Download Dynamics Notes - Blue Valley Schools

Dynamics
Review
Statics: The study of objects at equilibrium [Acceleration = Zero].
Rotational Equilibrium:
  0   net  Torque    0
Translational Equilibrium:
a  0  Fnet   Forces  0
 F  0, and
 F  0, and
F  0
x
y
z
You should have known (had in your head) all of the above!
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Dynamics: The study of the connection between force and motion.
“The study of why objects move.”
Newton’s Second Law:
Fnet = ma
Actually, defined in terms of acceleration …
Fnet
a
m
“acceleration is …
proportional to net force acting on an object, AND
inversely proportional to the mass of that object.”
Dynamics Example Problems:
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Example 4-3:
What average net force is required to bring a 1500-kg car to rest from a speed
of 100 km/h within a distance of 55 m?
m  ____________
x  ____________
vinitial  _________
v final  __________
a  ____________
t  _____________
F  ma  ________
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m 1500 kg
x   55 m
vinitial   28 m / s
v final  0 m / s
a   7.1 m / s / s
t  who cares!
F   1.1x10 4 N
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Example 4-6: Action-Reaction Force Pairs
Newton’s Third Law:
Whenever one object exerts a force on a second object, the second
object exerts a force in the opposite direction on the first.
Note:
Two different objects.
Equal but opposite.
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Fg = mg = (10.0 kg)(9.80 m/s/s) = 98.0 N
Determine FN in each case below:
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Example 4-7:
Let applied force to the crate be 100.0 N. Determine the acceleration of the
box.
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Example 4-19: Friction and Angled Forces:
A 10.0 kg box is pulled along a horizontal surface by a force of 40.0 N
applied at an angle of 30.0° [always assumed to be above the horizontal]. The
coefficient of kinetic friction of 0.30. Calculate the acceleration of the box.
Sketch the problem, showing all forces acting on the box.
[Hint: There are four forces acting on the box.]
Notes: The coefficient of friction.
  " mu" 
F friction
FNormal
 F f   FN
F f ,kinetic   k FN
F f ,static   s FN
 s   k  F f , s  F f ,k
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A 10.0 kg box is pulled along a horizontal surface by a force of 40.0 N
applied at an angle of 30.0°. The coefficient of kinetic friction of 0.30.
Calculate the acceleration of the box.
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Inclined Planes:
 Gravitational acceleration [g] is straight down,
thus, weight [FG] acts straight down!
(Gravity pulls toward the center of the Earth.)
 Normal Force is perpendicular to the surface!
 Thus, weight and normal force are NOT in the same direction!
The component of FG perpendicular to the plane is parallel to FN.
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 FG must be split into components.
 A suitable coordinate system must be chosen so as to determine FNet and a:
o Parallel to the plane and Perpendicular to the plane!
(Text shows x and y … I don’t like that convention … their x is parallel, y is perpendicular!)
Fnet ,   F
Fnet ,//   F//
ma  FN  Fg , ma//  Ff  Fg ,//
o NOT Horizontal and Vertical! (You could work it this way, but it would be a lot less fun.)
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Fnet ,   F
ma  FN  Fg ,  Fapplied,, if any
a  0  ma  0
Fg ,  Fg cos  mg cos


ma  FN  Fg ,
0 FN  mg cos  Fapplied,, if any
Fnet ,//   F//
ma//  F f  Fg ,//  Fapplied,//,if any 
Fg ,//  mg sin  
ma//    k FN  mg sin    Fapplied,//,if any 
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P(4): 52, 53: Inclined Planes.
52. The carton shown in Fig. 4–55 lies on a plane tilted at an angle   22 .0º to the
horizontal, with  k  0.12. (a) Determine the acceleration of the carton as it slides down
the plane. (b) If the carton starts from rest 9.30 m up the plane from its base, what will be
the carton’s speed when it reaches the bottom of the incline?
53. A carton is given an initial speed of 3.0 m s up the 22.0º plane shown in Fig. 4–55.
(a) How far up the plane will it go? (b) How much time elapses before it returns to its
starting point? Ignore friction. Rework, not ignoring friction.
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52.
(a)
Consider the free-body diagram for the carton on the surface.
There
is no motion in the y direction and thus no acceleration in the y direction.
Write Newton’s 2nd law for both directions.
 Fx  FN  mg cos  0  FN  mg cos 
F
x
FN
Ffr
y

 mg sin   Ffr  ma
x

mg
ma  mg sin    k FN  mg sin    k mg cos 
a  g  sin    k cos  

 9.80 m s 2
 sin 22.0
o

 0.12 cos 22.0o  2.58  2.6 m s 2
(b) Now use Eq. 2-11c, with an initial velocity of 0, to find the final velocity.

v 2  v02  2a  x  x0   v  2a  x  x0   2 2.58 m s 2
  9.30 m   6.9 m s
53.
(a)
Consider the free-body diagram for the carton on the
frictionless
surface. There is no acceleration in the y direction. Write Newton’s
2nd law for the x direction.
 Fx  mg sin  ma  a  g sin 
Use Eq. 2-11c with v0  3.0 m s and v  0 m s to find the distance
that it slides before stopping.
v 2  v02  2a  x  x0  
 x  x0  
v 2  v02
2a


0   3.0 m s 
FN
y
x


mg
2

2 9.8 m s 2 sin 22.0o
 1.2 m
The negative sign means that the block is displaced up the plane, which is the negative direction.
(b) The time for a round trip can be found from Eq. 2-11a. The free-body diagram (and thus the
acceleration) is the same whether the block is rising or falling. For the entire trip, v0  3.0 m s and
v  3.0 m s .
v  v0  at  t 
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v  v0
a

 3.0 m s    3.0 m s 
 9.8 m s  sin 22
2
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 1.6 s
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