Transcript AP Physics Review

Physics Review
What Are Newton's Laws of Motion?
Forces
 “Push or Pull” that acts between two bodies
–
Tension
– Gravitational force
– Frictional force
– Air resistance
– Electrostatic force
– Strong nuclear force
– Weak nuclear force
The SI unit for force is the
Newton (N). This unit is
equivalent to 1 kgm
2
s
Newton’s First Law
 “Law of Inertia”
 An object will continue in its state of
motion unless compelled to change by a
force impressed upon it.
 What net force is required to maintain a
5000 kg object moving at a constant
velocity of magnitude 7500 m/s?
–
Net force of 0  constant motion
Newton’s Third Law
 For every action, there is an equal, but
opposite, reaction.
Weight
 The weight of an object is the gravitational
force exerted on it by Earth (or whatever
planetary mass the object is on).
 What is the mass of an object that weighs
500 N?
=
Fw
mg
Fw = 500 N =
=
m
51 kg
g
9.8 ms2
Other Examples
 A book with a mass of 2 kg rests on a
table. Find the magnitude of the force
exerted by the table on the book.
FN
Fg = mg = FN
(
)
Fg = (2 kg) 9.8 sm2 = 20 N = FN
Fg
Note: This force is called the Normal force
because it acts perpendicular to the contact
surface of the object.
Other Examples
 A can of paint with a mass of 6 kg hangs
from a rope. If the can is to be pulled up to
+
a rooftop with a constant velocity of 1 m/s,
direction
what must the tension in the rope be?
FT
FT = Fg = mg
(
)
F T = 6 kg 9 .8 sm2 = 59 N
Fg
Friction
 Friction is a contact force that is parallel to the
contact surface and perpendicular to the normal
force.
 Static friction (Fs) occurs when a force tries
unsuccessfully to set a body in motion.
 Kinetic (sliding) friction (Fk) occurs when a force
acts on a body in motion.
 Generally Fs >Fk
Friction Equations
Fs
(max)
= m s FN
Fk = m k FN
 m represents the coefficient of friction - a
number that is related to the nature of the
surfaces in contact with each other.
 Fs has a range of values dependent on the
magnitude of the horizontal force being
applied.
Friction Problems
 A crate of mass 20 kg is sliding across a
wooden floor. mk between the crate and
the floor is 0.3. Determine the strength of
the force acting on the crate.
FN F k = m k FN = m k mg
F k = (0.3) (20 kg) ( 9.8
Fk
F
Fg
m
s2
)=
59 N
If the crate is being pulled
by a force of 90 N
(parallel to the floor), find
the acceleration of the
crate.
Fnet = F  F f = 90 N  59 N = 31N
Fnet
Fnet 31N
= ma a =
=
= 1.6 sm2
m
20kg
Friction Problems
 A crate of mass 100 kg rests on the floor.
ms is 0.4. If a force of 250 N (parallel to
the floor) is applied to the crate, what is the
magnitude of Fs on the crate?
= m s FN = m s mg
F s, max
FN
(
)
m
(
)
(
)
=
100
kg
9
.
8
= 390 N
F s, max 0.4
s2
Fs
Fg
F NOTE: Remember that Fs represents a range of
values. In this case, the applied force of 250N
is less than the maximum Fs so the actual
magnitude of Fs is 250N.
Inclined Planes
 When a mass sits on an inclined plane, its
weight has two components: normal and
parallel to the plane.

Weight mg

mg cos  Normal force
mg sin  Parallel force
Tension
T1
Fg 1
T2
T2
Fg 2
Pulleys
Pulleys are devices that change the direction of the
tension force in cords that slide over them.
In problems, we generally ignore the mass and
friction associated with pulleys.
Pulleys
Step 2
Step 1
Fnet 1 = m1g - T
Fnet 2 = T – m2g
m1
Step 3
m1g - T = m1a
T - m2g = m2a
Step 4
a = m1-m2 g
m1 + m2
T = m1m2 g
m1 + m2
m2
T
T
m1
m2
m1g
m2g
Uniform Circular Motion
 Describes an object that moves in a circle with a
constant speed
 An object accelerates without changing its speed.
 The direction continuously changes as an object moves
around a circle
Circular Motion
v=
circumfere nce 2r
=
period
T
Uniform circular motion is a particle moving at constant
speed in a circle.
How do we describe the position of the particle?
The angle  is the
angular position.
Again  is defined to be
positive in the counter-clockwise direction.
Angles are usually measured in
radians.
s is arc length.
r is the radius of the circle.
 ( radians ) 
s
r
Circular Motion
Radians
 ( radians ) 
s
r
For a full circle.
s 2r
 fullcircle  =
= 2
r
r
rad
1 rev = 3600 = 2
rad
3600
1 rad = 1 rad
2 rad
s = r
Circular Motion
Angular velocity
The angular displacement is
   f   i
Average angular velocity
  f  i
  
=
t
t f  ti
Instantaneous angular velocity
limit  d


 t  0 t
dt
We will worry about the direction
later.
Like one dimensional motion +will do. Positive angular velocity
is counter-clock=wise.
Kinematical Equations
Conversion : x   , v   , a  
1 2
(1)  =  0   0t  t
2
(2)  =  0  t
(3)  =   2 ( -  0 )
2
2
0
Note :  = constant
Centripetal acceleration
2
v
ac =
r
directed toward the
center of curvature
(center of circle)
An object’s natural path
is a straight line,
therefore to pull an
object out of its natural
path a net force was
applied
Centripetal Force
This force could be any of our everyday forces (weight, normal,
tension, applied, or friction) or a combination of those five.
It is the force or forces that maintain(s) the circle
Centripetal force
Fc = mac
Centripetal Force
v
Fc

2
v
Fc = mac = m
R
v = R
2
v
2
Fc = mac = m = mR
R