Download Chapter 12

ENGR 214
Chapter 12
Kinetics of Particles:
Newton’s Second Law
All figures taken from Vector Mechanics for Engineers: Dynamics, Beer
and Johnston, 2004
1
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 resultant and
in the direction of the resultant.
F  ma
• If particle is subjected to several forces:
 F  ma
• We must use a Newtonian frame of reference, i.e., one that is
not accelerating or rotating.
• If no force acts on particle, particle will not accelerate, i.e., it will
remain stationary or continue on a straight line at constant
velocity.
2
Linear Momentum of a Particle
dv
 F  ma  m dt
d
d
  mv  
L
dt
dt
 
L  mv
Linear momentum
Sum of forces = rate of change of linear momentum
If
F  0
F  L
linear momentum is constant
Principle of conservation of linear momentum
3
Equations of Motion


 F  ma
• Newton’s second law
• Convenient to resolve into components:






 Fx i  Fy j  Fz k  m a x i  a y j  a z k




 Fx  max  Fy  ma y  Fz  maz
 Fx  mx  Fy  my  Fz  mz
• For tangential and normal components:
F
F
t
t
 mat
dv
m
dt
F
F
n
n
 man
m
v2

4
Dynamic Equilibrium
• Alternate expression of Newton’s law:
ma
 F  ma  0
inertia vector
• If we include inertia vector, the system of
forces acting on particle is equivalent to
zero. The particle is said to be in
dynamic equilibrium.
• Inertia vectors are often called inertia
forces as they measure the resistance
that particles offer to changes in
motion.
5
Sample Problem 12.2
SOLUTION:
• Draw a free body diagram
• Apply Newton’s law. Resolve
into rectangular components
An 80-kg block rests on a horizontal plane. Find
the magnitude of the force P required to give the
block an acceleration of 2.5 m/s2 to the right. The
coefficient of kinetic friction between the block and
plane is mk = 0.25.
6
Sample Problem 12.2
 Fx  ma :
Psin30
P cos30  0.25 N  80  2.5 
Pcos30
 200
W  mg  80  9.81  785 N
F  mk N  0.25 N
 Fy  0 :
N  P sin30  785  0
Solve for P and N
N  P sin30  785
P cos30  0.25  P sin30  785  200
P  534.7 N
N  1052.4 N
7
Sample Problem 12.3
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.
8
Sample Problem 12.3
• Kinematic relationship: If A moves xA
to the right, B moves down 0.5 xA
O
x
y
xB  12 x A
aB  12 a A
Draw free body diagrams & apply Newton’s law:
F
F
x
 mAaA
T1  100  aA
y
 mB aB
mB g  T2  mB aB
F
y
300  9.81  T2   300  aB
T2  2940-  300  aB
 mC aC
2940-  300  aB  2T1  0
T2  2T1  0
2940-  300  aB  200aA  0
2940-  300  aB  2  200aB  0
2
aB  4.2 m / s 2 aA  8.4 m / s
T1  840 N
T2  1680 N
9
Sample Problem 12.4
Block
Wedge
The 12-lb block B starts from rest and slides
on the 30-lb 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.
10
Draw free body diagrams for block & wedge
N1sinq
N1
N1
N cosq
1
aBn
WBsinq
WB WBcosq
aBt
WB sin q  mB aBt
12  0.5 
12
aBt
32.2
aA
 aBt  16.1 ft / s 2
N1 cosq  WA  N 2
N1  WB cosq  mB aBn
But
aBn  a A sin q
30
0.5 N1 
aA
32.2
Same normal acceleration (to maintain contact)
N1  WB cosq  mB a A sin q
a A  5.08 ft / s 2
N1 sin q  mAa A
N1  10.39  
12  0.5
aA
32.2
aBn  2.54 ft / s 2
11
N1sinq
N1
aBn
aBt
N1
N1cosq
WBsinq
WB WBcosq
aA
aBx  aBt cosq  aBn sin q  12.67 ft / s 2
aBy  aBt sinq  aBn cosq  10.25 ft / s
aB / A   12.67i  10.25 j    5.08i 
 17.75i  10.25 j
2
aB / A  aB  a A
30°
20.5
12
Sample Problem 12.5
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.
13
Sample Problem 12.5
Resolve into tangential and normal components:
 Ft  mat :
mg sin 30  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.
mgsin30
an 
mgcos30
v2

v  an 
2 m 16.03 m s 2 
v  5.66 m s
14
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  = 400 ft
banked through an angle q = 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.
15
Sample Problem 12.6
• Resolve the equation of motion for
the car into vertical and normal
components.
R cosq  W  0
 Fy  0 :
W
R
cosq
 Fn  man : R sin q 
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
an
g
W
W v2
sin q 
cosq
g 
• Solve for the vehicle speed.
v 2  g tan q


 32.2 ft s 2 400 ft  tan 18
v  64.7 ft s  44.1 mi h
16
Angular Momentum
From before, linear momentum: L  mv
Now angular momentum is defined as the moment of momentum
H O  r  mv
H O is a vector perpendicular to the plane
containing r and mv
Resolving into radial & transverse components:
H O  mvq r  mr 2q
Derivative of angular momentum with respect to time:
H O  r  mv  r  mv  v  mv  r  ma
 r F
Moment of F about O
  MO
Sum of moments about O = rate of change of angular momentum
17
Equations of Motion in Radial & Transverse Components
 Fr  mar
 Fq  maq


2

 m r  rq
 mrq  2rq 
18
Central Force
When force acting on particle is directed
toward or away from a fixed point O, the
particle is said to be moving under a
central force.
O = center of force
Since line of action of the central force passes through O:
M
O
 HO  0
r  mv  H O  constant
19
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 q0 .
• Substitute known information into the
transverse equation to find an
expression for the force on the block.
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.
20
Sample Problem 12.7
r  rq 2
dvr dvr dr
dvr

 vr
r  vr 
dt
dr dt
dr
But vr  r
Write radial and transverse
equations of motion:
 Fr  m ar
Fq  m
aq


F  m  rq  2rq 
0  m r  rq 2
rq 2  vr
vr
dvr
dr
rq 2dr  vr dvr
r
2
v
dv

r
q
 r r  o dr
0
ro
vr2  q02  r 2  r02 
vr  q 0  r  r
F  2mq
2
0
r

2 12
0
2
2

2 12
0
r
21