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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