Download Newton`s Second Law

Projectile Motion
Dave put a laser sight on his bow so he can
see directly where his arrow is pointed. If he
puts the laser on the bull’s eye, where will his
arrows land?
What will the path of the arrow look like?
Newton’s Second Law
N.S.L.
N.S.L. works based on these direct
and inverse relationships. As 2 of
the variable change, ONE of them
must remain constant.
If the force is constant, the
acceleration and mass change as
shown above.
F(net)=ma
2F=m(2a)
3F=m(3a)
If we add a second dog pulling
with 100N just like the first
dog, we could pull the sled
with twice the acceleration,
provided the mass of the sled
was constant.
Putting it all together
a  FNET
1
a
m
3N
10 N
10 kg
FNET
a
 FNET  ma
m
FNET  Total Force   F
FNET  0
Magnitude of FNET= 7 N
Direction = RIGHT
Acceleration = 0.70 m/s/s
N.S.L Tips
Draw a free body diagram
2.
Break vectors into components if needed
3.
Find the NET force by adding and subtracting
forces that are on the same axis as the
acceleration.
4.
Set net force equal to “ma” this is called
writing an EQUATION OF MOTION.
NOTE: To avoid negative numbers, always subtract
the smaller forces from the larger one.
1.
Example
An elevator with a mass of 2000 kg rises with an
acceleration of 1.0 m/s/s. What is the tension in the
supporting cable?
T
mg
FNET  ma
Equation of Motion
T  mg  ma
T  ma  mg
T  (2000)(1)  (2000)(9.8)
T  21,600 N
Example
A 50 N applied force drags an 8.16 kg log to the right
across a horizontal surface. What is the acceleration
of the log if the force of friction is 40.0 N?
Fn
a
50 N
40 N
mg
FNET  ma
Fa  F f  ma
50  40  8.16a
10  8.16a
a
1.23 m/s/s
Example
A sled is being accelerated to the right at a rate of 1.5 m/s/s by a
rope at a 33 degree angle above the + x . Calculate the
Frictional Force if the mass of the sled is 66 kg and the tension
in the rope is 150 N.
FN
Tsinq
q
Tcosq
FNET  ma
T cos q  F f  ma
Ff
mg
T cos q  ma  F f
150 cos 33  (66)(1.5)  F f
Ff 
26.8 N
Friction & Inclined Planes
TWO types of Friction


Static – Friction that keeps an object at rest
and prevents it from moving
Kinetic – Friction that acts during motion
Force of Friction

The Force of Friction is F f  FN
  constant of proportion ality
directly related to the
Force Normal.
  coefficien t of friction
 Mostly due to the fact
Fsf   s FN
The coefficient of
that BOTH are surface
forces
Fkf   k FN
friction is a unitless
constant that is
specific to the
material type and
usually less than
one.
Note: Friction ONLY depends on the MATERIALS sliding against
each other, NOT on surface area.
Friction & N.F.L
If the coefficient of kinetic friction between a 35-kg crate and the floor is
0.30, what horizontal force is required to move the crate to the right at
a constant speed across the floor?
Fa  Ff
Fn
Fa
Ff  k FN
Fa  k FN
FN  mg
Ff
Fa  k mg
mg
Fa  (0.30)(35)(9.8)
Fa 
102.9 N
Friction & N.S.L.
Suppose the same 35 kg crate was not moving at a constant speed, but
rather accelerating at 0.70 m/s/s. Calculate the applied force. The
coefficient of kinetic friction is still 0.30.
FNET  ma
Fa  Ff  ma
Fn
Fa
Fa  k FN  ma
Fa  k mg  ma
Fa  ma  k mg
Ff
Fa  (35)(0.70)  (0.30)(35)(9.8)
mg
Fa 
127.4 N
Inclines
q
Ff
FN
mg cos q
q
q
mg q
mg sin q
q
q
Tips
•Rotate Axis
•Break weight into components
•Write equations of motion or
equilibrium
•Solve
Friction & Inclines
A person pushes a 30-kg shopping cart up a 10 degree incline with a force of 85
N. Calculate the coefficient of friction if the cart is pushed at a constant
speed.
Fa  Ff  mg sin q
Fa
Fn
Ff  k FN
Fa  k FN  mg sin q FN  mg cos q
Fa  k mg cos q  mg sin q
Fa  mg sin q  k mg cos q
mg cos q
q
Ff
mg
mg sin q
q
Fa  mg sin q
k 
mg cos q
85  (30)(9.8)(sin10)
k 

(30)(9.8)(cos10)
0.117
What steps do we take to
solve problems with objects
experiencing forces in
multiple dimensions?
Example
A 5-kg block sits on a 30 degree incline. It is attached to string that is thread
over a pulley mounted at the top of the incline. A 7.5-kg block hangs from
the string.
 a) Calculate the tension in the string if the acceleration of the system is 1.2
m/s/s
 b) Calculate the coefficient of kinetic friction.
T
FN
FNET  ma
m1 g  T  m1a
m2gcos30
30
T
m2g
m1
30
m2gsin30
m1g
T  ( Ff  m2 g sin q )  m2 a
Ff
FN  m2 g cos q
Example
FNET  ma
m1 g  T  m1a
m1 g  m1a  T
T  ( Ff  m2 g sin q )  m2 a
T  F f  m2 g sin q  m2 a
T  k FN  m2 g sin q  m2 a
T  m2 a  m2 g sin q  k FN
(7.5)(9.8)  (7.5)(1.2)  T T  m a  m g sin q
2
2
 k
T  64.5 N
FN
FN  m2 g cos q
T  m2 a  m2 g sin q
 k
m2 g cos q
64.5  (5)(1.2)  (5)(9.8)(sin 30)
 k
(5)(9.8)(cos 30)
k  0.80 N
Kinematics - Analyzing motion under
the condition of constant acceleration
Kinematic Symbols
x,y
Displacement
t
Time
vo
Initial Velocity
v
Final Velocity
a
Acceleration
g
Acceleration due to
gravity
Kinematic #1
v  vo
v
a

t
t
v  vo  at
v  vo  at
Kinematic #1
Example: A boat moves slowly out of a marina (so as to not
leave a wake) with a speed of 1.50 m/s. As soon as it
passes the breakwater, leaving the marina, it throttles up
and accelerates at 2.40 m/s/s.
a) How fast is the boat moving after accelerating for 5 seconds?
What do I
know?
vi= 1.50 m/s
a = 2.40 m/s/s
t=5s
What do I
want?
v=?
v  vi  at
v  (1.50)  (2.40)(5)
v  13. 5 m/s
Kinematic #2
2
1
x  vixt  at
2
b) How far did the boat travel during that time?
2
1
x  vixt  at
2
2
1
x  (1.5)( 5)  (2.40)( 5 )
2
x  37.5 m
Kinematic #3
v  v  2ax
2
2
o
Example: You are driving through town at 12 m/s when suddenly a ball rolls
out in front of your car. You apply the brakes and begin decelerating at
3.5 m/s/s.
How far do you travel before coming to a complete stop?
What do I
know?
vo= 12 m/s
a = -3.5 m/s/s
V = 0 m/s
What do I
want?
x=?
v 2  vo2  2ax
0  12 2  2(3.5) x
 144  7 x
x  20.57 m
Choosing the Right Equation
Equation
Missing Variable
v  vo  at
x
x  voxt  1 at
2
2
v 2  vo2  2ax
v
t
Kinematics for the VERTICAL Direction
All 3 kinematics can be used to analyze one
dimensional motion in either the X direction OR the y
direction.
v  vi  at  v y  viy  g t
2
2
1
1
x  vixt 
at  y  viyt 
gt
2
2
2
2
2
2
v  vix  2ax  v y  viy  2 gy
Examples
A pitcher throws a fastball with a velocity of 43.5 m/s. It is
determined that during the windup and delivery the ball covers
a displacement of 2.5 meters. This is from the point behind the
body to the point of release. Calculate the acceleration during
his throwing motion.
What do I
know?
vi= 0 m/s
What do I
want?
a=?
Which variable is NOT given and
NOT asked for? TIME
v  v  2ax
2
2
i
x = 2.5 m
V = 43.5 m/s
43.5  0  2a(2.5)
2
2
a  378.45 m / s
2
Examples
How long does it take a car at rest to cross a 35.0 m
intersection after the light turns green, if the acceleration
of the car is a constant 2.00 m/s/s?
What do I
know?
vo= 0 m/s
x = 35 m
a = 2.00 m/s/s
What do I
want?
t=?
Which variable is NOT given and
NOT asked for? Final Velocity
x  vixt  1 at 2
2
35  (0)  1 (2)t 2
2
t  5.92 s
Examples
A car accelerates from 12.5 m/s to 25 m/s in 6.0
seconds. What was the acceleration?
What do I
know?
vo= 12.5 m/s
v = 25 m/s
t = 6s
What do I
want?
a=?
Which variable is NOT given and
NOT asked for?
DISPLACEMENT
v  vo  at
25  12.5  a(6)
a  2.08 m / s 2
Examples
A stone is dropped from the top of a cliff. It is
observed to hit the ground 5.78 s later. How high
is the cliff?
What do I
know?
v = 0 m/s
oy
g = -9.8 m/s2
t = 5.78 s
What do I
want?
y=?
Which variable is NOT given and
NOT asked for?
Final Velocity
y  viyt  1 gt 2
2
y  (0)( 5.78)  4.9(5.78)2
y  163.7 m
h  163.7 m
What is projectile?
Projectile -Any object which projected by some
means and continues to move due to its own
inertia (mass).
Projectiles move in TWO dimensions
Since a projectile
moves in 2dimensions, it
therefore has 2
components just
like a resultant
vector.
 Horizontal and
Vertical
Vertical “Velocity” Component

Changes (due to gravity), does NOT cover
equal displacements in equal time periods.
Both the MAGNITUDE and DIRECTION change. As
the projectile moves up the MAGNITUDE
DECREASES and its direction is UPWARD. As it
moves down the MAGNITUDE INCREASES and the
direction is DOWNWARD.
Horizontal “Velocity” Component

NEVER changes, covers equal displacements in
equal time periods. This means the initial
horizontal velocity equals the final horizontal
velocity
In other words, the horizontal
velocity is CONSTANT. BUT
WHY?
Gravity DOES NOT work
horizontally to increase or
decrease the velocity.
Combining the Components
Together, these
components produce
what is called a
trajectory or path. This
path is parabolic in
nature.
Component Magnitude
Direction
Horizontal
Constant
Constant
Vertical
Changes
Changes
Horizontally Launched Projectiles
Projectiles which have NO upward trajectory and NO initial
VERTICAL velocity.
vox  vx  constant
voy  0 m / s
Horizontally Launched Projectiles
To analyze a projectile in 2 dimensions we need 2
equations. One for the “x” direction and one for
the “y” direction. And for this we use kinematic #2.
2
1
x  voxt  at
2
x  voxt
Remember, the velocity is
CONSTANT horizontally, so
that means the acceleration
is ZERO!
y  1 gt 2
2
Remember that since the
projectile is launched
horizontally, the INITIAL
VERTICAL VELOCITY is
equal to ZERO.
Horizontally Launched Projectiles
Example: A plane traveling with
a horizontal velocity of 100
m/s is 500 m above the
ground. At some point the
pilot decides to drop some
supplies to designated
target below. (a) How long is
the drop in the air? (b) How
far away from point where it
was launched will it land?
y  1 gt 2  500  1 (9.8)t 2
2
2
102.04  t 2  t  10.1 seconds
What do I
know?
What I want to
know?
vox=100 m/s
t=?
x=?
y = 500 m
voy= 0 m/s
g = -9.8 m/s/s
x  voxt  (100)(10.1) 
1010 m
Vertically Launched Projectiles
NO Vertical Velocity at the top of the trajectory.
Vertical
Velocity
decreases
on the way
upward
Vertical Velocity
increases on the
way down,
Horizontal Velocity
is constant
Component Magnitude
Direction
Horizontal
Vertical
Constant
Changes
Constant
Decreases up, 0
@ top, Increases
down
Vertically Launched Projectiles
Since the projectile was launched at a angle, the
velocity MUST be broken into components!!!
vox  vo cos q
vo
q
vox
voy
voy  vo sin q
Vertically Launched Projectiles
There are several
things you must
consider when doing
these types of
projectiles besides
using components. If
it begins and ends at
ground level, the “y”
displacement is
ZERO: y = 0
Vertically Launched Projectiles
You will still use kinematic #2, but YOU MUST use
COMPONENTS in the equation.
vo
q
vox
voy
x  voxt
y  voyt  1 gt 2
2
vox  vo cos q
voy  vo sin q
Example
A place kicker kicks a football with a velocity of 20.0 m/s
and at an angle of 53 degrees.
(a) How long is the ball in the air?
(b) How far away does it land?
(c) How high does it travel?
vox  vo cos q
vox  20 cos 53  12.04 m / s
q  53
voy  vo sin q
voy  20sin 53  15.97 m / s
Example
A place kicker kicks a
football with a
velocity of 20.0 m/s
and at an angle of 53
degrees.
(a) How long is the ball
in the air?
What I know
What I want
to know
vox=12.04 m/s
t=?
x=?
voy=15.97 m/s
y=0
g = - 9.8
m/s/s
y  voy t  1 gt 2  0  (15.97)t  4.9t 2
2
2
15.97t  4.9t  15.97  4.9t
t  3.26 s
ymax=?
Example
A place kicker kicks a
football with a
velocity of 20.0 m/s
and at an angle of 53
degrees.
(b) How far away does it
land?
What I know
vox=12.04 m/s
voy=15.97 m/s
y=0
g = - 9.8
m/s/s
x  voxt  (12.04)(3.26) 
What I want
to know
t = 3.26 s
x=?
ymax=?
39.24 m
Example
A place kicker kicks a
football with a velocity
of 20.0 m/s and at an
angle of 53 degrees.
(c) How high does it
travel?
What I know
What I want
to know
vox=12.04 m/s
t = 3.26 s
x = 39.24 m
ymax=?
voy=15.97 m/s
y=0
g = - 9.8
m/s/s
y  voy t  1 gt 2
2
CUT YOUR TIME IN HALF! y  (15.97)(1.63)  4.9(1.63) 2
y  13.01 m