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KINEMATICS OF PARTICLES: FORCE AND ACCELERATION
State Newtons Laws of Motion and Gravitational attraction and to be able to define mass and weight
Analyze accelerated motion of a particle using the equation of motion
1. Newtons Second Law of Motion 2. The Equation of Motion 3. Equation of Motion for a System of Particles 4. Equations of Motion: Rectangular Coordinates 5. Equations of Motion: Normal and Tangential
Coordinates 6. Equations of Motion: Cylindrical Coordinates
Second Law: A particle acted upon by an unbalanced force F experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force.
Newtons Law of Gravitational Attraction A law governing the mutual attractive gravitational force acting between them
221
r
mmGF
Newtons Law of Gravitational Attraction Mass is a property of matter
Mass of the body is specified in kilograms
Weight is calculated using the equation of motion, F = ma
mgW
W = mg (N)
(g = 9.81 m/s2)
Equation of motion is written as Consider P of mass m subjected to the
action of two forces, F1 and F2 From free body diagram, the
resultant of these forces produces the vector ma
Represented graphically on the kinetic diagram
FR = F = 0, acceleration is zero Such a condition is called static equilibrium,
Newtons First Law of Motion
maF
Inertial Reference Frame
Acceleration of the particle is measured with respect to a reference frame that is either fixed or translates with a constant velocity
Such a frame of reference is known as a Newtonian or inertial reference frame,
The free body diagram for the its particle are shown. Applying equation of motion yields
F = ma; Fi + fi = miai If equation of motion is applied to each of
the other particles, these equations can be added together vectorially,
Fi + fi = miai
The summation of internal forces will be equal to zero where
Fi = miai
If rG is a position vector which locates the center of mass G of the particles, we have
mrG = miri
Differentiating twice w.r.t time yields maG = miai
Therefore, F = maG
The sum of the external forces acting on the system of particles is equal to the total mass of the particles times the acceleration of its center of mass G
When a particle is moving relative to an inertial x, y, z frame of reference,
F = ma
Fxi + Fyj + Fzk = m(axi + ayj + azk)
The three scalar equations:
zz
yy
xx
maF
maF
maF
Procedure for Analysis Free-Body Diagram Select inertial coordinate system Draw particles free body diagram (FBD) and
provides a graphical representation that accounts for all forces (F)
Direction and sense of the particles acceleration a is also be established
Acceleration is represented as ma vector on the kinetic diagram
Identify the unknowns in the problem
Procedure for Analysis Equation of Motion Apply equations of motion on FBD in their
scalar component form Cartesian vector analysis can be used for the
solution Kinematics Apply kinematics equations once the
particles acceleration is determined from F = ma
If acceleration is a function of time, use a = dv/dt and v = ds/dt
Procedure for Analysis
Kinematics
When acceleration is a function of displacement, integrate a ds = v dv to find velocity as a function of position
If acceleration is constant, use
tavv c 02
002
1tatvss c
020
2 2 ssavv c
The 50-kg crate rests on a horizontal plane for which the coefficient of kinetic friction is k = 0.3. If the crate is subjected to a 400-N towing force, determine the velocity of the crate in 3 s starting from rest.
Solution
Free-Body Diagram Weight of the crate is W = mg = 50 (9.81) = 490.5 N. The frictional force is F = kNC and acts to the left, There are 2 unknowns, NC and a.
Equations of Motion
Solving we get
030sin4005.490;
503.030cos400;
Cyy
Cxx
NmaF
aNmaF
2/19.5 , 5.290 smaNNC
Solution
Kinematics Acceleration is constant. Velocity of the crate in 3s is
sm
tavv c
/6.15
)3(19.50
0
A 10 kg projectile is fired vertically upward from the ground, with an initial velocity of 50 m/s. Determine the maximum height to which it will travel if (a) atmospheric resistance is neglected, and (b) atmospheric resistance is measured as F = 0.01v2 N, where v is the speed of the projectile at any instant measured in m/s.
The weight W = mg = 10(9.81) = 98.1 N
Equation of Motion
Kinematics
Initially zo = 0, vo = 50 m/s At max height, z = h, v = 0 a is constant
2 2
0 0
2
( ) 2 ( )
0 50 2( 9.81)( 0)
127
v v a z z
h
h m
2; -98.1 10 9.81 m/sz zF ma a a
Equation of motion
Kinematics
a is not constant since FD depends on velocity. Since a = f(v) we can relate a to position using
Initially, zo=0, v0=50 m/s
Finally, z=h, v=0
2 2; -0.01 -98.1 10 (0.001 9.81)z zF ma v a a v
2( ) - (0.001 9.81)adz vdv v dz vdv
2( ) - (0.001 9.81)adz vdv v dz vdv
00
2
2
0 50 50
500ln( 9810)0.001 9.81
114
hvdv
dz vv
h m
The baggage truck A has a weight of 3600 N and tows a 2200 N cart B and a 1300 N cart C. For a short time the driving frictional force developed at the wheels is FA = (160t) N where t is in seconds. If the truck starts from rest, determine its speed in 2 seconds. What is the horizontal force acting on the coupling between the truck and cart B at this instant?
Solution
Free-Body Diagram We have to consider all 3 vehicles.
Equations of Motion Only horizontal motion is considered.
ta
at
maF xx
221.0
81.9
130022003600160
;
Solution Kinematics The velocity of the truck is obtained using a = dv/dt with the initial condition that v0 = 0 at t = 0,
Free-Body Diagram Equations of Motion When t = 2 s, then
smtvdttdvv
/442.01105.0;)221.0(2
0
22
00
NTT
maF xx
8.157)2(221.081.9
3600)2(160
;
The 100-kg block A is released from rest. If the masses of the pulleys and the cord are neglected, determine the speed of the 20-kg block B in 2 s.
Solution
Free-Body Diagrams Since mass of the pulleys is neglected, then for pulley C, ma = 0.
It can be seen that T = 490.5 N for A to be static and T = 196.2 N for B to be static.
Assume both blocks accelerate downward, in the direction of +sA and +sB
Solution
Equations of Motion Block A:
Block B:
Kinematics It is seen that
Differentiating this expression twice w.r.t time yield
Byy aTmaF 202.196;
Ayy aTmaF 1002981;
lss BA 2
BA aa 2
where l is constant and represents the total vertical length of cord.
Solution
Kinematics Solving the three equations yields
Since aB is constant, velocity in block B in 2 s is
The negative sign indicates that block B is moving upward.
22 /54.6 , /27.3 , 0.327 smasmaNT BA
smtavv B /1.130
Equation of motion for the particle may be written in the tangential, normal and
bi-normal directions
Since the particle is constrained to move along the path, there is no motion in the
bi-normal direction
0b
nn
tt
F
maF
maF
at (=dv/dt) represents the time rate of change in the magnitude of velocity
an (=v2/) represents the time rate of
change in the velocitys direction
The 3 kg disk D is attached to the end of a cord as shown. The other end of the
cord is attached to a ball-and-socket joint located at the center of a platform.
If the platform is rotating rapidly, and the disk is placed on it, and released from
rest as shown, determine the time it takes for the disk to reach a speed great
enough to break the cord. The maximum tension the cord can sustain is 100 N,
and the coefficient of kinetic friction between the disk and the platform is mk= 0.1.
13 ;
31.0 ;
043.29 ;0
43.29)81.9(3
1.0
2vTmaF
aNmaF
NF
NW
NNF
nn
tDtt
Db
DDkm
Equations of motion
Setting T = 100 N and solve the equations, we get
ND = 29.43 N at = 0.981 m/s2 vcr = 5.77 m/s
vcr = critical velocity
Kinematics
Since at is constant, the time needed to break the cord is
vcr = v0 + att
5.77 = 0 + 0.981t
t = 5.89 s
Determine the banking for the race track so that the wheels of the racing cars will not have to depend upon friction to prevent any car from sliding up or down the track. Assume the cars have negligible size a mass m, and travel around the curve of radius with a speed v.
Solution
Free-Body Diagrams No frictional force acting on the car.
NC represents the resultant of the ground on all four wheels.
Unknown are NC and .
Solution
Equations of Motion Using the n and b axes,
Solving the 2 equations, eliminating NC and m,
0cos;0
sin;2
mgNF
vmNmaF
Cb
Cnn
g
v
g
v 212
tantan