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Transcript
```What causes motion
In our everyday life we observe that objects often
change their state of motion. A lot that happens in the
Universe results from a change in motion.
These changes are produced by forces and in our
everyday life there are just two forces we can sense
directly.
Gravity acts on mass
F = Gm1m2/r2
F = kq1q2/r2
Electromagnetic forces act on electric charge
Week 3,
Physics 214
Spring 2010
1
FORCES
Forces also bind objects together, and can even
change the composition of mattter – for example the
electromagnetic force is responsible for all of
chemistry and for the structure of atoms.
Electromagnetic forces cause opposite electric
charges to attract each other, as in the hydrogen atom
where the positive proton (which is the nucleus)
attracts the orbiting electron and binds it into the atom.
F = kq1q2/r2
Week 3,
Physics 214
Spring 2010
2
Newtons Second and First Law
Second Law
The acceleration of an object is directly
proportional to the magnitude of the
imposed force and inversely
proportional to the mass. The
acceleration is in the same direction as
the force
F = ma F and a are vectors
unit is a Newton (or pound) 1lb = 4.448N
First Law
An object remains at rest or in uniform
motion in a straight line unless it is
acted on by an external force.
F = 0 a = 0 so v = constant
Week 3,
Physics 214
Spring 2010
3
Newtons Second and First Law
Second Law
F = ma F and a are vectors
First Law is just a special case of the
Second Law
An object remains at rest or in uniform
motion in a straight line unless it is
acted on by an external force.
F = 0 : a = 0 so v = constant
NOTE: if a = 0, you can conclude that
the NET force on the object is ZERO.
The NET force is the vector sum of ALL
forces acting on the object.
Week 3,
Physics 214
Spring 2010
4
Ch 4 E6 & quiz
A 6kg block is being
pushed with a force P
and has an
acceleration of
3.0m/s2
a) What is the net force?
b) If P is 20N what is Ff?
Note red arrows are NOT
in proportion!!
P
Ff
A 3N, B 6N ,
a) F = ma = 6 x 3 = 18 N
b) F = P – Ff
Week 3,
a = 3m/s2
6 kg
D 18 N
,
C 12 N
E 24 N
Ff = 2N
Physics 214
Spring 2010
5
Mass and weight
Newtons second law enables us to measure relative mass. If we
apply the same force to two objects and measure the
accelerations then.
F = m1a1 and F = m2a2 so m1/m2 = a2/a1
We then need to have one mass as a calibration. A kilogram is
defined as the mass of a special piece of platinum held in Paris.
Measuring acceleration is a more difficult way of measuring mass
than using a spring or a balance.
The balance exploits the fact that gravity acts proportional to
mass, so that at a given place all masses accelerate equally.
Week 3,
Physics 214
Spring 2010
6
Mass and weight
Since gravity acts proportional to mass
(this is Newton’s law of Universal Gravitation)
then the force near the earths surface is
F = mg
This force is known as the
weight of an object so if we compare
F1 = m1g and F2 = m2g then
weight1/weight2 = m1/m2
An equal-arm balance has equal weights on each side when it is
“balanced”. Unequal-arms use the fact of leverage to measure
relative weights.
Week 3,
Physics 214
Spring 2010
7
Mass and weight
A spring exerts a force proportional to the force pulling
it. So a mass hung on a spring will stretch the spring
proportional to its weight, W, which is proportional to
its mass: W = mg
This method is cheap and quick
and typically less accurate than
comparing two masses with a
balance.
m
mg
Week 3,
Physics 214
Spring 2010
8
Inertia
Inertia = tendency of an object to resist changes in its
velocity.
Since F = ma
and a = Δv/t then Ft = mΔv : impulse
So if a force acts for a time INTERVAL t the change in
velocity will be smaller for larger masses. So it is
mass that determines inertia.
In particular if t is very small and m is large, then F can
be large while Δv is very small.
NOTE: for a given impulse (mΔv) if t (actually, Δt) is
really small, then F must be very large. PADDING!:
Week 3,
Physics 214
Spring 2010
9
Friction
In our everyday world any object which is moving feels a force
opposing the motion --- this is friction.
 An object which is sliding
The air resistance on your car
These types of friction result in energy being lost. So
minimizing friction is very important.
But
Friction is also useful and essential since with no friction
a car would not move but just spin it’s wheels
a car would not be able to turn a corner
we would not be able to walk
objects would slide off surfaces unless perfectly horizontal
Week 3,
Physics 214
Spring 2010
10
Force diagrams
When we are analyzing
a particular object we
have to take into account ALL the forces acting on the body both
in magnitude and direction. (i.e. vector sum of all Forces)
The acceleration of the object is equal to F/m in the direction of F
where F is the NET force acting.
As in the example above we know that, if we want to move an
object --- there is a force called friction which opposes what we
want to do.
In the case shown the 10N force is + and the 2N force (-) so the
net force is 8N and 8N = 5kg x a
so a = 1.6[N/kg or m/s2]
Check our work: a
Week 3,
has units m/s/s and N = kg m/s/s Looks OK
Physics 214
Spring 2010
11
Friction
In the case shown the 10N force is + and the 2N force - so the net force is 8N
So: 8N = 5kg x a
There can be sliding friction and there can be static (“stuck”) friction. Stuck
(static) friction opposes other forces, UP TO A LIMIT. Beyond which it “breaks”.
In the present example, since the NET force is not zero, the object must be
accelerating and the friction must be sliding friction.
There are cases where the other forces aren’t big enough to break the static
friction: then the static friction adjusts itself to just oppose the other forces, and
the NET force is zero. Hence no acceleration and no motion.
The maximum static friction is proportional to the size of the force pushing the
object towards the surface, and to the “coefficient of friction” : small c.o.f. = slick,
big c.o.f. = good traction.
Week 3,
Physics 214
Spring 2010
12
Friction
T
f = μk(W - Tvertical)
Thorizontal
W = mg
The maximum static friction is proportional to the size
of the force pushing the object towards the surface,
and to the “coefficient of friction” : small c.o.f. = slick,
big c.o.f. = good traction.
a = (Thorizontal – f )/m
It’s better to pull upwards than to push downwards on
a box or a suitcase you’re trying to slide sideways !!
Week 3,
Physics 214
Spring 2010
13
Terminal velocity
As an object moves through the atmosphere the air
exerts a frictional retarding force (drag) which increases
with velocity. An object that is dropped from a great
height first accelerates at 9.8m/s2. As the speed
increases, the drag increases and cancels more and
more of the gravity force. At TERMINAL velocity, the
drag force just balances the weight (gravity) force: NET
force is zero, acceleration is zero, so the velocity does
not change any further. Hence “terminal” velocity.
Ff
mg
v
Mg – Ff =ma
Similarly when a car is moving at constant speed on the
highway the forward force of the road on the tires is
balanced by the frictional force of the air on the car.
Fair
Ftire
Freaction – Fair = ma
Freaction
Week 3,
Physics 214
Spring 2010
14
Re-action : Newtons Third Law
For every force that acts on a body there is an equal
and opposite reaction force (by the body ON the
agent.)
block pulls you –F
NET on block
you pull block
F
F
F
f
floor on block
*
*
net
-Ff
block on floor
You pull on the block; the block pulls back on you
The floor exerts a frictional force holding the block back
The block exerts a frictional force on the floor trying to move the
floor to the right.
To find effect on block, use ALL, and ONLY, Forces* acting ON the block
The block accelerates providing F > Ff i.e. Fnet >0
Week 3,
Physics 214
Spring 2010
15
Force analysis & quiz
To analyze the motion of an object we need to draw a diagram and
put in all the forces that are acting ON the object.
We will only deal with problems that have an acceleration along a
single axis. The net force along an axis perpendicular to this axis
is zero
A
B
Ff
Fnet
D
F
-Ff
C
We can choose + in the direction of the acceleration. We now can
use the equations
v = v0 + at
d = v0t + ½ at2
Fnet = ma
Which one of the above force vectors did we ignore?
Week 3,
Physics 214
Spring 2010
16
Reaction forces
IN THESE DIAGRAMS, THE NET FORCES ON OBJECTS ARE ZERO
N
For the book, magnitudes
Nt = mg
Nt, table on book, is vector
pointing opposite to mg
For the table, magnitude
Nearth = mg + mtableg
mg
mtableg mg
Ne
Here the situation is static, that is, no acceleration (and no
motion). The “Normal” forces are support forces, and ADJUST
themselves to prevent motion – up to the point where something
breaks. Static friction is a little bit like this too – it adjusts itself
up to the point where the object breaks free and starts to move.
Week 3,
Physics 214
Spring 2010
17
Reaction forces
N
For the book, magnitudes
Nt = mg
Nt, table on book, is vector
pointing opposite to mg
For the table, magnitudes
Nearth = mg + mtableg
mg
mtableg mg
Ne
For a stack of plates or a brick wall and even a mountain each
layer has to support the weight of everything higher. So for a
stack of 48 plates the force on the second plate from the top is
mg. For the bottom plate it is 47mg. Of course each plate has a
NET Normal force of +mg to balance it’s own weight, -mg.
Week 3,
Physics 214
Spring 2010
18
I-clicker Registration
Would the following students please register your
Quizzes can be entered?
Bic
Blackwell
Chan
O’Keefe
Walter
Week 3,
Physics 214
Spring 2010
19
Forces in an accelerating elevator
W = mg = true weight with no acceleration +
N = apparent weight = “scale on object”
N
N – mg is the net force, where + means up
N – mg = ma is the equation of motion
g
If N > mg a is positive and the apparent
weight is > than the true weight
If N < mg a is negative and the apparent
weight is less than the true weight
mg
If N < 0, scale & floor aren’t supporting and
object is in free fall at 1 g of acceleration downward
IRRESPECTIVE OF THE DIRECTION OR MAGNITUDE
OF THE VELOCITY
Week 3,
Physics 214
Spring 2010
20
Connected objects
T
Both objects have the same acceleration
The coupling pulls back on the 4kg mass but accelerates the
2kg mass.
30N – 8N – T = 4kg a
T – 6N = 2kg a
Two equations, and two unknowns (T and a).
You can eliminate T between the two eqns., then solve for a.
(Just add the 2 eqns. This is equivalent to treating the two
blocks as one system of mass 6 kg, with total friction 14 N, and
a common acceleration.)
30N – 8N – 6N = 6kg a ;
a = 16/6 m/s2 ; T = 6N+ 16/3N
T = 111/3 N
[N/kg=m/s2]
Week 3,
Physics 214
Spring 2010
21
Connected objects
If we have a freight train with 100 cars each of mass
m, and the train is accelerating at a rate a , then
for the coupling between the engine and the first car
T1 = 100ma
And between the last two cars T99 = ma or 100
times less since that tension has to accelerate only
ONE car.
Week 3,
Physics 214
Spring 2010
22
Motion of a car
Fair
Ftire
Freaction – Fair = ma
Freaction
Generally Fair is proportional to v2 . And energy lost to air drag = Force x
Distance. So the difference in energy lost to air resistance between 55mph
and 80mph is a factor of 2.16. Air resistance is however only part of the
losses, which include friction in the drive train, friction in the wheel bearings,
and friction in the tires as they flex, among other things. Assuming these
other frictional losses ALSO go as v2, (which may be an overstatement) you
use (80/55)2 = 2.16 more gasoline to cover the same distance.
Travel from Indianapolis to Lafayette a distance of 60 miles
Car does 30miles/gallon at 55mph gas costs \$3/gallon
At 55mph use 60mi/(30mi/gal) = 2 gallons
cost \$6
time = 60mi/(55mi/hr) = 1.0909 hr = 65.45 minutes
At 80 mph use 4.32 gallons cost \$12.96
time = 60mi/(80mi/hr) = 0.75 hr = 45 minutes
Cost of saving 20.45 minutes = \$6.96
Week 3,
Physics 214
Spring 2010
23
1H-03 CO2 Rocket
Imagine that you are sitting in a cart with a pile of bricks.
ROCKET PROPULSION !
How could you use the bricks to get yourself and the cart to
move ?
What would happen if you throw a brick out of the cart ?
Then you throw out another .
What if you throw smaller bricks faster and more frequently ?
Now, if the bricks were the size of molecules. . .
What happens when the fire extinguisher rapidly “throws” out
CO 2 molecules ?
THIS
IS DIFFERENT
THAN THE FAN CART. CO2 IS EXPELLED AT HIGH VELOCITY AND IN
Because
the repulse
TERMS OF FORCES THE REACTION FORCE CAUSES THE CART TO MOVE IN THE OPPOSITE
DIRECTION. THE QUANTITY CALLED MOMENTUM IS CONSERVED AND THE MASS OF THE
CO2 X AVERAGE SPEED = TOTAL MASS OF THE CART X AVERAGE VELOCITY. THIS IS HOW A
ROCKET IS ACCELERATED AND IT ALSO WORKS IN OUTER SPACE.
Week 3,
Physics 214
Spring 2010
24
Summary of Chapter 4
Forces are responsible for all physical phenomena
Gravitation and the electromagnetic force are responsible for all
the phenomena we normally observe in our everyday life.
Newton’s laws F = ma where F is net force on m
d = v0t + ½ at2
d = ½(v + v0)t
v = v0 + at
Every force produces an equal and opposite reaction
Fa on b = -Fb on a
Weight = mg where g = 9.8m/s2 locally
Apparent “weight” in an elevator depends on the acceleration
a up : “weight” is higher
lower
a down: “weight” is
If your “weight” becomes zero it’s time to worry because you are in
free fall!!
Week 3,
Physics 214
Spring 2010
25
Ch 4 E16
g
N
A vertical force of 6N presses on a book.
a) What is the gravitational force?
b) What is the normal force on the book?
6N
0.4 kg
a) Gravitational Force = mg = 3.92 N
b) Upward Force = 6 + 3.92 = 9.92 N
Week 3,
Physics 214
Spring 2010
26
1C-04 Bocce Ball Tracks
Two balls are released with the same initial velocity. One
travels a flat path the other goes down a slope increasing
it’s speed and then climbs a slope and slows down before
reaching the end. The balls roll so friction is small.
V
V
V
Which ball will
reach the end
first ?
1
5
2
3
4
The time for the yellow ball is just d/v. The time for the blue ball
can be computed in 5 parts. Parts 1 and 5 are identical for both
balls but the blue ball is faster for part 3 and qualitatively one
would expect the blue ball to arrive first.
Week 3,
Physics 214
Spring 2010
27
1F-04 Brass Rod (Inertia)
How to remove
the paper
without toppling
the rod ?
One needs to remove the paper quickly so that the frictional
force only lasts for a short time and the inertia of the rod
prevents it from toppling over.
Week 3,
Physics 214
Spring 2010
28
1F-05 Coin, Hoop & Milk Bottle (Inertia)
How can you get
the coin into the
bottle without
touching it ?
This is actually a trick which depends on hitting the
ring so that the top deflects down and the coin is
free to drop
Week 3,
Physics 214
Spring 2010
29
1F-06 Inertial Ball
Which string
breaks first ?
Case 1: Place the aluminum rod in the
lower loop and pull SLOWLY downward.
Case 2: Use the wooden mallet to
strike a sharp blow to the aluminum rod.
IF IT IS DONE SLOWLY THE UPPER STRING BREAKS FIRST
BECAUSE THE TENSION IN THAT STRING WILL BE THE
WEIGHT OF THE BALL PLUS THE TENSION IN THE LOWER
STRING.
IF THE LOWER STRING IS STRETCHED SUFFICIENTLY
RAPIDLY, IT WILL REACH ITS BREAKING POINT BEFORE
THE BALL HAS A CHANCE TO MOVE APPRECIABLY.
Week 3,
Physics 214
Spring 2010
30
1F-02 Stack of Washers
This is a demonstration of Inertia where a washer can be
removed from a stack if the blow is fast.
Why does it
work less well
as the stack
gets shorter ?
Strike the stack quickly. So the friction will
be very short-lived and the stack will not
gain speed before the force is gone.
THIS TENDENCY TO RESIST CHANGES IN THEIR STATE OF
MOTION IS DESCRIBED AS INERTIA. MASS IS A MEASURE OF
THE AMOUNT OF INERTIA. SO FOR A FIXED FRICTIONAL
FORCE ACTING FOR A SHORT TIME. THE BIGGER THE MASS,
THE LESS IT WILL MOVE.
Week 3,
Physics 214
Spring 2010
31
1F-07 Table Cloth Jerk
Can the table cloth
be removed without
breaking any dishes ?
THERE IS A FORCE ACTING ON THE DISHES, BUT IT LASTS
FOR A VERY SHORT TIME. COMBINED WITH THE
RELATIVELY LARGE MASS OF THE DISHES, THIS FORCE IS
OVER SO QUICKLY AND IS SO SMALL THAT THE DISHES
HARDLY MOVE.
Week 3,
Physics 214
Spring 2010
32
1F-03 Egg drop
Is it possible to
get the eggs in
the beakers
without touching
them ?
IF THE PAN IS HIT SHARPLY A FORCE WILL ACT ON THE
EGGS FOR A VERY SHORT TIME AND THEY WILL NOT
MOVE HORIZONTALLY. THE PAN HAS TO BE HIT HARD
ENOUGH SO THAT HAS MOVED OUT OF THE WAY BEFORE
THE EGGS DROP ANY APPRECIABLE DISTANCE
Week 3,
Physics 214
Spring 2010
33
1H-02 Fan Cart (Action-Reaction)
Can a fan attached to a cart propel the cart?
What if
the sail is
removed?
Reaction
Action
In which
direction will
it move ?
What if the
sail is canted
at an angle ?
INTERNAL FORCES IN A SYSTEM CANCEL EACH OTHER WHEN THE SYSTEM AS A WHOLE
IS CONSIDERED. SO IF THE SAIL IS PERPENDICULAR THE FAN DOES NOT MOVE.
IF THE SAIL IS REMOVED THE FAN MOVES IN THE OPPOSITE DIRECTION TO WHICH IT
BLOWS AIR. THE FAN WOULD NOT MOVE IN OUTER SPACE.
Week 3,
Physics 214
Spring 2010
34
1H-04 Hero's Engine
A glass bulb emits steam from small nozzles
What happens
when the Glass
Bulb begins to
emit steam ?
Reaction = Bulb Spins
Action = Ejects Steam
Same Principle
causes a Lawn
Sprinkler to Turn.
THE REACTION FORCE TO THE EJECTION OF MASS CAUSES THE OBJECT TO SPIN.
THIS IS THE SAME AS THE CO2 ROCKET IN THAT MATERIAL IS EXPELLED AT HIGH
VELOCITY
Week 3,
Physics 214
Spring 2010
35
A Scale Measures the Force acting on it
What is the
the scale ?
NOW, What
is the
the scale ?
WALL
mg
mg
mg
T
T
T
T
mg
T = mg
mg
T = mg
mg
mg
THE TENSION IN THE CORD IS THE SAME FOR BOTH CASES. THE SCALE
MEASURES THE TENSION IN THE CORD. FOR EXAMPLE THE TENSION IN A ROPE
IS THE SAME IF TWO PEOPLE PULL ON EACH END WITH FORCE F OR IF ONE
PERSON PULLS WITH FORCE F TO A ROPE TIED TO A WALL.
Week 3,
Physics 214
Spring 2010
mg
36
Questions Chapter 4
Q8 A 3-kg block is observed to accelerate at a rate twice that of a
6-kg block. Is the net force acting on the 3-kg block therefore
twice as large as that acting on the 6-kg block? Explain.
The net force is the same
Q9 Two equal-magnitude horizontal forces act on a box as shown
in the diagram. Is the object accelerated horizontally? Explain.
-F
F
No the net force is zero
Q10 Is it possible that the object pictured in question 9 is moving,
given the fact that the two forces acting on it are equal in size but
opposite in direction? Explain.
Yes, constant velocity
Week 3,
Physics 214
Spring 2010
37
Ch 4 CP6
A 60kg person accelerating down at 1.4m/s2
a) What is the true weight?
b) What is the net force?
c) What is N?
d) What is the apparent weight?
e) a, b, c, d with 1.4m/s2 up?
a) True weight = mg
N
1.4 m/s2
Mg
= 60 x 9.8 = 588 N
b) Net Force = Ma = 84 N
c) N = 588 – 84 = 504 N
d) 504 N
e) ↑1.4 m/s2
Net Force = 84 N↑
Week 3,
N = 588 + 84 = 672 N
Physics 214
Spring 2010
W = 672 N
38
Q18 The acceleration due to gravity on the moon is approximately
one-sixth the gravitational acceleration near the earth’s surface. If
a rock is transported from the earth to the moon, will either its
mass or its weight change in the process? Explain.
It’s mass will not change but it’s weight wil be 6 times less
Q22 The engine of a car is part of the car and
cannot push directly on the car in order to
accelerate it. What external force acting on the
car is responsible for the acceleration of the car
on a level road surface? Explain.
Fair
Ftire
Freaction
It’s the reaction force between the tires and the road
Q23 It is difficult to stop a car on icy road surface. It is also
difficult to accelerate a car on this same icy road? Explain.
Because of a lack of friction the wheels will skid or spin
Week 3,
Physics 214
Spring 2010
39
Q25 When a magician performs the tablecloth trick, the objects on
the table do not move very far. Is there a horizontal force acting on
these objects while the tablecloth is being pulled off the table? Why
do the objects not move very far? Explain.
Yes but the force acts for a very short time and the objects
start to move, then when the cloth is gone friction stops them.
Q30 Two masses, m1 and m2, connected
by a string, are placed upon a fixed
frictionless pulley as shown in the
diagram. If m2 is larger than m1, will the
two masses accelerate? Explain.
Yes m2 will fall and m1 will rise
Week 3,
Physics 214
Spring 2010
•
m
m
2
1
40
Q31 Two blocks with the same mass are connected by a string
and are pulled across a frictionless surface by a constant force, F,
exerted by a string (see diagram).
A. Will the two blocks move with constant velocity? Explain.
B. Will the tension in the connecting string be greater than, less
than, or equal to the force F? Explain.
F
A. They will accelerate F = ma
B. The tension will be less
Week 3,
Physics 214
Spring 2010
41
Q33 If you get into an elevator on the top
floor of a large building and the elevator
begins to accelerate downward, will the
normal force pushing up on your feet be
greater than, equal to, or less than the force
of gravity pulling downward on you?
Explain.
N – mg = ma but a is negative so N is smaller than mg
The only force pulling you down is gravity so if you are
accelerating down the force due to gravity must be larger
than the reaction force N ( N is apparent weight)
Week 3,
Physics 214
Spring 2010
+
N
g
mg
a
42
Ch 4 E4
A 2.5kg block is pulled
with a force of 80N and
friction is 5N
a) What is the
acceleration?
Net force = 75 N
Week 3,
5N
2.5 kg
80 N
a = 75/2.5 = 30 m/s2
Physics 214
Spring 2010
43
Ch 4 E18
A 60kg person is in an elevator
With an upward acceleration of
1.2m/s2
a) What is the net force on her?
b) What is the gravitational force?
c) What is the normal force?
a) Net Force
m = 60 KG
a = 1.2 m/s2
mg
F = Ma = 60 x 1.2
= 72 N
b) mg = 60 x 9.8 = 588 N
c) N = 588 + 72 = 660 N
Week 3,
Physics 214
Spring 2010
44
Ch 4 CP4
A 60kg crate is lowered from a height of 1.4m
and the tension is 500N
a) Will the crate accelerate?
b) What is the acceleration?
c) How long to reach the floor?
d) How fast does the crate hit the floor?
a) Net Force
= 60 x 9.8 – 500
= 588 – 500
g
60 kg
= 88 N
b) Will accelerate down
a = 88/60 = 1.47 m/s2
c) d = 1/2 at2
t = 1.38s
d) v = v0 + at
v = 2.03 m/s
Week 3,
500 N
Physics 214
Spring 2010
45
Ch 4 CP4
An 80kg crate is accelerated upward by a motor pulling on the rope.
The breaking tension of the rope is 1,500N. What is the MAXIMUM
possible upward acceleration of the crate? (g = 9.8 m/s2)
A. 4.32 m/s2
B. 6.61 m/s2
C. 8.95 m/s2 D. 11.33 m/s2 E. 15.0 m/s2
Upward Fnet = 1500N - mg = 1500N – 784N = 716 N
F = ma
Week 3,
so
1,500 N
max.
g
80 kg
a = Fnet/m = 716 N / 80 kg = 8.95 m/s2
Physics 214
Spring 2010
46
Ch 4 E14
A 4kg rock is dropped and experiences
air resistance of 15N
a) What is the downward acceleration?
4 kg
15 N
Mg
F = 4 x 9.8 – 15
F = ma = 24.2 N
a = 24.2/4 = 6.05m/s2
Week 3,
Physics 214
Spring 2010
47
Review Chapters 1 - 4
-
d
+ x
Units----Length, mass, time SI units m, kg, second
Coordinate systems
Average speed = distance/time = d/t
Instantaneous speed = d/Δt
Vector quantities---magnitude and direction
Magnitude is always positive
Velocity----magnitude is speed
Acceleration = change in velocity/time =Δv/Δt
Force = ma Newtons
Week 3,
Physics 214
Spring 2010
48
Conversions, prefixes and
scientific notation
giga
1,000,000,000
109
billion
1 in
2.54cm
mega
1,000,000
106
million
1cm
0.394in
kilo
1,000
103
thousand
1ft
30.5cm
centi
1/100
0.01
10-2
hundredth
1m
39.4in
milli
1/1000
0.00
1
10-3
thousandth
1km
0.621mi
1mi
5280ft
1.609km
1lb
0.4536kg
g =9.8
1kg
2.205lbs
g=9.8
micro
1/1,000,000
1/106
10-6
millionth
nano
1/1,000,000,000
1/109
10-9
billionth
Week 3,
Physics 214
Spring 2010
3.281ft
49
Speed, velocity and acceleration
v = Δd/Δt
a = Δv/Δt
The magnitude of a is not
related to the magnitude of v
the direction of a is not
related to the direction of v
2 3
4
1
v = v0 + at constant acceleration
d = v0t + 1/2at2
d,v0 v,a can be + or d = 1/2(v + v0) t
Week 3,
Physics 214
Spring 2010
50
One dimensional motion and gravity
v = v0 + at d = v0t + 1/2at2
d = ½(v + v0)t
+
g = -9.8m/s2
+
Week 3,
At the top v = 0 and t = v0/9.8
At the bottom t = 2v0/9.8
Physics 214
Spring 2010
51
Equations
v = v0 + at
d = v0t + 1/2at2
d = ½(v + v0)t
Sometimes you have to use two equations. NO YOU
step.
v0 = 15m/s v = 50m/s What is h?
v = v0 + at
v0
50 = 15 + 9.8t t = 3.57 s
`
h = v0t + 1/2at2
g
h
2
v
Week 3,
h = 15 x 3.57 + 1/2x9.8x3.57
= 116m
h = ½(15 + 50) x 3.57 = 116m
Physics 214
Spring 2010
52
Projectile Motion
axis 1
axis 2
v1 = constant and d1 = v1t
vv = v0v + at
and d = v0vt + 1/2at2
v1
g
9.8m/s2
h
v
R
Use + down so g is + and h is +
h = v0vt + 1/2at2
v0v = 0, t2 = 2h/a
R = v1t
v = v0v + at
Week 3,
Physics 214
Spring 2010
53
Complete Projectile
v0v
v1
9.8m/s2
v1
v1
v0v
highest point the vertical velocity is zero
vv = v0v + at
so t = v0v/9.8 h = v0vt + 1/2at2
end t = 2v0v/9.8 and R = v1 x 2v0v/9.8
and the vertical velocity is minus v0v
Week 3,
Physics 214
Spring 2010
54
Newton’s Second and First Law
Second Law F = ma unit is a Newton (or pound)
First Law
F = 0 a = 0 so v = constant
Third law For every force there is an equal and opposite
reaction force
N
Weight = mg
mg
Ff
F
F
Ff
F = ma
Week 3,
v = v0 + at
d = v0t + ½ at2 d = ½(v + v0)t v2 = v02 + 2ad
Physics 214
Spring 2010
55
Examples
+
T
N
g
30 – 8 – T = 4a
T – 6 = 2a
30 – 8 – 6 = 6a
mg
N – mg = ma
a + N > mg
a – N < mg
Week 3,
Physics 214
Spring 2010
56
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