendicular distance from the curve to the center of curvature at that point.
The position is the distance, s, along the curve from a fixed reference point.

71. THE n-t COORDINATE SYSTEM
VELOCITY IN
THE n-t COORDINATE SYSTEM
The velocity vector is always tangent to the path of motion (t-direction).
The magnitude is determined by taking the time derivative of the path function, s(t).
v = vut where v = s = ds/dt .

72. THE n-t COORDINATE SYSTEM
ACCELERATION IN
THE n-t COORDINATE SYSTEM
Acceleration is
a = dv/dt = d(vut)/dt = vut + vut .
Here v represents the change in the magnitude of velocity and ut change in the direction of ut. . .
a = vut + (v2/r)un = atut + anun.
The acceleration vector can be expressed as:

73. ACCELERATION IN THE n-t COORDINATE SYSTEM (continued)
The acceleration vector is
a = at ut + an un
• The tangential component is tangent to the curve and in the direction of increasing or decreasing velocity.
at = v or at ds = v dv .
• The normal or centripetal component is always directed toward the center of curvature of the curve. an = v2/r

74. SPECIAL CASES OF MOTION
There are some special cases of motion to consider.
1) The particle moves along a straight line.
r  => an = v2/r = => a = at = v .
The tangential component represents the time rate of change in the magnitude of the velocity.
2) The particle moves along a curve at constant speed.
at = v = => a = an = v2/r .
The normal component represents the time rate of change in the direction of the velocity.

75. SPECIAL CASES OF MOTION (continued)
3)The tangential component of acceleration is constant, at =(at)c.
s = so + vot + (1/2)(at)ct2
v = vo + (at)ct
v2 = (vo)2 + 2(at)c(s – so)
4) The particle moves along a path expressed as y = f(x).
The radius of curvature, r, at any point on the path can be calculated from
r = ________________
]3/2
(dy/dx)2 1 [ + 2
d2y/dx

76. EXAMPLE PROBLEM
Given: Starting from rest, a motorboat travels around a circular path of r = 50 m at a speed v = (0.2 t2) m/s.
Find: The magnitudes of the boat’s velocity and acceleration at the instant t = 3 s.

77. EXAMPLE (Solution)
The velocity vector: v = v ut , v = (0.2t2) m/s.
At t = 3s: v = 0.2t2 = 0.2(3)2 = 1.8 m/s
2) The acceleration vector: a = atut + anun = vut + (v2/r)un. .
Tangential component: at = v = d(.2t2)/dt = 0.4t m/s2
At t = 3s: at = 0.4t = 0.4(3) = 1.2 m/s2 .
Normal component: an = v2/r = (0.2t2)2/(r) m/s2
At t = 3s: an = [(0.2)(32)]2/(50) = m/s2
Magnitude is: a = [(at)2 + (an)2]0.5 = [(1.2)2 + (0.0648)2]0.5 = 1.20 m/s2

78. GROUP PROBLEM SOLVING
Given: A jet plane travels along a vertical parabolic path defined by the equation y = 0.4x2. At point A, the jet has a speed of 200 m/s, which is increasing at the rate of 0.8 m/s2.
Find: The magnitude of the plane’s acceleration when it is at point A.

79. GROUP PROBLEM SOLVING (Solution)
1) The tangential component of acceleration is .
2) Determine the radius of curvature at point A (x = 5 km):
3) The normal component of acceleration is
4) The magnitude of the acceleration vector is
= m/s2

80. End of 12.7
Let Learning Continue

81. ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES (Section 12.9)
Objectives:
To relate the positions, velocities, and accelerations of particles undergoing dependent motion.

82. APPLICATIONS
The cable and pulley modify the speed of B relative to the speed of the motor. It is important to relate the various motions to determine the power requirements for the motor and the tension in the cable.
If the speed of the A is known, how can we determine the speed of block B?
Note: always start from a fixed datum.

83. APPLICATIONS (continued)
Rope and pulley arrangements are used to assist in lifting heavy objects. The total lifting force required from the truck depends on the acceleration of the cabinet A.
How can we determine the acceleration and velocity of A if the acceleration of B is known?

84. COMPATABILITY EQUATIONS
DEPENDENT MOTION
COMPATABILITY EQUATIONS
In many kinematics problems, the motion of one object will depend on the motion of another.
The blocks are connected by an inextensible cord wrapped around a pulley. If block A moves downward, block B will move up.
sA and sB define motion of blocks. Each starts from a fixed point, positive in the direction of motion of block.

85. DEPENDENT MOTION (continued)
sA and sB are defined from the center of the pulley to blocks A and B.
If the cord has a fixed length, then: sA + lCD + sB = lT
lT is total cord length and lCD is the length of cord passing over arc CD on the pulley.

86. DEPENDENT MOTION (continued)
Velocities can be found by differentiating the position equation. Since lCD and lT remain constant, so dlCD/dt = dlT/dt = 0
dsA/dt + dsB/dt = 0 => vB = -vA
The negative sign indicates that as A moves down (positive sA), B moves up (negative sB direction).
Accelerations can be found by differentiating the velocity expression. Prove: aB = -aA .

87. DEPENDENT MOTION EXAMPLE
sA and sB are defined from fixed datum lines, measured along the direction of motion of each block.
sB is defined to the center of the pulley above block B, since this block moves with the pulley.
The red colored segments of the cord and h remain constant in length

88. DEPENDENT MOTION EXAMPLE (continued)
The position coordinates are related by 2sB + h + sA = l
Where l is the total cord length minus the lengths of the red segments.
Velocities and accelerations can be related by two successive time derivatives:
2vB = -vA and 2aB = -aA
When block B moves downward (+sB), block A moves to the left (-sA).

89. DEPENDENT MOTION EXAMPLE (continued)
The example can also be worked by defining the sB from the bottom pulley instead of the top pulley.
The position, velocity, and acceleration relations become
2(h – sB) + h + sA = l
and 2vB = vA aB = aA
Prove that the results are the same, even if the sign conventions are different than the previous formulation.

90. EXAMPLE PROBLEM
Given: In the figure the cord at A is pulled down with a speed of 8 m/s.
Find: The speed of block B.

91. EXAMPLE (Solution) :
Define the position coordinates one for point A (sA), one for block B (sB), and one relating positions on the two cords (pulley C). sA sC sB
DATUM
Coordinates are defined as +ve down and along the direction of motion of each object.

92. 2vA + 4vB = 0 => vB = - 0.5vA = - 0.5(8) = - 4 m/s
EXAMPLE (continued) sA sC sB
DATUM
If l1=length of the first cord, minus any segments of constant length and l2 for the second:
Cord 1: 2sA + 2sC = l1
Cord 2: sB + (sB – sC) = l2
Eliminating sC, 2sA + 4sB = l1 + 2l2
Velocities are found by differentiating: (l1 and l2 constants):
2vA + 4vB = => vB = - 0.5vA = - 0.5(8) = - 4 m/s

93. GROUP PROBLEM SOLVING
Given: In this system, block A is moving downward with a speed of 4 m/s while block C is moving up at 2 m/s.
Find: The speed of block B.

94. GROUP PROBLEM SOLVING (Solution)
A datum line is drawn through the upper, fixed, pulleys.
Define sA, sB, and sC
Differentiate to relate velocities:

95. End of 12.9
Let Learning Continue

96. RELATIVE MOTION ANALYSIS (Section 12.10)
Objectives:
Understand translating frames of reference.
Use translating frames of reference to analyze relative motion.

97. APPLICATIONS
The boy on the ground is at d = 3m when the ball was thrown to him from the window. If the boy is running at a constant speed of 1.2m/s, how fast should the ball be thrown?
If aircraft carrier travels at 50km/hr and plane A takes off at 200 km/hr (in reference to water), how do we find the velocity of plane A relative to the carrier? the same for B?

98. RELATIVE POSITION
The absolute position of two particles A and B with respect to the fixed x, y, z reference frame are given by rA and rB. The position of B relative to A is represented by
rB/A = rB – rA
Therefore, if rB = (10 i + 2 j ) m
and rA = (4 i + 5 j ) m,
then rB/A = (6 i – 3 j ) m.

99. RELATIVE VELOCITY
The relative velocity of B with respect to A is the time derivative of the relative position equation:
vB/A = vB – vA or
vB = vA + vB/A
vB and vA are absolute velocities and vB/A is the relative velocity of B with respect to A.
Note that vB/A = - vA/B .

100. RELATIVE ACCELERATION
The time derivative of the relative velocity equation yields a similar vector relationship between the absolute and relative accelerations of particles A and B.
aB/A = aB – aA or
aB = aA + aB/A

101. Solving Problems Two ways of solution:
1. The velocity vectors vB = vA + vB/A could be written as Cartesian vectors and the resulting scalar equations solved for up to two unknowns (for 2D motion) or three unknowns (for 3D motion).
2. Or can be solved “graphically” by use of trigonometry. This approach is not recommended. Although it is easy for 2D, it is complicated for 3D (needs descriptive geometry).
Thus way 1 presents a unified approach that could be extended (tensor) to nD models.

102. LAWS OF SINES AND COSINES
GRAPHICAL SOLUTION
LAWS OF SINES AND COSINES a b c C B A
Since vector addition or subtraction forms a triangle, sine and cosine laws can be applied to solve for relative or absolute velocities and accelerations.
Law of Sines: C c B b A a
sin =
Law of Cosines: A bc c b a
cos 2 - + = B ac C ab

103. EXAMPLE
Given: vA = 600 km/hr
vB = 700 km/hr
Find: vB/A

104. vB/A = vB – vA = (- 1191.5 i + 344.1 j ) km/hr
EXAMPLE (continued)
Solution:
vA = 600 cos 35 i – 600 sin 35 j
= (491.5 i – j ) km/hr
vB = -700 i km/hr
a) Vector Method:
vB/A = vB – vA = ( i j ) km/hr hr km v A B 2 .
1240 ) 1
344 ( 5
1191 / = +
where - 16
tan q

105. EXAMPLE (continued) b) Graphical Method: ° 145 vB = 700 km/hr vB/A
vA = 600 km/hr
vB/A
Note that the vector that measures the tip of B relative to A is vB/A.
Law of Cosines: - + =
cos )
600 )(
700 ( 2 / A B v q hr km .
1240
Law of Sines: q
sin )
145°
sin( / A B v = or 1 . 16

106. GROUP PROBLEM SOLVING Given: vA = 10 m/s vB = 18.5 m/s at)A = 5 m/s2y x
Given: vA = 10 m/s
vB = 18.5 m/s
at)A = 5 m/s2
aB = 2 m/s2
Find: vA/B
aA/B
Solution:
The velocity of Car A is:
vA =

107. GROUP PROBLEM SOLVING (continued)
The velocity of B is:
vB =
The relative velocity of A with respect to B is (vA/B):
vA/B =

108. GROUP PROBLEM SOLVING (continued)
aA = (at)A + (an)A = [5 cos(45)i – 5 sin(45)j]
+ [-( ) sin(45)i – ( ) cos(45)j]
aA = 2.83i – 4.24j (m/s2)
102
100
aB = 2i (m/s2)
The relative acceleration of A with respect to B is:
aA/B = aA – aB = (2.83i – 4.24j) – (2i) = 0.83i – 4.24j
aA/B = (0.83)2 + (4.24)2 = m/s2
b = tan-1( ) = 78.9° b
4.24
0.83

109. End of 12.10
Let Learning Continue