chapter 2 0 mechanic jan11
TRANSCRIPT
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CHAPTER 2
Mechanics1
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OBJECTIVE
Coplanar forces in static equilibrium
Distance, time, velocity and acceleration
Projectile motions
Basic Newton’s law Centripetal forces
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SCALARS & VECTORS
Scalar quantity – has only magnitude and is completelyspecified by a number and a unit (eg. Mass, volume,frequency)
Vector quantity – has both magnitude and direction
(eg. Displacement, velocity, force, voltage, current) Vector is represented by an arrow whose length is
proportional to a certain vector quantity and whosedirection indicates the direction of the quantity
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EXAMPLE
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10m, to east
20m, 400
North of East
400
15m, 350
south of west
350
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EXAMPLE
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2
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FORCE - EXAMPLE
Find their resultant force for the figure below:
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80N
100N110N
160N
30o
20o
45o
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CONT..
Method 1: graphically Method 2: rectangular component method
Magnitude X-component Y-component
80 80 cos 0 = 80 80 sin 0 = 0
100 100 cos 45 = 70.71 100 sin 45 = 70.71
110 -110 cos 30 = -95.26 110 sin 30 = 55
160 -160 cos 20 = -150.35 -160 sin 20 = - 54.72
total -94 71
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CONT..
Resultant force, R total = √ (x2 + y2) = √ (-94)2 +(71)2 = 117.80 N
Resultant angle = tan-1 (y/x)
= 2nd
quadrant (from the coordinate)= 143o
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Sum of vector (sum of force) = 117.8N, 1430
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CONT..
Method 3: polar form to
rectangular form
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EXAMPLE
A woman walks eastward for 5 km and then
northward for 10 km.
How far is she from her starting point? (distance)
If she had walked directly to her direction, in what
direction would she have headed? (displacement)
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N
E
S
W
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EXAMPLE
Let F have a magnitude of 300N and make angle,
θ=30o with the positive x direction. Find Fx and Fy
If F=300N and θ=145o (2nd Quadrant), find Fx andFy
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EXAMPLE
The man in figure below exerts a force of 100
N on the wagon at an angle of 30 degrees
above horizontal. Find the horizontal and
vertical components of this force
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VECTOR ADDITION
Components
Simply add together the x-components, y-
components and z-components separately and the
sums are now the x, y and z components of the
resultant.
A = Ax + Ay + Az
B = Bx + By + Bz
C = Cx + Cy + Cz
Resultant, R = Rx +Ry +Rz
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VECTOR ADDITION
where Rx = Ax +Bx +Cx
Ry = Ay +By +Cy
Rz = Az +Bz +Cz
To avoid confusion unit vectors are introduced.
A = Ax i + Ay j + Az k
B = Bx i + By j + Bz k
R = A + B = ( Ax +Bx ) i + ( Ay +By ) j +
( Az + Bz ) k15
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SUBTRACTION OF VECTORS
Using Components
A = Ax i + Ay j + Az k
B = Bx i + By j + Bz k
R = A - B = ( Ax - Bx ) i + ( Ay - By) j + (Az -
Bz) k
Subtract the coefficients of i, j and k separately.
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MULTIPLICATION BY A SCALAR
If a vector R is multiplied by a scalar k the direction
of the vector remains unaltered but the magnitude
is now k R. The resulting vector is kR.
The magnitude of each component is multiplied by
k.
Example: If R = 2i + 3 j + 5k
Then: 2R = 4i + 6 j + 10k
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VECTOR PRODUCT ( "CROSS PRODUCT")
The vector or cross-product C of 2 vectors A B
has:
A
B = i j k Ax Ay Az
Bx By Bz
R= i ( AyBz - AzBy) - j ( AxBz - AzBx ) + k ( AxBy- AyBx )
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MOTION IN ONE DIMENSION
Motion continuous change in the position of an
object
3 types of motion
Translational: car moving down a highway
Rotational: earth’s spin on its axis
Vibrational: back-and-forth movement of a pendulum
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DISTANCE, D
Is a scalar quantity
Is a path length transverse in moving from 1
location to another
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DISPLACEMENT, S
In the study of translational motion, moving
object can be viewed as a particle
regardless of its size
Displacement is a vector quantityDisplacement is defined as distance or the
change in particle’s position,
Ds=sf –si where si initial positions
sf are and final positions21
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EXAMPLE
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Initial position is 0m
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SPEED, V
Scalar quantity
defined as the total distance traveled
divided by the total time it takes to travel
that distance Instantaneous speed, V= distance /t
Average speed is the rate of change of
distance
V = Δd / Δt
= change of distance / time interval24
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VELOCITY, V
Vector quantity Instantaneous velocity, V = s/t
(S refer to the displacement on constant period)
Average velocity is the rate of change ofdisplacement
V = Δs / Δt= change of displacement/time interval
Its direction is in the direction of the
displacementObject moving in uniform velocity if
ds/dt=constant25
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ACCELERATION, A
Vector quantity
Instantaneous acceleration, a = v/t
Average acceleration is the rate of change of
velocitya = Δv / Δt
= change of velocity / time interval
Its direction is in the direction of motion
Acceleration is uniform when magnitude ofvelocity change (dv/dt) at constant rate and fixdirection
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GRAPH REPRESENTATION
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LINEAR MOTION WITH CONSTANT
ACCELERATION
Uniform acceleration, a = (v – v0) / t
v = v0 + at
Displacement, s =1/2 (v0 + v)t
= area of v vs t graph s = v0t +1/2 at
2
v2 = v02 + 2as
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v0 = u = initial velocity
V = final velocity
a = acceleration
T= time
S= displacement
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FREE FALL
Vertical motion with constant acceleration, g undergravitational field without air resistance
g=9.81m/s2; direction toward center of the earth(downward)
a=-g v = v0 – gt v2 = v0
2 - 2gs
s = v0t -1/2 gt2
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*Note:V=0 when particle reach max height
If the free fall just show onedirection, we can assumea=g=9.81m/s2 for easier calculation
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PROJECTILE MOTIONS
Consists of two motions
Vertical component (y-comp): motion under constant
acceleration, -g
Horizontal component (x-comp): motion under constant
velocity
Path followed by a projectile is called trajectory
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CONT..
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CONT..
Initial condition (point A)
X-comp: Vox= Vocos θ constant all time
Y-comp: Voy= Vosin θ varies
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CONT..
Point B and point D X-comp: V1x= V2x=Vox = Vocos θ
(Point B) Y-comp: V1y= Voy – gt1
(Point D) Y-comp: V2y= Voy – gt2
Velocity: Magnitude at (B), V1= √ (V1x)2 + (V1y)2
Direction, θ1= tan-1 (V1y / V1x)
Displacement (S):
X = V0x t
Y = V0y t - ½ gt2
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CONT..
Point C X-comp: Vx= Vox = Vocos θ Y-comp: Vy= 0
Displacement (S):
V2 = u2 + 2as Vy2 = v0y2 - 2gsy 0 = (V0sin θ)2 – 2gH
H= (V02 sin2 θ) / 2g V= u +at
vy = v0y
– gt t = (V0sin θ)/ g
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EXAMPLE
A projectile is fired with an initial velocity of 80 m/s
at an angle of 30oabove the horizontal. Find:
Its position and velocity after 6s
Time required to reach the maximum height
Maximum height
The horizontal range, r (before it strike on floor)
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FORCES
Force is something capable of changing an object’sstate of motion
4 types:
Gravitational force: involve attraction between massive
bodies, long-range force, weakest force in nature
Electromagnetic force: attraction and repulsive force
between electric charges, long-range force
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CONT..
Strong nuclear force: attraction force bonds theneutrons and protons together in nucleus, short-rangeforce, strongest force in nature
Weak nuclear force: cause unstable condition foratomic nucleus and for radioactive decay, short-range
force, 12 time weaker than electromagnetic force Force is vector quantity
F= ma ; kgms-2 / N
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NEWTON’S SECOND LAW
aF m=∑
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The rates of change of linear momentum of amoving body is proportional to the resultantforce and is in the same direction as force acton it
States that the acceleration, a of an object inthe direction of a resultant force, F is directlyproportional to the magnitude of the force andinversely proportional to the mass
1N = 1kgms-2
And the force of gravity or weight,
gFg m
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NEWTON’S THIRD LAW
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Every action must produce an equal and
opposite reaction (not act on same object)
Or whenever one object exerts a forces, F12 on
a second object, the second exerts an equaland opposite force, F21 of the first object
F12 = – F21Someone climb a ladder
Rung must have same but opposite force on the
foot to avoid collapse
Ffoot = - Frung
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EQUILIBRIUM
Occur when the resultant of all external
force is zero
A body in equilibrium must be either at rest
or in motion with constant velocity∑Fx = 0
∑Fy = 0
A body is in transitional equilibrium if andonly if the vector sum of the forces acting on
it is zero.42
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FREE-BODY DIAGRAM
Is a diagram that drawn with all known quantities
are labeled. Then a force diagram indicating all
forces and their components is constructed.
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Circular Motion
Motion along the perimeter of a circle
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Definition of Angular Displacement
1) When a rigid body rotates about a fixed axis, the angulardisplacement is the angle Δθ swept out by a line passing
through any point on the body and intersecting the axis ofrotation perpendicularly.
2) By convention, the angular displacement is positive if it is
counterclockwise and negative if it is clockwise.
SI unit o f Angular Displacement : radian (rad) *
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Angular Displacement
• Angular displacement if often expressed in one ofthree units. The first is the familiar degree, and itis well known that there are 360°in a circle.
•The second unit is the revolution (rev), onerevolution representing one complete turn of 360°.
•The most useful unit from a scientific viewpoint,
however, is the SI unit called the radian (rad).
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Angular Displacement
As the disc rotates, the point traces out and arc of length (s),which is measured along a circle of radius (r). The angle θ will be
in radians:-
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Example 1
Synchronous or “stationary” communication
satellites are put into an orbit whose radius is r =
4.23 x 107 m. The orbit is in the plane of the
equator, and two adjacent satellites have anangular separation of θ = 2.0°. Find the arc length
(s) that separates the satellites.
Solut ion:
Step 1 – Convert degree into radians,
2.0°= (2.0 degrees) (2π radians/ 360 degrees) = 0.0349
radians
Step 2 – Calculation of the arc length,
S = r θ = (4.23 x 107
m) (0.0349 rad) = 1.48 x 106
m
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Angular Velocity
Example 2
A gymnast on a high bar swings through two revolutions
(clockwise) in a time of 1.90 s. Find the average angular
velocity (rad/s) of the gymnast.
Solut ion:
Δθ = -2.0 revolutions (2π radians / 1 revolution)= -12.6 radians
Where the minus sign denotes that the gymnast
rotates clockwise. The average angular velocity is:-
ω= (Δθ / Δt) = (-12.6 rad / 1.90 s) = - 6.63 rad/s
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Definition of Angular Acceleration
In linear motion, a changing velocity means that an
acceleration is occurring. Changing angular velocity means
that an angular acceleration is occuring. The angular
acceleration is defined as :-
The SI unit for angular acceleration is rad/s2
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Angular Acceleration
Example 3
A jet awaiting for takeoff is momentarily stopped on the
runway. As seen from the front of one engine, the fan blades
are rotating with an angular velocity of -110 rad/s, where the
negative sign indicates a clockwise rotation. As the plane
takes off, the angular velocity of the blades reaches – 330
rad/s in a time of 14 s. Find the angular acceleration,
assuming it to be constant.
Solut ion:
α = (ω-ω0)/(t-t0) = (-330 rad/s) – (-110rad/s) / (14 s)
= -16 rad/s2
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The Equations of Rotational Kinematics
Centripetal Acceleration and Tangential Acceleration
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Centripetal Acceleration and Tangential Acceleration
1) When an object picks up speed as it moves around a circle, it
has a tangential acceleration.
2) In addition, the object also has a centripetal acceleration.3) Even when the magnitude of the tangential velocity is
constant, an acceleration is present, since the direction of the
velocity changes continually.
4) Because the resulting acceleration points toward the center of
the circle, it is called CENTRIPETAL acceleration.
ac= VT2 / r
Subscript T is the reminder
for tangential speed.
The centripetal acceleration can
be expressed in terms of angular speed ω:-
ac= (r ω)2 / r
= r ω2
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UNIFORM CIRCULAR MOTION
Movement of object in a circular path atuniform speed
∑Fc = mac
Angular velocity, W (rad/s)Tangential/linear velocity, V = rW
Angular acceleration, α = (Wf – Wi) /tac = centripetal acceleration
= V2 / r = r W2
Centripetal force, F= m(V2/r) = mW2r
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**1 rev = 2π rad
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REFERENCES
Hugh D. Young et al., University
Physics: With Modern Physics, 13th
Edition, 2012.
Giancoli, D.C. Physics for Scientistsand Engineers with Modern Physics.
4th Edition. Pearson, 2009.
Knight, R.D., Jones, B., Field, S.,College Physics. 2nd Edition.
Pearson/Addison Wesley, 2010.