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MID TERM REVIEW PART IV
NOTES FOR MID-TERM EXAM
Please review these formulas in preparation for the mid-term exam. The table that follows summarizes the formulas
and gives an example of each formula.
Motion
v =d/t = vaverage
d = vt
t = d/v
vaverage = (vi + vf)/2
a = v/t = (vf-vi)/t
vf = vi +at
vf2 = vi2 + 2ad
d = vit + ½at2
Projectile Motion
tup = tdown
vtop = 0 = vV – gtup
tup = vV/g
ttot = 2tup = 2vV/g
vAV to top = (vV + vtop)/2 = vV/2
dH = vHttot = 2vVvH/g
dV = vav*tup = ½vVtup = ½vV2/g
Vertical Distance as a Function of Horizontal Distance
vVtop = 0m/s
Vertical Distance Dv
20.0
dV
time to top of arc = ttop = vV/g
total time from launch to landing = ttot = 2tup = 2vV/g
vV
average vertical velocity from launch to top = vAV = (vV + vVtop)/2 = (1/2)vV
dV = vAV*ttop = (1/2)vV*vV/g = 1/2(vV)2/g
distance traveled horizontally = dH = vH*ttot = 2vHvV/g
-
dH
12
.0
18
.0
24
.0
30
.0
36
.0
42
.0
48
.0
54
.0
60
.0
66
.0
72
.0
78
.0
84
.0
90
.0
96
.0
10
2.
0
10
8.
0
11
4.
0
12
0.
0
-
6.
0
vH
Horizontal Distance Dh
Forces
Fnet = 0
W = mg
Fgrav = Gm1m2/r2
Newton’s 1st Law
Weight
Law of Gravitation
[R13/R23] = [T12/T22]
Kepler’s 3rd Law
Fslid = slid*Fnormal
F = -kx
Fnet = ma
F1 on 2 = – F2 on 1
g = GMearth/rearth2
Fstatic = static*Fnormal
Sliding Friction
Fcentripetal = mv2/r
Hooke’s Law (Spring Restoring Force)
Newton’s 2nd Law
Newton’s 3rd Law
Acceleration due to
gravity on Earth
Static Friction
Centripetal Force
Formula
Distance/Velocity/Acceleration
v = d/t
The symbol “” means “change
in” the quantity the immediately
follows. Sometimes this is
simply written as:
Definitions and Units
Application
Examples
v = velocity
d = distance
t = time
This formula is used when
only the total distance
traveled is known and the
total time it took to move
that distance. This formula
gives only an average
velocity.
1) Car moves 500 meters in 100
seconds. What is the average
speed or velocity?
v = 500m/100s or
v = 5 meters/sec
v = d/t
Metric system units:
v: meters/sec (m/s)
d: meters (m)
t: seconds (s or sec)
and it is understood that d refers
to the distance traveled and t is
the time from an initial time t =
0
Velocity (or speed) is the
change in distance (d)
divided by the elapsed
time (t).
vaverage = (vi + vf)/2
vaverage = average velocity
vf = final velocity
vi = initial velocity
a = v/t = (vf-vi)/t
Metric system units:
v: meters/sec (m/s)
a = acceleration
vf = final velocity
vi = initial velocity
t = time
Metric system units:
a: meters/sec2 (m/s2)
v: meters/sec (m/s)
t: seconds (sec, s)
vf2 = vi2 + 2ad
Acceleration is the
change in speed or
velocity (v) divided by
the time (t) it took to
change the speed or
velocity.
a = acceleration
vf = final velocity
vi = initial velocity
t = time
d = distance
Metric system units:
a: meters/sec2 (m/s2)
v: meters/sec (m/s)
t: seconds (sec, s)
d: meters (m)
Knowing any two terms
yields the 3rd:
v = d/t
d = vt
t = d/v
This formula is used when
an object is accelerated or
decelerated and only the
initial and final velocities
are known.
2) Car has an average speed of
50m/s and sustains this for 60
seconds, how far did it travel?
d = (50m/s)*(60s) = 3000m
3) Car travels 500 meters with an
average velocity of 50m/s. How
long did this take?
t = 500m/50m/s = 10s
Car starts at 10m/s and speeds up
to 30m/s. The average speed is:
vaverage = (10m/s + 30m/s)/2 =
20m/s
This formula is used when
both initial and final
velocities are known and
the elapsed time is known.
1) Initial speed of a car is 20m/s
and the final speed is 40m/s and it
takes 10s to do this. Find a;
a = (40m/s – 20m/s)/10s = 2m/s2
Knowing any two terms
yields the 3rd:
2) The initial speed of a car is
20m/s and it has an acceleration of
5m/s2 for 5s. What is the final
speed?
vf = 20m/s + (5m/s2)*5s = 45m/s
a = v/t
v = (vf-vi) = at or
vf = vi + at
t = v/a
This formula is used as a
shortcut to find the final
speed or velocity when the
initial speed is known as
well as the acceleration and
the time. Knowing any three
of the terms allows us to
find the 4th:
vf2 = vi2 + 2ad
vi2 = vf2 – 2ad
d = (vi2 – vf2)/2a
a = (vi2 – vf2)/2d
3) A car changes its speed by
30m/s when it has an acceleration
of 6m/s2. How long did this take?
t = (30m/s)/(6m/s2) = 5s
1) A runner’s initial speed is 5m/s
and he accelerates at 5m/s2 over a
distance of 7.5 meters. What is his
final speed?
vf2 = (5m/s)2 + 2*(5m/s2)(7.5m) =
100m2/s2 or vf = 10m/s
2) Same problem except the final
speed was 10m/s. What was the
initial speed?
vi2 = (10m/s)2 – 2*(5m/s2)(7.5m) =
25m2/s2 or vi = 5m/s
3) A car’s initial speed was 3m/s
and its final speed was 5m/s and it
traveled a distance of 8m. What
was the acceleration?
a = [(5m/s)2 – (3m/s)2]/(2*8m)
a = (25m2/s2 – 9m2/s2)/16m =
1m/s2
4) Same problem except a is given
as 1m/s. What was the distance?
d = [(5m/s)2 – (3m/s)2]/(2*1m/s2)
d = (25m2/s2 – 9m2/s2)/2m = 8m
Formula
d = vit + ½at2
Definitions and Units
a = acceleration
vi = initial velocity
t = time
d = distance
Metric system units:
a: meters/sec2 (m/s2)
v: meters/sec (m/s)
t: seconds (sec, s)
d: meters (m)
Application
This formula is used to find
the distance when the time,
acceleration and initial
speed are known. Knowing
any three of the terms
allows us to find the 4th:
½at2
d = vit +
vi = d/t – ½at
a = 2(d – vit)/t2
t = is the solution to a messy
quadratic equation and
rarely used. Here is the
solution:
Examples
1) A rock is dropped downwards
with an initial speed of 5m/s and
the acceleration due to gravity is
10m/s2. How fare did it fall in 3
seconds?
d = (5m/s)*(3s) + ½(10m/s2)*(3s)2
d = 15m + 45m = 60m
2) Same problem except d = 60m
is given. What was the initial
velocity?
vi = (60m/3s) – ½(10m/s2)(3s)
vi = 20m/s – 15m/s = 5m/s
t = [-vi +/- (vi2 + 2ad)1/2]/a
3) Same problem except vi = 5m/s
and d = 60m. What was the
acceleration if the time was 3
seconds?
a = 2[60m – (5m/s)*3s]/(3s)2 =
(2*45m)/(9s2) = 10m/s2
4) Same problem. What was the
time if the distance as 60m, initial
speed was 5m/s and the
acceleration was g = 10m/s2?
t = [-5m/s +/- (25m2/s2
+2*10m/s2*60m)1/2]/10m/s2
t = [-5m/s +/- 35m/s]/10m/s2 =
(30m/s)/10m/s2 = 3 seconds
Formula
Projectile Motion
tup = tdown
Definitions and Units
Application
Examples
tup = time from launch to
the maximum height
This formula is used in
projectile motion. Questions
can be phrased as: (a) what is
the time to maximum height?
(b) what is the total time if
the time to the top is t
seconds? (c) if total time from
launch to landing is t seconds,
what is the time to the top?
1) time to top for a projectile is 5
seconds; what is the time from the
top to landing?
and
ttot = 2tup
Metric system units:
t: seconds (sec, s)
tup = tdown
tup = 5 seconds, so tdown = 5 seconds
2) time to top for a projectile is 10
seconds; what is total flight time?
tup = tdown so ttot = 2ttop
ttot = 2*10seconds = 20 seconds
3) total time of flight is 30 seconds;
what is the time to the top of the
flight path?
vtop = 0 = vV – gtup
and
tup = vV/g
ttot = 2vv/g
vtop = vertical velocity at
maximum height
vv = initial vertical velocity
g = acceleration due to
gravity
tup – time to maximum
height
Metric system units:
vtop: meters/sec (m/s,
m/sec)
vv: meters/sec (m/s, m/sec)
g: meters/sec2 (m/s2,
m/sec2)
tup: seconds (sec, s)
vAV to top = (vV + vtop)/2
= vV/2
dH = vHttot = 2vVvH/g
vtop = vertical velocity at
maximum height
vv = initial vertical velocity
tup – time to maximum
height
Metric system units:
vtop: meters/sec (m/s,
m/sec)
vv: meters/sec (m/s, m/sec)
vAV: meters/sec (m/s,
m/sec)
vH = horizontal velocity
vv = initial vertical velocity
dH – horizontal distance
g = acceleration due to
gravity
Metric system units:
vH: meters/sec (m/s, m/sec)
vv: meters/sec (m/s, m/sec)
dH: meters (m)
g: meters/sec2 (m/sec2, m/s2)
This formula is used to find
the time to the top. At the
maximum height, the
projectile’s velocity is
precisely 0m/sec. Knowing
that we can then calculate the
time to the top given the
initial speed, vv.
The second formula, derived
from the first, states that the
time to the top is the initial
vertical velocity divided by
the acceleration due to
gravity.
The third formula is derived
from the time up/total time
relations; it states that total
time of flight is twice time
up, using the initial velocity
and g for the calculation.
This formula allows for a
simpler calculation for
distance to the top of the
projectile path. The average
velocity is, in reality, a
constant velocity. That means
we can use the average
velocity in a simpler
distance/time – velocity
formula.
This formula requires the
intermediate step of finding
the total time. Once that is
known, the horizontal
distance is found from the
product of a constant
horizontal velocity and the
total time of flight.
tup = (1/2)*ttot
tup = (1/)*30 seconds = 15 seconds
1) If the initial vertical velocity is
30m/s and g = 10m/s2, what is the
time to the top?
tup = vv/g = (30m/s)/(10m/s2) = 3sec
2) If the time to the top is 5 seconds
and g = 10m/s2, what was the initial
vertical velocity?
vv = g*tup = (10m/s2)*(5sec) = 50
meters
3) If the initial vertical velocity is
20m/s and g = 10m/s2, what is the
total time of flight?
ttot = 2vv/g = 2*(20m/s)/(10m/s2)
ttot = 4 seconds
1) The initial vertical velocity of a
projectile is 50m/s. What is the
average velocity to the top of the
projectile’s path?
vAV = (vV + vtop)/2 = vV/2
vAV = (50m/s)/2 = 25m/s
1) What is dH if vH = 40m/s and the
time to the top of the path is 10
seconds?
dH = vH*ttot = vH*2*tup
dH = (40m/s)*2*(10sec) = 800m
Formula
Projectile Motion
dV = vav*tup
dV = ½vVtup
dV = ½vV2/g
Definitions and Units
Application
Examples
vAV = average velocity
vv = initial vertical velocity
dv = horizontal distance
g = acceleration due to
gravity
tup = time to top
This formula is shown in
three variations, all
equivalent. Each formula is
found by substituting in the
formulas for vAV and tup.
1) What is the vertical distance from
launch to highest point if the
average velocity in the vertical
direction is 10m/s and the time to
the top is 10 seconds?
dV = vav*tup = (10m/s)*(10s) = 100m
Metric system units:
vAV: meters/sec (m/s, m/sec)
vv: meters/sec (m/s, m/sec)
dv: meters (m)
g: meters/sec2 (m/sec2,
m/s2)
tup: seconds (sec, s)
2) If the initial vertical velocity is
20m/s and the time to the top is 10
seconds, what is the maximum
vertical height?
dV = ½vVtup = ½ (20m/)*10s = 100m
3) If the initial vertical velocity is
20m/s and g = 10m/s2, what is the
maximum height reached?
Special case when object
lands at height different
from launch height
d = vvttot + ½ gttot2
vv = initial vertical velocity
d = distance above or
below original launch
height. Always take
original launch height = 0.
g = acceleration due to
gravity
ttot = total time of flight
Use the quadratic formula to
fund the total time of flight.
To do this one should execute
the following steps:
First, re-write the distance
formula as follows:
½ gttot2 + vvttot - d = 0
Metric system units:
vv: meters/sec (m/s, m/sec)
d: meters (m)
g: meters/sec2 (m/sec2,
m/s2)
ttot: seconds (sec, s)
Second, divide through by the
coefficient of t2 which is ½g
ttot2 + 2(vv//g)ttot - 2d/g = 0
dV = ½vV2/g = ½ (20m/s)2/10m/s2 =
100m
1) An object is launched with an
initial vertical speed of 20m/sec and
an initial horizontal speed of 20
m/sec. It lands on a valley floor
which is 100 meters below the
launch height. What is the total time
of flight?
ttot =
{-(vv/g)[(vv/g)2+(2d/g)]1/2)}
ttot = (20/10)  {(20/10)2 +
(2*100/10)}1/2
ttot = 2  {4 + 20}1/2
Third, in the quadratic
formula (x =( -b  [b2 –
4ac]1/2)/2a) we now have a =
1, b = 2vv/g and c = -2d/g.
Plug in and get this solution
for ttot:
ttot ={-(2vv/g)[(2vv/g)2 -4(2d/g)]1/2)}/2
Fourth, carry out the division
by 2 in the result:
ttot =
{-(vv/g)[(vv/g)2+(2d/g)]1/2)}
ttot = 2  4.90 = 6.90 seconds
The positive root gives the total
time. The negative root gives the
time when the object would have
been launched from the landing
height (time before actual launch)
with the same speed as the object
has when it hits the ground.
If the object landed at the same
height as the launch height, then d =
0 and our formula reduces to;
ttot = {-(vv/g)[(vv/g)2]1/2)}
Note that the minus sign in
the lead term vanishes since g
and vv point opposite to each
other, i.e., g is negative.
The lead term is the time to
the top. The term in the
square root adjusts it by
adding the extra time to get to
the landing height. When the
landing height = launch
height (d = 0) then we get the
two familiar solutions of ttot =
0 sec and ttot = 2ttop = 2vv/g.
ttot = 2vv/g = 2ttop
which is 4 seconds.
Formula
Vectors
F
FV

FH
FH = Fcos()
FV = Fsin()
The key triangles you
should remember which
may possibly appear on
the mid-term and will
certainly be found on the
PSAT, SAT and maybe the
HSPAs:
3-4-5 
5x
4x
All vectors in physics can
be broken down into two
perpendicular
components.
A vector requires both a
magnitude and a direction
to specify it. A magnitude
is a number; a direction is
something like an angle, or
North-South-East-West.
Metric system units:
Both the resultant vector
and the component vectors
have the same units.
3x
5-12-13 
13x
12x
5x
30o-60o-90o 
2x
Definitions and Units
F is called the resultant
force vector and FH and FV
are called the components
of the force vector F. The
force vector is pointed at
an angle  with respect to
the horizontal.
60o
1x
30o
F: force vectors all have
Newtons (N) as units:
v: velocity vectors all have
m/s (meters/sec, meters/s)
as units
a: acceleration vectors all
have meters/sec2 (m/s2,
meters/s2) as units
d: distance vectors all have
meters (m) as units
x3
45o-45o-90o 
x2
45o
1x
45o
Application
The vectors we have
encountered this far in this
course are:
1) Force
2) Acceleration
3) Velocity
4) Distance
Of particular note: the
magnitude of the velocity
vector is called the speed.
We use components of
vectors to simplify the
analysis of forces and
motion. We break down
vectors into the x and y or
horizontal and vertical
components. Once a vector
has been broken down into its
components, we can analyze
the motion or the forces
independently in each
direction.
For each of the four special
s in column 1 as well as for
the trigonometry relations for
vectors, you can verify that
the relations hold by using
the Pythagorean Theorem.
For each I wrote out the
Pythagorean Theorem and
verified that the ratio of the
sides in each triangle is
correct:
Vectors are often written
in bold type or with an
arrow above them in texts:
Trig relation:
F or F
F2 = (Fsin(q))2 + (Fcos(q))2
F2 = F2*(sin2(q) + cos2(q))
F2 = F2 *(1) = F2
The rules for drawing
vectors in diagrams:
Special s:
1x
In all vector s we use the
Pythagorean Theorem:
F2 = FH2 + FV2 where F is
the resultant vector and FV
and FH are the vertical and
horizontal coordinates of
the resultant vector.
a) each vector is
represented as an arrow:
(5x)2 = (3x)2 + (4x)2
25x2 = 9x2 + 16x2 = 25x2
25x2 = 25x2
b) the length of the arrow
is proportional to the
magnitude of the vector;
c) the arrow points in the
direction the vector
operates;
d) all vectors representing
the same kind of quantity
(force or velocity, etc)in
the same diagram must use
the same scale.
e) resultant vectors and
their components are
drawn “head to tail” as in
the vector diagrams in
column 1.
(13x)2 = (5x)2 + (12x)2
169x2 = 25x2 + 144x2 =
169x2
169x2 = 169x2
(2x)2 = (1x)2 + x3 )2
4x2 = x2 + 3x2 = 4x2
4x2 = 4x2
(x2 )2 = (1x)2 + (1x)2 = 2x2
2x2 = 2x2
Examples
1) the horizontal component of a
force is 30N and the vertical
component is 40N, what is the
resultant force vector?
For those who have had some
geometry, these two components
are the legs of a 3-4-5 triangle,
magnified by a common factor of
10 (3*10 = 30, 4*10 = 40). Hence,
the resultant vector is 50N (5*10).
The angle it operates at is found
from the inverse sine function on a
calculator:
sin-1(40N/50N) =  = 53.13o.
(The other triangles you might need
to know are shown in column 1.)
2) a force of 100N is applied at an
angle of 30o above the horizontal.
What are the horizontal and vertical
components?
FH = Fcos(30o) = 100N*0.866 =
86.6N
FV = Fsin(30o) = 100N*(0.5) = 50N
3) additional applications include
the inclined plane problem which
will be done below in the force
section.
Notes for Mid-term Exam
Formula
Newton’s First Law or the
Law of Inertia
Definitions and Units
“An object will stay at rest
or move in a straight line
with a constant speed
unless acted upon by an
unbalanced force”
Application
This law is the basis of
classical mechanics. By
definition, if an object either
does any of the following:
a) moves when it had
previously been at rest;
b) changes direction; or
c) speeds up or slows down;
then a force acts on it.
Newton’s Second Law:
F = the net force
m = the mass of the object
a = acceleration of the
object
If a net force is acting on an
object, then the acceleration
the object experiences is
proportional to the force and
inversely proportional to the
mass of the object.
F = ma
Metric system units:
F: Newtons
m: kilograms
a: m/s2 (meters/sec2,
meters/s2)
Examples
1) a car stops suddenly and a
passenger without a seatbelt strikes
the windshield;
2) The net force of an object at rest
on a table is 0 Newtons.
3) An object on a string and swung
in a circle at a constant speed still
has a force acting on it since the
object’s direction changes every
instant.
4) The net force acting on a car
moving in s straight line at a
constant speed is 0 Newtons.
1) A brick of mass = 5kg is
accelerated at 5m/s2. What is the
force acting on the brick?
F = ma = (5kg)*(5m/s2) = 25N
The three formulas are;
2) A mass of 10kg experiences a
force of 150N. What is the
acceleration?
F = ma
a = F/m
m = F/a
a = F/m = 150N/10kg = 15m/s2
3) A force of 200N on an object
results in its acceleration of 20m/s2.
What is the mass of the object?
m = F/a = 200N/20m/s2 = 10kg
W = mg
W = weight of an object
g = acceleration due to
gravity = 9.8m/s2 and on
tests and quizzes, we use
10m/s2
m = mass of an object
Metric system units:
W: Newtons
m: kilograms
g: m/s2 (meters/sec2,
meters/s2
This is a specific example of
Newton’s 2nd Law. Any
mass, m, has a weight equal
to mg whether the object is
moving or not.
1) A brick has a mass = 5kg.What is
its weight?
W = ma = (5kg)*(10m/s2) = 50N
The three formulas are;
2) A mass of 10kg has a weight of
150N. What is the acceleration due
to gravity?
W = mg
g = F/m
m = W/g
g = W/m = 150N/10kg = 15m/s2
NOTE: the values we use for g are
only valid on the surface of the
earth. At other locations – on the
moon, another planet, the Sun the
value for g changes. It can be higher
or lower. (See the section on
Gravitation.)
3) An object has a weight 200N
What is the mass of the object?
m = W/g = 200N/10m/s2 =210kg
Formula
Newton’s Third Law:
F1 on 2 = – F2 on 1
Definitions and Units
Both forces use Newtons
as the units in the metric
system.
The force of object 1 on
object 2 is equal in
magnitude but opposite in
direction to the force of
object 2 on object 1.
The Normal Force, Fnormal,
is the force a table or
surface exerts back on an
object in contact with the
table or surface. The object
can be at rest or moving.
There can be other forces
acting. Fnormal is equal to
the net downward force
acting on the object and
this net downward force is
the force the table or
surface experiences. It
pushes back on the object
with a force equal to the
net downward force.
The minus sign in
Newton’s Third law
indicates the forces are
opposite in direction.
A Free Body Diagram is a
diagram of all the forces
acting on an object.
Free Body Diagrams
Fnormal =
10N
Ffric =
5N
W = mg = 10N
Fpull =
5N
Each force arrow is drawn
in the proper direction and
the length of each arrow is
proportional to the
magnitude of the force.
Each force must be labeled
with the proper name and
the magnitude, if known,
should be shown as well.
Application
This law is often
misunderstood. To understand
it properly one must draw
“free body diagrams” which
are simple pictures showing
all forces acting on a single
body or object. (See next
topic.)
The key is to understand that
one needs to consider all the
forces acting on a single
object. To understand an
object’s motion we need only
know the forces acting on the
object, not the forces it exerts
on another object or objects.
Examples
1) A mass of 10kg sits on a table at
rest. What is the force of the table
on the object?
The objects weight is W = mg =
(10kg)*(10m/s2) = 100N
The object is at rest and its weight
presses down on the table. The table
presses back with a force equal in
magnitude, 100N, but opposite in
direction, i.e., up.
The two forces acting on the object,
then, are its weight, acting down,
and the force of the table up on the
object, which, in this example,
equals the weight. This latter force
is the Fnormal or normal force.
2) A baseball is hit by a bat with a
force of 500N. What is he force of
the baseball on the bat?
Fbatonball = 500N
= – Fballonbat
Fballonbat = 500N acting opposite in
direction to the Fbatonball force.
Free Body Diagrams are
essential for depicting forces
on an object. From the
diagram one then notes
separately all of the
horizontal forces and then all
of the vertical forces. This
includes finding the
horizontal and vertical
components of forces that act
at angles.
By doing that, one can then
write equations to find the
NET horizontal and NET
vertical forces acting on an
object. This takes the form of
Newton’s 2nd law for the
horizontal and vertical
directions.
A 1kg mass is at rest on a table.
Then a force Fpull of 5N is applied
to the right. A friction force
opposing motion of 5N results.
Draw the Free Body Diagram and
calculate the net forces acting on
the object. The picture in column 1
shows this example.
Since the mass is at rest, the forces
in the up and down directions must
balance. So W equals the normal
force of the table on the object.
Vertical forces:
W = mg = (1kg)*(10m/s2) = 10N,
down
W = Fnormal = 10N, up
Net FV = W – Fnormal = 10N – 10N =
0N
Horizontal forces:
Fpull = 5N to the right
Ffric = 5N to the left
Net FH = Fpull – Ffric = 5N – 5N =
0N
Formula
Fstatic = static*Fnormal
Fslid = slid*Fnormal
Definitions and Units
Friction forces act opposite
in direction to the motion
of an object. A frictional
force by itself cannot
create motion, but it can
act to impede it. Friction is
also needed to create
traction for car tires,
runners’ shoes, and almost
anything that must move
along a surface.
Friction is always
proportional to the normal
force between two
surfaces.
Fstatic – the force of static
friction
Fslid = the force of sliding
friction
 = the coefficient of
friction
Fnormal = the normal force
between the two surfaces
static > slid
And  always ≤ 1.0
Metric system units:
Fslid, Fstatic and Fnorm:
Newtons (N)
: no units.  is the ratio of
two forces and has no
units.
Application
The procedure to solve
friction problems is as
follows:
1) Draw a free Body Diagram
and carefully note all forces
acting vertically and
horizontally.
2) Calculate the normal force.
To do this, remember that the
normal force is the force that
makes the total net force in
the vertical direction equal to
zero. If, before you calculate
the normal force, you find
that the net vertical force is
already positive, that means
the object is moving away
from the surface and there is
then no normal force.
3) Calculate the appropriate
friction force. If the object is
stationary, then the
appropriate force is the static
friction force. If the
horizontal forces are
sufficient to overcome the
static force, then the
appropriate force is the
sliding friction force.
Examples
1) A 10kg mass is on a table. A
force, Fdown, of 5N also pushes
down on the mass from above. A
force of 10N pulls to the right. The
coefficient of static friction = 0.4. Is
this pulling force sufficient to move
the object?
Find the normal force. The forces
acting down on the table are the
weight of the object W and the
extra 5N force:
Fnormal = W + Fdown =
(10kg)*(10m/s2) + 5N = 105 N
Fstatic = static*Fnormal = (0.4)*(105N)
= 42N
Fstatic > Fpull so the object does not
move.
2) An object with weight = 50N is
on a table. A force, Fup, of 10N is
pulling up on the object. The
coefficient of sliding friction = 0.3.
A force, Fpull, of 20N pulls to the
right. What is the net force on the
object in the horizontal direction?
Find the normal force.
Fnormal = W – Fup = 50N – 10N =
40N and this force acts up on the
object.
Fslid = slid*Fnormal = (0.3)*(40N) =
12N
Net FH = Fpull – Fslid = 20N – 12N =
8N
Formula
Gravitation
Fgrav = Gm1m2/r2
g = GMearth/rearth2
Definitions and Units
Application
Examples
Fgrav is the gravitational
force of attraction between
two masses m1 and m2
separated by a distance
equal to r
Any two objects are attracted
to each other via the
gravitational force. The force
is one of attraction.
1) What is the gravitational force
between mass 1 = 100kg and mass
2 = 500kg if the distance between
them is 1000 meters?
It is an inverse square law
and because of that, when the
distance between the two
masses doubles, the force is
reduced to one fourth of its
previous value.
Fgrav = Gm1m2/r2 =
(6.67x10-11Nm2/kg2)*(100kg)*
(500kg)/(1000)2 =
(6.67x10-11)*(5x104)/(1*106) =
3.335x10-12 N
Fgrav – gravitational force
m1 = mass of first object
m2 = mass of second object
R = distance between the
two objects
G = universal constant of
gravitation = 6.67x10-11
Nm2/kg2
Metric system units:
F: Newtons (N)
m1, m2: kilograms (kg)
G: Nm2/(kg)
r: meters (m)
The gravitational force is a
very weak force but it does
operate over vast distances,
galaxies and beyond.
The second equation defines
the acceleration due to
gravity, g, at the surface of
the earth in terms of G, the
radius of the earth and the
mass of the earth.
When gravity holds an object
in orbit, such a the moon
around the earth, the force of
gravity is also a centripetal
force and can be equated to
the centripetal force formula,
mv2/r, to find the tangential
velocity of the object in orbit.
2) If the distance between the two
objects doubles, what happens to
the force?
Fnew = Gm1m2/(2r)2 = ¼ Gm1m2/r2
3) Let mass m1 be 50kg and let two
masses, m2 = 50kg and m3 = 200kg.
Let the distance between m1 and m2
be 50m and between m1 and m3 be
100m. Are the two gravitational
forces acting on m1 equal?
r12
m2
r13
m1
m3
Fg2on1 = Gm1m2/r122
Fg30n1 = Gm1m3/r232
We want to know if these forces are
equal and this really is the question
whether m2/r122 = m3/r122 since all
other factors cancel out (G and m1
cancel out).
So, does m2/r122 = m3/r132 or does
(50kg)/(50m)2 = (200kg)/(100m)2
Divide both sides by 50kg:
1/(50m)2 = 4/(100m)2
Multiply both sides by (100m)2:\
which leaves only the number 4 on
the right hand side of the equation
(100m)2/(50m)2 = (100m/50m)2 =
22 = 4
So they are equal!
Formula
Circular motion
Fcentripetal = mv2/r
f = #revolutions/time
elapsed
T = 1/f
v = r
v = Circumference/Period
v = 2r/T
v = r
 = 2f
v = r
r
Definitions and Units
Application
Examples
Fcent is the Centripetal
Force. Centripetal force is
an inward-directed force,
i.e., towards the center of
the circle, acting on an
object moving in a circle.
These formulas completely
describe circular motion with
a constant angular or constant
tangential velocity.
1) a mass of 10kg moves in a circle
of radius r = 3 meters at a tangential
speed of 9m/s. What are (a) the
centripetal force, (b) angular
velocity, (c) frequency, (d) period.
Fcent = centripetal force
v = tangential or linear
velocity
m = mass of the object
moving in a circle
r = radius of the circle
f = frequency which is the
number of revolutions an
object makes per second.
T = the period of the
circular motion and T = the
inverse of the frequency
Fcent acts towards the
center; v is tangent to
the circle. Hence v and
Fcent are always
perpendicular to each
other.
 = angular velocity and it
is related to the frequency
by a factor of 2.
Metric system units:
F: newtons (N)
v: m/s (meters/s,
meters/sec)
m: kilograms (kg)
r: meters (m)
f: revs/s (revs/sec)
T: seconds (sec, s)
: radians/s (rads/s,
rads/sec)
The centripetal force is the
force that causes an object
too move in a circle and,
therefore, it can be generated
by any of the following:
a) gravitational force as
between the Earth and the
Moon for example;
b) tension, as in a string;
c) normal force, as in a wall
pushing against an object;
d) friction, as in the friction
on a road allowing a car to
move in a circle.
A commonly used word is
centrifugal force which is not
a real force. What we
experience in circular motion
is the sensation of being
pushed to the outside, as in a
car going around a curve.
This is actually your inertia
wanting to continue moving
in a straight line but you are
prevented from doing so by a
seatbelt or the side of a car.
(a) Fcent = mv2/r =
(10kg)*(9m/s)2/(3m) = 270N
(b) v = r so w = v/r = (9m/s)/(3m)
= 3 radians/sec
(c ) f = /2 = (3m/s)/ = 0.955
revs/sec
(d) T = 1/f = 1/0.955 = 1.05 seconds
2) a mass of 2kg on a string is spun
in a circle of radius 4m. What is the
tangential or linear velocity if the
centripetal force is 18N?
Fcent = mv2/r or v2 = r*F/m
v2 = (4m)*(18N)/(2kg) 36m2/s2
v = 6m/s
3) what is the radius of the circle an
object of mass = 100kg is moving
in if the force causing the circular
motion is the force of friction is
200N and the tangential speed is
10m/s?
Fcent = Ffric = mv2/r so r = mv2/Ffric
r = 100*(10m/s)2/200N = 50m
4) Let m1 be a mass orbiting a
second mass, m2, at a distance r.
What is the tangential velocity in
terms of m1, m2, r and G the
gravitational constant?
Fgrav = Fcent
Gm1*m2/r2 = m1v2/r and cancel a
m1/r on both sides
Gm2/r = v2
v = (Gm2/r)1/2
The velocity of m1 is independent
of the objects mass and depends on
the m2, r and G.
Formula
Bracket analysis
Fcentripetal = mv2/r
Fnew = [ ]*[ ]2 * Forig
[ ]
F = centripetal force
Each set of brackets[ ]
stands in for one of the
parameters in the centripetal
force formula
First pair of brackets stands
in for the mass
Second pair stands in for the
velocity
Third pair stands in for the
radius
Definitions and Units
Procedure
Fcent is the Centripetal
Force. Problems are posed
as follows:
Mass increases by 50%
Speed decreases 50%
Radius doubles
Application
Examples
Instead of carrying out messy
calculations, it is far easier to
manipulate simple fractions
to find the new force as a
multiple of the original force.
1) Mass increases by 50%
Speed decreases 50%
Radius doubles
What is the new
Centripetal Force as a
MULTIPLE of the original
Centripetal Force
Step 1: mass multiple = 1.5 or 3/2
Speed multiple = 0.5 or ½
Radius multiple = 2
Step 3: Plug in the numbers:
Fnew = [1.5]*[0.5]2 * Forig
[2]
Fnew = (3/16)*Forig = 0.1875*Forig
Step 1: Determine the
multiple of each of the
original parameters
Step 2: Plug in the
multiples into the
appropriate set of brackets
Step 3: Carry out the
simple math
2) Mass remains the same
Speed triples
Radius is cut in half
Step 1: mass multiple = 1
Speed multiple = 3
Radius multiple = ½
Step 3: Plug in the numbers:
Fnew =
[1]*[3]2 * Forig
[1/2]
Fnew = 18*Forig
Same analysis is used for
gravitational force problems
Fgrav is the Gravitational
Force. Problems are posed
as follows:
Fgrav = Gm1m2/r2
Fnew = [ ]*[ ] * Forig
[ ]2
F = centripetal force
Each set of brackets[ ]
stands in for one of the
parameters in the centripetal
force formula
First pair of brackets stands
in for the first mass
Second pair stands in for the
second mass
Third pair stands in for the
distance between the masses
Mass increases by 50%
Speed decreases 50%
Radius doubles
What is the new
Gravitational Force as a
MULTIPLE of the original
Gravitational Force
Instead of carrying out messy
calculations, it is far easier to
manipulate simple fractions
to find the new force as a
multiple of the original force.
1) Mass1 increases by 50%
Mass2 decreases 50%
Distance doubles
Step 1: mass1 multiple = 1.5 or 3/2
Mass2 multiple = 0.5 or ½
Distance multiple = 2
Step 3: Plug in the numbers:
Fnew = [1.5]*[0.5] * Forig
[2]2
Fnew = (3/16)*Forig = 0.1875*Forig
Step 1: Determine the
multiple of each of the
original parameters
Step 2: Plug in the
multiples into the
appropriate set of brackets
Step 3: Carry out the
simple math
2) Mass1 remains the same
Mass2 triples
Distance is cut in half
Step 1: mass1 multiple = 1
Mass2 multiple = 3
Distance multiple = ½
Step 3: Plug in the numbers:
Fnew =
[1]*[3] * Forig
[1/2]2
Fnew = 12*Forig
Formula
Hooke’s Law
F = -kx
k = Hooke’s or spring
constant
x = distance spring stretches
or compresses
x cm
Definitions and Units
Application
Examples
F is the force exerted by a
spring when it is
compressed or stretched.
The force the spring exerts
is always opposite to the
stretch or compression
force.
Springs are used in scales.
When properly calibrated
they can be used to measure
weights. Springs are common
in grocery stores as well as
the more common bathroom
scales.
1) What force is needed to stretch a
spring a distance of 5cm if the
spring constant is 6N/cm?
1) If a spring is stretched
by pulling on it, then the
spring is pulling back.
F = kx
k = F/x
x = F/k
2) If the spring is
compressed by a force,
then the spring pushes
against the force.
Springs can be combined in
series or in parallel.
Metric system units:
F: Newtons (N)
x: cm or meters
k: N/m or N/cm depending
on type of problem
Series Springs
Series springs are springs
connected in a line. If the
spring constants are known,
then we look for a single
spring that can replace the
two springs and have the
same effect, i.e., the same net
stretch distance.
(1/k*) = (1/k1) + (1/k2)
k* = (k1*k2)/(k1+k2)
Parallel Springs
Parallel springs are springs
that have one weight
connected to them and all are
stretched the same distance.
F = kx = 6*5 = 30N
2. What is the spring constant if a
force of 10N stretches a spring scale
a distance of 3.5cm?
k = F/x = 10/3.5 = 2.857N/cm
3. What distance will a spring
stretch if it has a Hooke’s constant
= 12N/cm and a mass of 8.5kg is
hung from the spring?
x = F/k = (mg)/x = 85/12 =
7.083cm
4. What force is needed to stretch
two springs “in series” a distance of
8cm if k1 = 4N/cm and k2 = 5N/cm?
Ftotal = k*x
k* = (4*5)/(4+5) = 20/9
Ftotal = (20/9)*(8) = 160/9
Ftotal = 17.78N
5. What is the equivalent spring
constant if a force of 12N stretches
a pair of series springs a total
distance of 6cm? If one of the
Hooke’s constants is 4N/cm, what
is the second one?
Ftotal = k*x
k* = Ftotal/x = 12/6 = 2N/cm
(1/k*) = (1/k1) +(1/k2)
(1/2) = (1/4) + (1/k2)
(1/4) = 1/k2
k2 = 4N/cm
6. What force is needed to stretch
two springs “in parallel” a distance
of 8cm if k1 = 4N/cm and k2 =
5N/cm?
If the spring constants are
known, then we look for a
single spring that can replace
the two springs and have the
same effect, i.e., the
k* = k1 + k2
Ftotal = k*x = (4+5)*8 = 72N
2. What is the equivalent spring
constant if a force of 12.5N
stretches a pair of parallel springs a
distance of 2.5cm? If one of the
Hooke’s constants is 3.5N/cm, what
is the second one?
Ftotal = k*x
k* = Ftotal/x= 12.5/2.5 = 5N/cm
k* = k1 + k2
k2 = k* – k1 = 5 – 3.5 = 1.5N/cm
Formula
Kepler’s 3rd Law
[R13/R23] = [T12/T22]
R1 and R2 are the radii of
orbits of two planets or
satellites and T1 and T2 are
their respective periods of
orbit.
Definitions and Units
Application
Examples
Kepler’s 3rd law was
originally derived from the
elliptical orbits the planets
have as the move around
the Sun. A circle is a
special case of an ellipse.
The derivation using
ellipses is somewhat more
difficult.
Kepler’s 3rd Law allows
periods of orbit of distant
planets or satellites to be
found simply knowing the
radius of the orbit and the
period of a known orbiting
body. For example, the Moon
orbits the Earth and its
distance is known. Hence, a
satellite orbiting the Earth at
some known height (from the
center of the Earth) wil have
a period of orbit found from
Kepler’s 3rd law.
What is the distance from the center
of the Earth for a geosynchronous
satellite?
Using R1 as the radius of the
orbit of the satellite about the
Earth and R2 as the radius of
the orbit of the Moon about
the Earth and T1 as the period
the satellite and T2 as the
period the Moon’s orbit (27.3
days) then we have:
Solve for R1:
Metric system units:
These are ratios of like
quantities and hence the
ratios themselves do not
have any units. The radii
can be measured in any
consistent set of units –
meters, miles, kilometers.
The periods likewise must
be in consistent units such
as seconds, days, months,
years.
As long as both radii are in
the same units and both
periods are in the same
units, then the ratios work.
Use the Moon as the reference
orbit: R2 = 1 Moon distance = 1MU
= 3.8x108 meters. Use the unit
“MU” until the very end for
calculations. T2 for the Moon is
27.3 days. T1 for the satellite is 1
day since it remains stationary over
the Earth’s surface.
R13/R23 = T12/T22
R1 = R2*[T1/T2]2/3
(the exponent 2/3 means take the
cube root of the square of the
quantity in the brackets)
R1 = 1MU*[1/27.3]2/3
(T1/T2)2 = (R1/R2)3
R1 = 1MU*(0.1103) = 0.1103MUs
Then T1 is:
Plug in for MU:
T1 = T2*((R1/R2)3/2
R1 = (1.103x10-1)*(3.8x108)
Note the exponent 3/2 means
take the square root of the
cube of the quantity.
Hence, if R1 and R2 are
known, then one knows the
period the orbit of the
satellite about the Earth.
Knowing the period and the
circumference of the satellites
orbit (2R1) then the
tangential or orbital velocity
of the satellite is v = d/t =
2R1/T1 and use the known
value of R1 and the calculated
value for T1.
A specific case is when the
satellite is in geosynchronous
orbit about the Earth. That
means its period is the same
as the Earth’s time for one
rotation, or one day.
Knowing the Moon’s period
(27.3 days) and the satellite’s
period (1 day) one can find
the radius required for
geosynchronous orbit from
Kepler’s 3rd law.
R1 = 4.19x107meters
Since the radius of the Earth is
6.4x106 meters we can calculate the
height, H., of the satellite above the
Earth as:
H = R1 – RE = 41.9x106 – 6.4x106 =
35.5x106m = 35,500 kilometers.
This number, by the way, is a very
accurate estimate of the required
altitude above the Earth. (This is
about 22,000 miles.)
These satellites are used for cable
TV and satellite TV. Cable
operators receive programming
from satellites and re-broadcast
over their coax and fiber networks
it to their subscribers. DISH and
DirecTV deliver signals directly
from satellites to the home or
business location. All of these TV
satellites are in geosynchronous
orbit.