Engineering Mechanics
‫ عبد الرزاق طوقان‬.‫د‬
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)
Decomposition
 Bodies are



Rigid
Flexible
Fluids
 Forces are


Static
Dynamics


Kinematics
Kinetics
Degree of components (hierarchy of
creation)
Define laws governing mechanics
(versus theories)
 Laws:
constitutive relationships
 equilibrium equations
 compatibility equations

 Theories:




Based on:
Assumptions based on:
Available knowledge is constrained with:
True dynamics is
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
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)
Second (s)
Force (F)
Newton* (N)
Pound (Ib)
(1ft = 0.3048 m, 1lb = 4.448 N. 1slug = 1lb s²/ft = 14.59 kg)
*Derived unit
1 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
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
109
giga
G
106
mega
M
103
kilo
k
10-3
milli
m
10-6
micro
μ
10-9
nano
n
Multiple
Sub-multiple
Design methodology for solving problems
Input
1. Problem to be solved
2. Physics of problem
3. Mathematical model
Processing
1. Propose theory
2. Formulate equations
3. Solve equations
Output
1. Verify laws
2. Build engineering sense
3. Start a new cycle
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!!!)
CHAPTER 12
RECTILINEAR KINEMATICS: 1D MOTION
Objectives:
-find the kinematic
quantities of a
particle traveling
along a straight
path.
APPLICATIONS
The motion of large objects, such
as rockets …, can often be
analyzed as if they were particles.
Why?
Mechanism definition
 Motion of rigid bodies are



Translational:
Rotational
General
 Definition of a particle
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
The total distance, sT, is …
Scalar :  s = s’ - s
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
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
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.
Position:
Velocity:
v
t
v
s
 dv =  a dt or  v dv =  a ds
s
t
 ds =  v dt
vo
o
vo
so
so
o
• so and vo are the initial position and velocity at t = 0.
CONSTANT ACCELERATION
(particular case)
Constant acceleration (e.g. gravity ac =g= -9.81 m/s2 ),
then
v
t
 dv =  a dt
c
vo
o
s
t
 ds =  v dt
so
v
v = vo + act
yields
s = s o + v ot + (1/2)a ct 2
yields
v2 = (vo )2 + 2ac(s - so)
o
s
 v dv =  ac ds
vo
yields
so
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.
EXAMPLE (solution)
1) Determine the velocity.
a = dv / dt => dv = a dt
v
t
vo
o
 dv =  (-6t )dt
=> v – vo = -3t2 => v = -3t2 + vo
2) Find time to stop (v = 0). Use vo = 27 m/s.
0 = -3t2 + 27 => t = 3 s
3) Calculate distance using so = 0:
t
v = ds / dt => ds = v dt =>s
 ds =  (-3t 2 + vo)dt
=> s – so = -t3 + vot
so
o
=> s – 0 = -(3)3 + (27)(3) => s = 54 m
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.
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
ERRATIC MOTION
Objectives:
Determine position,
velocity, and
acceleration using
graphs.
APPLICATION
Having a v-s graph, can
we determine a at s =
300m?
How?
GRAPHING
-Better to handle complex motions difficult to
describe with formulas.
-Graphs provide a visual description of motion.
-Graphs are true meaning of dynamics
S-T GRAPH
V-t: Find slope of s-t (v = ds/dt)
at various points.
V-T GRAPH
a-t: Find slope of v-t (a = dv/dt)
at various points.
A-T GRAPH
v-t is the area under the a-t
curve.
We need initial velocity of the
particle.
A-S GRAPH
Area under a-s curve= one half
difference in the squares of the
speed (recall ∫a ds = ∫ v dv ).
s2
½ (v1² – vo²) = a ds
s1
We need initial velocity
V-S GRAPH
Knowing velocity v and the
slope (dv/ds) at a point,
acceleration is.
a = v (dv/ds)
EXAMPLE
Given: v-t graph for a train moving between two stations
Find: a-t graph and s-t graph
EXAMPLE (Solution:)
a0-30 = dv/dt = 40/30 = 4/3 m/s2
a30-90 = 0
a90-120 = -4/3 m/s2
a(m/s2)
4
3
t(s)
-4
3
EXAMPLE (continued)
GROUP PROBLEM SOLVING
Given: The v-t graph shown
Find: The a-t graph,
average speed,
and distance traveled
for the 30 s interval
GROUP PROBLEM SOLVING
a(m/s²)
t(s)
GROUP PROBLEM SOLVING (continued)
s0-10 =  v dt =
= 400/3 m
s10-30 =  v dt =
=
s0-30 =
= 1133.3 m
vavg(0-30) = total distance / time
=
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.
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:
1 distance + 2 angles
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?
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
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.
ACCELERATION
Average acceleration is: aavg = v/t = (v’ - v)/t
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?
RECTANGULAR COMPONENTS: POSITION
The position can be defined as
r=xi+yj+zk
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
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.
RECTANG. COMPONENTS: ACCELERATION
Acceleration vector is:
a = dv/dt = d2r/dt2 = axi + ayj + azk
where
•
•
•
••
ax = vx = x = dvx /dt, ay = vy = y
= dvy /dt,
••
••
az = vz = z = dvz /dt
Magnitude is
a = [(ax)2 + (ay)2 + (az)2 ]0.5
What about direction of a?
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.
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 = 0 => t = 2 or – 1/3 s
- So, t = 2s. Substituting yields: xA = xB = 6 m, yA = yB = 0
EXAMPLE (continued)
Velocity vectors:
.
.
vA = drA/dt =.xA i + yA j = [3i + (18 – 18t)j] m/s
At t = 2 s: vA = [3i – 18j] m/s
•
•
vB = drB/dt =xBi + yBj = [(6t – 6)i + 3j] m/s
At t = 2 s: vB = [6i + 3j] m/s
Speed is
vA = (32 + 182) 0.5 = 18.2 m/s
vB = (62 + 32) 0.5
= 6.71 m/s
GROUP PROBLEM SOLVING
Given: A particle travels along a path y = 0.5x2. The xcomponent 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.
GROUP PROBLEM SOLVING (Solution)
1) x-components:
•
Velocity:
vx = x = dx/dt =
x
Position:
 dx
0
t
=

0
••
•
Acceleration:ax = x = vx =
2) y-components:
Position:
y=
Velocity:
vy = dy/dt =
Acceleration: ay = •vy =
GROUP PROBLEM SOLVING (continued)
3) The displacement from the origin is:
r =xi+yj=
At t = 1 s, r =
Distance: d = r =
The magnitude of the acceleration is:
Acceleration vector: a =
Magnitude: a =
= 37.8 m/s2
MOTION OF A PROJECTILE
(Section 12.6)
Objectives:
Analyze the free-flight
motion of a projectile.
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 ?
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!
KINEMATIC EQUATIONS:
HORIZONTAL MOTION
Since ax = 0, vox remains constant and the position in the
x direction can be determined by:
x = xo + (vox)(t)
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?
Example 1
Given: vo and θ
Find: The equation that defines
y as a function of x.
vx = vo cos θ
Solution: Using
We can write: x = (vo cos θ)t
or
vy = vo sin θ
and
t =
x
vo cos θ
y = (vo sin θ)t – ½ g(t)2
By substituting for t:
y = (vo sin θ)
(
x
vo cos θ
) ( )(
–
g
2
2
x
vo cos θ
)
Example 1 (continued):
Simplifying the last equation, we get:
g x2
2θ)
y = (x tanθ) –
(1
+
tan
( 2v 2 )
o
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!
Example 2:
Given: Snowmobile is going 15
m/s at point A.
Find: The distance it travels (R)
and the time in the air.
GROUP PROBLEM SOLVING
Given:Skier leaves the
ramp at θA = 25o and
hits the slope at B.
Find: The skier’s initial speed vA.
GROUP PROBLEM SOLVING (continued)
Solution:
Motion in x-direction:
Motion in y-direction:
vA = 19.42 m/s
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.
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?
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, naxis is perpendicular to the t-axis and +ve toward the center
of curvature of the curve.
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.
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
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.
The acceleration vector can be
expressed as:
.
a = vut + (v2/r)un = atut + anun.
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
SPECIAL CASES OF MOTION
There are some special cases of motion to consider.
1) The particle moves along a straight line.
.
2
r
 => an = v /r = 0 => 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 = 0 => a = an = v2/r
The normal component represents the time rate of change
in the direction of the velocity.
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
2 ]3/2
[
1
(dy/dx)
+
r = ________________
d2y/dx 2
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.
EXAMPLE (Solution)
1) The velocity vector: v = v ut , v = (0.2t2) m/s.
2) 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) = 0.0648 m/s2
Magnitude is: a = [(at)2 + (an)2]0.5 = [(1.2)2 + (0.0648)2]0.5 = 1.20 m/s2
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.
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
= 0.921 m/s2
ABSOLUTE DEPENDENT MOTION ANALYSIS
OF TWO PARTICLES (Section 12.9)
Objectives:
To relate the positions, velocities,
and accelerations of particles
undergoing dependent motion.
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.
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?
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.
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.
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 .
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
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).
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
2aB = aA
Prove that the results are the same, even if the sign
conventions are different than the previous formulation.
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.
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).
DATUM
sA
sC
sB
Coordinates are defined as +ve
down and along the direction of
motion of each object.
EXAMPLE (continued)
DATUM
sA
sC
sB
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 = 0 => vB = - 0.5vA = - 0.5(8) = - 4 m/s
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.
GROUP PROBLEM SOLVING
(Solution)
A datum line is drawn through the upper, fixed, pulleys.
Define sA, sB, and sC
Differentiate to relate
velocities:
RELATIVE MOTION ANALYSIS (Section 12.10)
Objectives:
a) Understand translating
frames of reference.
b) Use translating frames of
reference to analyze
relative motion.
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?
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.
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 .
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
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.
GRAPHICAL SOLUTION
LAWS OF SINES AND COSINES
C
b
a
B
Since vector addition or subtraction
forms a triangle, sine and cosine laws
can be applied to solve for relative or
absolute velocities and accelerations.
A
c
Law of Sines:
a
sin A
=
b
sin B
=
c
sin C
Law of Cosines: a 2 = b 2 + c 2 - 2 bc cos A
b = a + c - 2 ac cos B
2
2
2
c = a + b - 2 ab cos C
2
2
2
EXAMPLE
Given:
Find:
vA = 600 km/hr
vB = 700 km/hr
vB/A
EXAMPLE (continued)
Solution:
a) Vector Method:
vA = 600 cos 35 i – 600 sin 35 j
= (491.5 i – 344.1 j ) km/hr
vB = -700 i km/hr
vB/A = vB – vA = (- 1191.5 i + 344.1 j ) km/hr
(1191. 5 )2 + ( 344.1 )2 = 1240. 2 km
hr
- 1 344 .1
q
(
) = 16.1°
tan
q
=
where
1191 . 5
vB /A =
EXAMPLE (continued)
b) Graphical Method:
Note that the vector that measures
the tip of B relative to A is vB/A.
vB = 700 km/hr
q
145 °
Law of Cosines:
2
vB/A2 = ( 700 ) 2 + ( 600 ) - 2 ( 700)(600 )cos 145 °
vB/A = 1240 . 2 km
hr
Law of Sines:
vB/A
vA
=
sin(145° )
sin q
or
q = 16 . 1 °
GROUP PROBLEM SOLVING
Given:
Find:
y
x
Solution:
The velocity of Car A is:
vA =
vA = 10 m/s
vB = 18.5 m/s
at)A = 5 m/s2
aB = 2 m/s2
vA/B
aA/B
GROUP PROBLEM SOLVING (continued)
The velocity of B is:
vB =
The relative velocity of A with respect to B is (vA/B):
vA/B =
GROUP PROBLEM SOLVING (continued)
aA = (at)A + (an)A = [5 cos(45)i – 5 sin(45)j]
+ [-( 102 ) sin(45)i – ( 102 ) cos(45)j]
100
100
aA = 2.83i – 4.24j (m/s2)
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 = 4.32 m/s2
b = tan-1( 4.24 ) = 78.9°
0.83
b