Download Chapter 2 KINETICS OF PARTICLES: NEWTON`S SECOND LAW

Chapter 2 KINETICS OF PARTICLES: NEWTON’S SECOND LAW
• Newton’s first and third laws are sufficient for the study of bodies at
rest (statics) or bodies in motion with no acceleration.
• When a body accelerates (changes in velocity magnitude or direction),
Newton’s second law is required to relate the motion of the body to the
forces acting on it.
Newton’s Laws of Motion
• First Law: Inertia
– a body will continue in its state of motion unless acted upon by a net force.
• Second Law: ΣF = ma
– Acceleration is proportional to the net force acting on a body.
• Third Law: Action-reaction
– For every force there is an equal and opposite force.
• Force
– Effect that one body has on another
– A push or a pull applied to an object
– That required to change the state of motion of an object (i.e., that which causes
acceleration)
Force (Cause)
Acceleration (Effect)
Characteristics of Force
• Force is a vector quantity:
– Magnitude & direction
• A third, unique characteristic:
– The point of application (especially important relative to the determination of
moments or torques)
• Therefore, to completely understand the influence of a net force, we must have
knowledge of all three characteristics
Classification and Behavior of Forces
A practical classification scheme for forces
repulsive — describes a force that tends to push the two participating objects apart
attractive — describes a force that tends to pull the two participating objects together
oblique — describes a force that acts at some other angle, one that is not a direct
repulsion or attraction
normal force — the force that keeps two objects from occupying the same space
static friction — a friction force between surfaces that are not slipping past each other
kinetic friction — a friction force between surfaces that are slipping past each other
fluid — a gas or a liquid
fluid friction — a friction force in which at least one of the object is is a fluid
spring constant — the constant of proportionality between force and elongation of a
spring or other object under strain
Newton's Third Law
Two magnets exert forces on each other
Two people's hands exert forces on each other.
Newton's third law does not mean that forces always cancel out
so that nothing can ever move. If these two figure skaters,
initially at rest, push against each other, they will both move.
Rockets work by pushing exhaust gases out the back.
Newton's third law says that if the rocket exerts a
backward force on the gases, the gases must make an
equal forward force on the rocket. Rocket engines can
function above the atmosphere, unlike propellers and
jets, which work by pushing against the surrounding air.
A runner exerts a force against the ground. This creates an equal and opposite
reaction force which moves the body over the ground.
Newton's third law states that all forces have an equal and opposite reaction forces
on the other object, like in this elastic exercise. The –150-N (FA) force applied by the
person on the elastic cord coincides with a 150-N reaction force (FB) exerted on the
person by the cord.
A major consequence of Newton's third law is that the forces we exert on an object with larger
inertia often create motion in the direction opposite of those forces. In running, the downward
backward push of the foot on the ground (FA) late in the stance (a) creates a ground reaction force
which acts forward and upward, propelling the runner through the air. A defensive player trying to
make a tackle (FT) from a poor position (b) may experience reaction forces (FR) that create
eccentric muscle actions and injurious loads.
Single DNA manipulation experiment using optical tweezers
One end of the DNA is attached to a glass surface
What happens when a force is applied?
Think about Newton’s laws of motion...
Introduction
Newton’s Second Law of Motion
Linear Momentum of a Particle
Systems of Units
Equations of Motion
Dynamic Equilibrium
Angular Momentum of a Particle
Equations of Motion in Radial &
Transverse Components
Conservation of Angular Momentum
Newton’s Law of Gravitation
Trajectory of a Particle Under a Central
Force
Application to Space Mechanics
Kepler’s Laws of Planetary Motion
Newton’s Second Law of Motion
If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to
the magnitude of the resultant and in the direction of this resultant force.
F1
a1
F1 F2 F3


 ....  constant  m
a1 a2 a3
Denoting by m the mass of a particle, by  F the sum, or resultant, of the forces acting on the particle,
and by a the acceleration of the particle relative to a newtonian frame of reference, we write
Linear Momentum of a particle. Rate of change of linear momentum
 F  ma  m
d d m 

since mass is constant
dt
dt
L=m is the linear momentum or momentum. (directions of velocity and momentum are the same)
The resultant of the forces acting on a particle is equal to the rate of change of the linear momentum of
the particle.
If F is zero, L is constant in both magnitude and direction (Conservation of linear momentum)
Systems of Units
SI Units (absolute system of units: length, mass and time can be used anywhere on the earth)
Length: meter(m)
Mass:
kilogram (kg)
Time:
second(s)
1 N= (1 kg)(1 m/s2)=1 kg.m/s2
Weight : W : a force, W=mg
m=kg.m/s
US Customary (gravitational system of units: length, force and time)
Length: foot(ft)
Force:
pound (lb)
Time:
second(s)
1 lb= (1 slug)(1 ft/s2)=1 kg.m/s21 slug=1 lb.s2/ft
m=W/g
m=lb.s
g=32.2 ft/s2
Force and motion do not always act
in the same direction.
This free-body diagram of the forces
and resultant force (FR) on a
basketball before release
illustrates how a skilled player applies
a force to an object
(Fh) that combines with the force of
gravity (Fg) to create the desired
effect.
The acceleration (motion ?) of the
ball will be in the direction of FR.
The velocity vector may not be in the
direction of that force
Equations of Motion
• Newton’s second law provides


F

m
a

• Solution for particle motion is facilitated by resolving
vector equation into scalar component equations, e.g.,
for rectangular components,
• For tangential and normal components,
Friction forces acting on ice skaters during push-off and gliding. Newton' Second Law of
Motion applied in the horizontal direction will determine the horizontal acceleration of the
skater.
he three external forces that determine the acceleration of the centre of mass
(COM) of a sprinter: ground reaction force (GRF), gravitational force
equivalent to body weight (BW), and wind resistance.
ay
y
P
az
To solve a problem involving the
ax
x
z
y
an
at
motion of a particle, F = ma should
be replaced by equations containing
scalar quantities. Using rectangular
components of F and a, we have
Using tangential and
normal components,
P
x
O
Using radial and transverse components,
a
r
O

P
ar
x
Whole-body human tolerance to vehicle acceleration based on impact duration.
Apply Newton’s 2nd law
The Wayne State Tolerance Curve for head injury.
Apply Newton’s 2nd law
Dynamic Equilibrium: (D’Alembert Principle)
F  ma  0
F
x
0
-ma : inertia vector, inertia force (resistance that a particle offer when we try to set
them in motion or when we try to change the conditions of their motion)
F
y
0
Including inertia force or vector
Atalet kuvveti
Eylemsizlik kuvveti
Inertial forces may be conceptually useful but are not like the contact and gravitational forces found in
statics.
Sample Problem 12.1
90 kg
SOLUTION:
• Resolve the equation of motion for the
block into two rectangular component
equations.
A 90-kg block rests on a horizontal
plane. Find the magnitude of the force
P required to give the block an acceleration of 3 m/s2 to the right. The coefficient of kinetic friction between the
block and plane is k 0.25.
• Unknowns consist of the applied force
P and the normal reaction N from the
plane. The two equations may be
solved for these unknowns.
Sample Problem 12.1
SOLUTION:
• Resolve the equation of motion for the block into
two rectangular component equations.
Fx  ma:
y
O
x
Fy 0:
• Unknowns consist of the applied force P and the
normal reaction N from the plane. The two
equations may be solved for these unknowns.
N  Psin30  90*9.81
P cos30  0.25Psin30  90*9.81  270
P  662N
If the body is already moving with constant velocity, 662 N force is enough to access
3 m/s^2 acceleration.
But if it is standing on a surface with friction larger than coeffiecient of 0.25 , the
External force should be high enough to exceed static friction force.
Assume s = 0.35, then what happens:
Fs=0.35 N instead of 0.25 N
Both P and N change, after exceeding this number, P can be reduced to 662 N.
P>662 N
662 N
Sample Problem 12.3
SOLUTION:
• Write the kinematic relationships for the
dependent motions and accelerations of
the blocks.
• Write the equations of motion for the
blocks and pulley.
• Combine the kinematic relationships
with the equations of motion to solve for
the accelerations and cord tension.
The two blocks shown start from rest.
The horizontal plane and the pulley are
frictionless, and the pulley is assumed
to be of negligible mass. Determine
the acceleration of each block and the
tension in the cord.
 x A  2 yB  constant
O
x
SOLUTION 1:
• Write the kinematic relationships for the dependent
motions and accelerations of the blocks.
yB   12 x A
aB  12 a A
y
• Write equations of motion for blocks and pulley.
 Fx  m A a A :
T1  100 kg a A
 Fy  mB a B :
m B g  T2  m B a B
300 kg 9.81 m s 2  T2  300 kg a B
T2  2940 N - 300 kg a B
 Fy  mC aC  0 :
T2  2T1  0
• Combine kinematic relationships with equations of
motion to solve for accelerations and cord tension.
O
x
y
y B  12 x A
a B  12 a A
T1  100 kg a A
T2  2940 N - 300 kg a B

 2940 N - 300 kg  12 a A

T2  2T1  0
2940 N  150 kg a A  2100 kg a A  0
a A  8.40 m s 2
a B  12 a A  4.20 m s 2
T1  100 kg a A  840 N
T2  2T1  1680 N
• 2nd way:
y
x
SOLUTION: 2
• Write the kinematic relationships for the dependent
motions and accelerations of the blocks.
2 yB  x A  c
2a B  a A  0
• Write equations of motion for blocks and pulley.
 Fx  m A a A :
T1  100 kg a A
 Fy  mB a B :
T2  mB g  mB a B


T2  300 kg  9.81 m s 2  300 kg a B
T2  300 kg a B  2940 N
 Fy  mC aC  0 :
2T1  T2  0
y
x
• Combine kinematic relationships with equations of
motion to solve for accelerations and cord tension.
2 yB  xA  0
T1  100 kg a A
2a B  a A  0
T2  300 kg a B  2940 N


 300 kg   12 a A  2940 N
2T1  T2  0
2100 kg a A  150 kg a A  2940 N  0
a A  8.40 m s 2
a B   12 a A  4.20 m s 2
T1  100 kg a A  840 N
T2  2T1  1680 N
• 2nd way:
SOLUTION: 3 Dynamic equilibrium-D’Alembert’s
Principle
• Write the kinematic relationships for the dependent
motions and accelerations of the blocks.
2 yB  x A  c
2a B  a A  0
• Write equations of motion for blocks and pulley.
y
x
W
-mAaA
0
T1
=
N
T2
0
=
W
-mBaB
F
x
 0:
F
y
 0:
T1  m A a A  0
T1  100 kg a A  0
T2  mB g  mB a B  0


T2  300 kg  9.81 m s 2  300 kg a B  0
T2  300 kg a B  2940 N
F
y
 0:
2T1  T2  0
a A   8 . 40 m s 2
aB  
1
2
a A   4 . 20 m s 2
T1  100 kg a A  840 N
T 2   2T1   1680 N
x
m1
m1
m1 a
T1  T2  m1 a
T2  T3  m1 a
T3  m1 a
T3
T3
m1
m1
F  T1  m1 a
m1
T2
T2
m1
T1
T1
m1
F
F
T1
m1
F
In statics: this is a 2-F member. The forces should be along the same line and
They should be equal in magnitude but in opposite directions
In dynamics: they do not have to be on the same line and have equal magnitude
The resultant force causes an acceleration
Sample Problem 12.4
SOLUTION:
• The block is constrained to slide down
the wedge. Therefore, their motions are
dependent. Express the acceleration of
block as the acceleration of wedge plus
the acceleration of the block relative to
the wedge.
The 6 kg block B starts from rest and
slides on the 15 kg wedge A, which is
supported by a horizontal surface.
Neglecting friction, determine (a) the
acceleration of the wedge, and (b) the
acceleration of the block relative to the
wedge.
• Write the equations of motion for the
wedge and block.
• Solve for the accelerations.
SOLUTION:
• The block is constrained to slide down the
wedge. Therefore, their motions are dependent.



aB  a A  aB A
• Write equations of motion for wedge and block.
 Fx  m A a A :
y
x
• Solve for the accelerations.
a A  0.15 m s 2
aB
A
 5 m s2
Sample Problem 12.5
SOLUTION:
• Resolve the equation of motion for the
bob into tangential and normal
components.
• Solve the component equations for the
normal and tangential accelerations.
The bob of a 2-m pendulum describes
an arc of a circle in a vertical plane. If
the tension in the cord is 2.5 times the
weight of the bob for the position
shown, find the velocity and acceleration of the bob in that position.
• Solve for the velocity in terms of the
normal acceleration.
SOLUTION:
• Resolve the equation of motion for the bob into
tangential and normal components.
• Solve the component equations for the normal and
tangential accelerations.
mg sin 30  mat
 Ft  mat :
at  g sin 30
 Fn  man :
at  4.9 m s 2
2.5mg  mg cos 30  man
an  g 2.5  cos 30
an  16.03 m s 2
• Solve for velocity in terms of normal acceleration.
an 
a  16 . 76 m / s
2
v2

v  an 
2 m 16.03 m s 2 
v  5.66 m s
Sample Problem 12.6
SOLUTION:
• The car travels in a horizontal circular
path with a normal component of
acceleration directed toward the center
of the path.The forces acting on the car
are its weight and a normal reaction
from the road surface.
Determine the rated speed of a
highway curve of radius  = 120 m
banked through an angle  = 18o. The
rated speed of a banked highway curve
is the speed at which a car should
travel if no lateral friction force is to
be exerted at its wheels.
• Resolve the equation of motion for
the car into vertical and normal
components.
• Solve for the vehicle speed.
• Resolve the equation of motion for
the car into vertical and normal
components.
R cos  W  0
 Fy  0 :
R
W
cos
 Fn  man : R sin   man
SOLUTION:
• The car travels in a horizontal circular
path with a normal component of
acceleration directed toward the center
of the path.The forces acting on the
car are its weight and a normal
reaction from the road surface.
W
v2
sin   m

cos 
• Solve for the vehicle speed.
v 2  g tan 


 9.81 m s 2 120 m  tan 18
v  20 m s  72 km h
Angular momentum of a particle. Rate of change of angular momentum
HO
y
mv

O
z
r
The angular momentum HO of a particle about point O is
defined as the moment about O of the linear momentum mv of
that particle.(moment of momentum or angular momentum)
P
x
We note that HO is a vector perpendicular to the plane
containing r and mv and of magnitude
Resolving the vectors r and mv into rectangular components, we express the angular momentum HO in
determinant form as
(m)(kg.m/s)=kg.m2/s
(ft)(lb.s)=ft.lb.s
y
mv
HO
O
i
HO = x
mvx
mv

 r
mvr
P
j
k
y
z
mvy mvz
x
z
In the case of a particle moving in the xy plane, we have z = vz = 0. The angular
momentum is perpendicular to the xy plane and is completely defined by its
magnitude
HO = rmv sin = rmv
.
.
=rmr=mr2
.
Computing the rate of change H
Radial and transverse components
of the angular momentum HO,
and applying Newton’s second law
O
HO = Hz = m(xvy - yvx)
.
MO = HO
which states that the sum of the moments about O of the forces acting on a particle is equal to the rate
of change of the angular momentum of the particle about O.
mv
If linear momentum is constant, there is no force acting on the particle.
the speed is constant,
the direction is also constant.
v: constant, direction: constant
Linear momentum is not constant
the speed is constant
the direction is not constant
Since the linear momentum is not constant, there should be a force
1. in the direction of motiontangential acceleration: tangential force
2. in the perpendicular direction of the motionnormal acceleration: normal force
mv
r
If angular momentum is constant, there is no moment acting on the particle.
But there may be a force acting through the center of rotation
As a result the angular momentum stays constant
ax
ay
The gravitational force between earth and the sun does not affect the angular
momentum of the earth. Since the force passes through one of the centers of the
ellipse.
er
ar
a=0
The gravitational force between earth and the sun does not affect the angular
momentum of the earth. Since the force passes through one of the centers of the
ellipse.
en
at
an
The gravitational force between earth and the sun does not affect the angular
momentum of the earth. Since the force passes through one of the centers of the
ellipse.
Angular momentum is not constant
the speed is not constant
There should be a force in a different direction than the radial direction
F
r
.
MO = HO
Eqs of Motion in Radial & Transverse Components
• Consider particle at r and , in polar coordinates,
• This result may also be derived from conservation
of angular momentum,
ME402-Kettering33motioncenterofmass
Motion under a central force. Conservation of angular momentum
mv

P
mv0
r
0
O
r0
P0
When the only force acting on a particle P is a force
F directed toward or away from a fixed point O, the
particle is said to be moving under a central force.
Since  MO = 0 at any given instant, it follows that
HO = 0 for all values of t, and
.
HO = constant
We conclude that the angular momentum of a particle moving under a central force is constant, both in
magnitude and direction, and that the particle moves in a plane perpendicular to HO .
mv

P
mv0
r
0
O
r0
Recalling that HO = rmv sin , we have, for points PO and P
rmv sin = romvo sin o
for the motion of any particle under a central force.
P0
.
.
Using polar coordinates and recalling that v = r and HO = mr2, we have
.
r2 = h
where h is a constant representing the angular momentum per unit mass Ho/m, of the particle.
mv

P
mv0
r
0
O
r0
.
r2 = h
P0
The infinitesimal area dA swept by the radius
vector OP as it rotates through d is equal to r2d/2
and, thus, r2 represents twice the areal velocity
dA/dt of the particle. The areal velocity of a particle
moving under a central force is constant.
.
r d
dA
d
F
O
P
• Radius vector OP sweeps infinitesimal area
dA  12 r 2 d

• Define
dA 1 2 d 1 2 
 2r
 2 r   areal velocity
dt
dt
Newton’s Law of Gravitation
r
m
F
-F
M
• Gravitational force
exerted by the sun on a
planet or by the earth on a
satellite is an important
example of gravitational
force.
An important application of the motion under a central force is
provided by the orbital motion of bodies under gravitational attraction.
According to Newton’s law of universal gravitation, two particles at a
distance r from each other and of masses M and m, respectively, attract
each other with equal and opposite forces F and -F directed along the
line joining the particles. The magnitude F of the two forces is
where G is the constant of gravitation. In the case of a body of mass m
subjected to the gravitational attraction of the earth, the product GM,
where M is the mass of the earth, is expressed as
where g = 9.81 m/s2 = 32.2 ft/s2 and R is the radius of the earth.
F G
Mm
r2
G  constant of gravitation
 66.73  10
12
m3
kg  s
2
 34.4  10
9
ft 4
lb  s 4
• For particle of mass m on the earth’s surface,
W m
MG
R2
 mg
g  9.81
m
s2
 32.2
ft
s2
Sample Problem 12.7
SOLUTION:
• Write the radial and transverse
equations of motion for the block.
• Integrate the radial equation to find an
expression for the radial velocity.
A block B of mass m can slide freely on
a frictionless arm OA which rotates in a
horizontal plane at a constant rate 0 .
Knowing that B is released at a distance
r0 from O, express as a function of r
a) the component vr of the velocity of B
along OA, and
b) the magnitude of the horizontal force
exerted on B by the arm OA.
• Substitute known information into the
transverse equation to find an
expression for the force on the block.
• Integrate the radial equation to find an
expression for the radial velocity.
SOLUTION:
• Write the radial and transverse
equations of motion for the block.


 Fr  m ar : 0  m r  r 2
 F   m a : F  mr  2r 

vr2  02 r 2  r02

• Substitute known information into the
transverse equation to find an expression
for the force on the block.

F  2m02 r 2  r02

12
Sample Problem 12.8
SOLUTION:
• Since the satellite is moving under a
central force, its angular momentum is
constant. Equate the angular momentum
at A and B and solve for the velocity at B.
A satellite is launched in a direction
parallel to the surface of the earth
with a velocity of 18820 mi/h from
an altitude of 240 mi. Determine the
velocity of the satellite as it reaches it
maximum altitude of 2340 mi. The
radius of the earth is 3960 mi.
SOLUTION:
• Since the satellite is moving under a
central force, its angular momentum is
constant. Equate the angular momentum
at A and B and solve for the velocity at B.
v B  12550 mi h
Trajectory of a Particle Under a Central Force
• Second expression is equivalent to r 2  h  constant, from which,
• After substituting into the radial equation of motion and simplifying,
• If F is a known function of r or u, then particle trajectory may be
found by integrating for u = f(), with constants of integration
determined from initial conditions.
r

A
A particle moving under a central
force describes a trajectory defined
by the differential equation
O
d 2u + u = F
mh 2u 2
d2
where F > 0 corresponds to an attractive force and u = 1/r. In the
case of a particle moving under a force of gravitational attraction,
we substitute F = GMm/r2 into this equation. Measuring  from the
axis OA joining the focus O to the point A of the trajectory closest
to O, we find
1
GM
= u = 2 + C cos 
r
h
1
GM
= u = 2 + C cos 
r
h
r

A
This is the equation of a conic of
eccentricity  = Ch2/GM. The
conic is an ellipse if  < 1, a
parabola if  =1, and a hyperbola
if  > 1. The constants C and h can
be determined from the initial conditions; if the particle is
projected from point A with an initial velocity v0 perpendicular to
OA, we have h = r0v0.
O
The values of the initial velocity corresponding, respectively, to
a parabolic and circular trajectory are
2GM
vesc =
r0
GM
vcirc =
r0
• Trajectory of earth satellite is defined by
Application to Space Mechanics
• Integration constant C is determined by conditions
at beginning of free flight,  =0, r = r0 ,
• Satellite escapes earth orbit for
  1 or C  GM h 2  GM r0 v0 2
vesc  v0 
2GM
r0
• Trajectory is elliptic for v0 < vesc and becomes
circular for  = 0 or C = 0,
• Recall that for a particle moving under a central
force, the areal velocity is constant, i.e.,
dA 1 2  1
 2 r   2 h  constant
dt
• Periodic time or time required for a satellite to
complete an orbit is equal to area within the orbit
divided by areal velocity,
where
r

O
A
2GM
vesc=
r0
GM
vcirc=
r0
vesc is the escape velocity, which is
the smallest value of v0 for which
the particle will not return to its
starting point.
The periodic time  of a planet or satellite is defined as the time
required by that body to describe its orbit,
2ab
=
h
where h = r0v0 and where a and b represent the semimajor and
semiminor axes of the orbit. These semiaxes are respectively
equal to the arithmetic and geometric means of the maximum
and minimum values of the radius vector r.
Sample Problem 12.9
SOLUTION:
• Trajectory of the satellite is described by
Evaluate C using the initial conditions
at  = 0.
A satellite is launched in a direction
parallel to the surface of the earth
with a velocity of 36,900 km/h at an
altitude of 500 km.
Determine:
a) the maximum altitude reached by
the satellite, and
b) the periodic time of the satellite.
• Determine the maximum altitude by
finding r at  = 180o.
• With the altitudes at the perigee and
apogee known, the periodic time can
be evaluated.
SOLUTION:
• Trajectory of the satellite is described by
1 GM
 2  C cos
r
h
Evaluate C using the initial conditions
at  = 0.
1 GM
C  2
r0 h

1
6.87  10 m
6

 65.3  109 m -1
398  1012 m3 s 2
70.4 m2 s2
• Determine the maximum altitude by finding r1
at  = 180o.
max altitude  66700 - 6370 km  60300 km
• With the altitudes at the perigee and apogee known,
the periodic time can be evaluated.
  70.3  103 s  19 h 31min
Kepler’s Laws of Planetary Motion
• Results obtained for trajectories of satellites around earth may also be
applied to trajectories of planets around the sun.
• Properties of planetary orbits around the sun were determined
astronomical observations by Johann Kepler (1571-1630) before
Newton had developed his fundamental theory.
1) Each planet describes an ellipse, with the sun located at one of its
foci.
2) The radius vector drawn from the sun to a planet sweeps equal
areas in equal times.
3) The squares of the periodic times of the planets are proportional to
the cubes of the semimajor axes of their orbits.
Any difference ?
g
g
60 kg
160 kg
100 kg
200 kg
F
 m a1
(40)9.81  160 a1
F
 ma 2
(40)9.81  360 a 2
g
g
T
160 kg
T
(100)9.81 N
F
 ma
T  (60)9.81  60 a
(100)9.81  T  100 a
(60)9.81 N
200 kg
F
 ma
T  (160)9.81  160 a
(200) * 9.81  T  200 a
g
150 kg
100 kg
100 kg
150 kg
F=500 N
F=500 N
Any difference ?
g
150 kg
100 kg
100 kg
F=500 N
50 kg
F=500 N
Any difference ?
g
150 kg
100 kg
100 kg
W=500 N
150 kg
W=500 N
g
m2=50 kg
m3= 75 kg
m1=25 kg
F
F+m1g+m2g-m3/2*g=((50+25)+75/2)4
2(F+m1g+m2g)-m3*g=((50+25)*2+75)8
T5
75a
=
75*9.81 N
T5  75 * 9.81  75 a  75 * 2
T3
T4
T2=T3=T4=T5/2
T5
T2
=
m2=50 kg
T1
m2g
T1  m 2 g  T2  m 2 a
m1a
T1
=
F
m1g
m1=25 kg
m1a
F  m1 g  T1  m1 a
clear all
clc
format long g
m1=25;m2=50;m3=75;a1=4;a2=4;a3=2;g=9.81;
km2=[1 -1 0 0;
0 1 -1 0;
0 0 2 -1;
0 0 0 1];
F  T1  m1 a  m1 g
st2=[m1*(a1-g);m2*(a2-g);0;m3*(g+a3)];
T1  T2  m 2 a  m 2 g
unknowns2=inv(km2)*st2
2T2-T5=0
T5  m 3 * a 3  m 3 g
A=solve('F-T1=25*(4-9.81)','T1-T2=50*(4-9.81)','2*T2T5=0','T5=75*(9.81+2)');
bilinmeyenler1=eval([A.F A.T1 A.T2 A.T5]')
mw g  F  mw a
OR
m1 g  m w g  T1  m1 a  m w a
F
T1
=
=
mwg
F
m1g
m1a
m1=25 kg
F  m1 g  T1  m1 a
mwa
B=solve('mw*(9.81-4)-T1=25*(4-9.81)','T1-T2=50*(4-9.81)','2*T2T5=0','T5=75*(9.81+2)');
bilinmeyen3=eval([B.mw*9.81 B.T1 B.T2 B.T5]')
km4=[(9.81-4) -1 0 0;
0
1 -1 0;
0
0 2 -1;
0
0 0 1];
st4=[m1*(a1-g);m2*(a2-g);0;m3*(g+a3)];
unknowns4=inv(km4)*st4;
g
100 kg
100 kg
m=50 kg
F=500 N
150 kg
150 kg
F=500 N
F
n
T
 man
T  mg cos 60o  man
60o
t
mg sin 60o  mat
mg sin   mat
 g sin   at
vdv  at ds
mg
n
vdv  at Ld
v
0o
0
60o
 vdv    g sin  Ld
v2
v
0o
 gL cos  60o
2
v 2  2 gL  cos 0o  cos 60o 
0
 1
v 2  2  9.81 1  
 2
v  2  9.81
1
 3.132m/s
2
Fs   s N
  s mg
  0.4 10  9.81  39.24N
Fk  k N
 k mg
  0.310  9.81  29.43N
Assume Fapplied  Fs  0.01N
 F  ma
F  Fk  ma
39.24  0.01  29.43
 9.82m/s 2
a
10
It the truck accelerates there will be a force namely friction force between the crate
and the truck. If the force is static friction force, there is no motion. If we have
Kinetic friction force, this means friction.
So if the friction force due to acceleration of the truck which is the inertia force
is larger than the static friction then the crate slides and drops.
W
Fs  s N  ma
a
s mg
 s g
m
 0.4 g  3.924m/s2
Fs
N
 F  ma
F  Fk  ma
F  Fk  ma
 k mg  ma
 30.43N
F  Fs  s mg
  0.410 9.81
 39.24N
i
j
k
HO  m rx
ry
rz
vx vy
vz
 ry
 mi
 vy

rx ry 
rx rz
j
k

vz
vx vy 
vx vz
rz
NA
s NA
s NB
W
NB
•
A mass is falling from a height of 11 m. At the bottom, there is a
spring with k = 1000 N/m and free length of 1 m. What is the
maximum velocity of the mass after it hits the spring.
h=11 m
l=1 m
•
A mass (m=10 kg) is falling from a height of 11 m. At the bottom,
there is a spring with k = 1000 N/m and free length of 1 m. What is
the maximum velocity of the mass after it hits the spring.
W
F=ma
W=ma
mg=ma
a=g is valid until collision
h=11 m
W
l=1 m
F=ma
W-k(x=0.001)=ma
mg-k0.001=ma
a=?
W
F=ma
W-k(x=0.002)=ma
mg-k0.002=ma
a=?
Find the location maximum velocity?
F=ma
W-k(x)=ma
mg-kx=ma
a=0
x=(mg-0)/k
Find magnitude of the maximum velocity?
g
ay
ar
a

at
an
ax
ax=0
g
ay= -9.81 m/s2
ar
a

anay
at