Transcript Part23 - FacStaff Home Page for CBU

More situations with
Newton’s Second Law
Consider a problem that includes an air
resistance term: Fair resistance = bv2 , where
b is a coefficient that depends on the shape
of the object and the density of the air. The
direction of the air resistance term is
opposite the motion.
Air Resistance – the coefficient
For medium speeds, (not real slow and not real fast),
the coefficient for air resistance, b, can be
expressed as:
b = (1/2)*C*Across-section*ρ
Where C is a constant that depends on the shape (for
a sphere, C = ½), A is the cross-sectional area
perpendicular to the velocity, and ρ is the density
of the material through which the object is
moving. For air at sea level, ρ = 1.2 kg/m3.
Air Resistance and Gravity
Consider an object dropped from a high place:
• we have Fgravity = W = mg directed down,
and with air resistance (AR)
• we have FAR = bv2 directed up.
Newton’s Second Law gives (for 1-D):
-mg + bv2 = ma , or -mg + bv2 = m(dv/dt)
which is a differential equation with v(t)
being the solution.
Air Resistance and Gravity
-mg + bv2 = m (dv/dt)
This differential equation can be solved, but it
is beyond the level of this course.
We can get an approximate solution, however,
by using a numerical (discrete) technique
rather than the differential equation
(continuous function) technique.
Air Resistance and Gravity
-mg + bv2 = m (dv/dt)
The first step is to calculate the forces based on
initial conditions. Let’s say we drop an object that
has a mass of 2 kg and an air resistance coefficient
0.03 Nt-s2/m2 from a helicopter 1,000 meters
above the ground.
Fgravity = mg = 2 kg * 9.8 m/s2 = 19.6 Nt.
FAR = bv2 = 0.03 Nt-s2/m2 * [0 m/s]2 = 0 Nt.
Using Newton’s Second Law, we can now calculate
the initial acceleration:
Air Resistance and Gravity
F = ma gives: -19.6 Nt + 0 Nt = 2 kg * a,
or a = -9.8 m/s2; we now assume that the
acceleration does not change much over the first
second, and calculate the velocity at one second:
v1 = vo + ao*(1 sec)
v1 = 0 m/s + (-9.8 m/s2)*(1 sec) = -9.8 m/s.
x1 = xo + ½(vo+v1)*(1 sec)
x1 = 1,000 m + ½ (0 m/s + -9.8 m/s)*(1 sec) =
995.1 m.
Air Resistance and Gravity
We now repeat the whole process with the
new position and velocity providing new
initial conditions:
Fgravity = mg = 2 kg * 9.8 m/s2 = 19.6 Nt.
(gravity remains the same)
FAR = bv2 = 0.03 Nt-s2/m2 * [-9.8 m/s]2 = 2.88 Nt.
(air resistance increased with increased speed)
Air Resistance and Gravity
F = ma  -19.6 Nt + 2.88 Nt = 2 kg * a,
or a = -8.36 m/s2; we now assume that the
acceleration does not change much over the
second second, and calculate the velocity at
two seconds: v2 = v1 + a1*(1 sec)
v2 = -9.8 m/s + (-8.36 m/s2)*(1 sec) = -18.16 m/s.
x2 = x1 + ½ (v1+v2)*(1 sec)
x2 = 995.1 m + ½(-9.8 m/s + -18.16 m/s)*(1 sec) = 981.12 m.
Air Resistance and Gravity
We simply keep this process up until x
becomes zero.
Normally this would be a lot of steps, but we can use
either a computer program to do this or a
spreadsheet. We can then plot the graph of either
v versus t or x versus t to see what the motion
looks like. (See the Excel spreadsheet FallAR.xls
which can be downloaded from my PHYS 150
web page.)
Air Resistance and Gravity
To see the effects of different air resistance
coefficients, we simply change the value of b and
run the program again or recalculate the
spreadsheet.
To see the effects of different air resistance
functions, such as F=-bv, simply change the F-air
resistance function in cell E-5 and copy this
change into all the following E cells.
This is the same iterative type process that is
used in the Computer Homework Vol. 1 #8 on Projectiles and in Vol. 1 - #9 on
Gravity.
Terminal Velocity
As we can see from the spreadsheet, the velocity
does not continue to increase in a linear fashion
(as it would without air resistance), but instead it
approaches a terminal velocity.
This is because as the speed increases due to gravity,
the air resistance also increases which acts against
gravity.
As the air resistance approaches the strength of
gravity, the net force approaches zero and so the
acceleration also approaches zero, and the velocity
levels off.
Terminal Velocity
We can calculate the terminal velocity without
having to set up the spreadsheet by recognizing
that there will be no acceleration and hence no
increase in speed when the air resistance equals
the gravity: mg = bv2; solving for the terminal
speed gives:
vterminal = [mg/b]1/2.
Using the numbers we used earlier,
vterminal = [2 kg * 9.8 m/s2 / .03 Nt-s2/m2]1/2 =
25.56 m/s = 57 mph.
Air Resistance
b = (1/2)*C*Across-section*ρ
For a sphere, C = ½; for air, ρ = 1.2 kg/m3;
for a tennis ball, r = 3.25 cm, m = 56 grams;
for a golf ball, r = 2.1 cm, m = 45 grams;
for a baseball, circumference = 23 cm, m = 145 grams.
From these values, we should be able to get the b value
and hence the terminal speed for these three balls:
golf ball: 73 mph; tennis ball: 53 mph; baseball: 75
mph.
Note that these values do NOT take into account any
aerodynamic lift due to the seams or dimples and
spin.
Ramps and Pulleys
We use ramps and pulleys to make it easier to
do certain things.
We’ll first look at ramps, and then pulleys.
Ramps
Why is it easier to push something up a ramp
than it is to lift it?
Let’s look at a picture of the situation:
Fc
P
W=mg
Ramps
In the static case with no friction, Newton’s
2nd law gives:
F// = P - mg sin() = ma//
F = Fc - mg cos() = ma = 0
Fc
P

W=mg
Ramps
In the absence of friction, to balance gravity
we only need the Pull, P to be
Pbalance = mg sin().
Since sin() is always
less than or equal to 1, Pbalance will always
be less than or equal to mg: Pbalance mg
Fc
P
θ
W=mg
Ramps
Pbalance = mg sin()
The smaller the angle, the less P is required.
But what price do we pay for decreasing the
pull? The price we pay is for an increased
distance, L, to pull the object: h/L = sin(),
or L = h/sin(). Fc
P
L
θ
W=mg
h
Ramps
If we now include friction, Newton’s 2nd law
gives: F// = P - mg sin() - Ff = ma//
F = Fc - mg cos() = ma = 0
Fc
Ff =Fc
θ
P
W=mg
Ramps
From the perpendicular equation, we see that
Fc = mg cos(), and since Ff = Fc we have
Ff = mg cos(), and so the parallel
equations, with a// = 0 (just balancing weight
and friction) gives:
P = mg sin() +  mg cos()
Fc
Ff =Fc
θ
P
W=mg
Ramps
P = mg sin() +  mg cos() or
P = mg [sin() +  cos()]
As long as  and  are such that
sin() +  cos() < 1, the pulling force, P, will
be less than the weight, which means that
the ramp will make it easier to raise the
object up.
Ramps
Let’s let: f() = sin() +  cos()
(the fraction of mg that the pull needs to be).
To find the minimum (or maximum) of this
function with respect to angle, we solve:
df()/d = 0 = cos() -  sin()
Which leads to the condition for minimum (or
maximum): tan() = 1/ .
Ramps
The condition for minimum (or maximum) pull is:
tan() = 1/ .
Let’s look at an example: Suppose  = 0.3.
Then m = arctan (1/.3) = 73.3o. The fraction, f, with
 = 73.3o and  = 0.3 gives a value:
f(73.3o) = sin(73.3o) + 0.3 cos(73.3o) = 1.044 .
If we use a different angle, say 45o, we get
f(45o) = sin(45o) + 0.3 cos(45o) = 0.919 .
Thus the angle 73.3o gives a maximum! Of course,
with a value greater than 1 at the critical angle of
73.3o, we realize it would be easier to just lift it up
than slide it at this angle!
Pulleys
Pulleys can do two different things for us:
1. Pulleys can change the direction of a
force - sometimes it is easier to pull down
than to lift up.
2. Pulleys can “add ropes” to an object to
reduce the tension in the rope, and hence
reduce the pull we need to apply.
Pulleys
Consider the single pulley below:
All the pulley does is change the direction of
the rope so we can pull down on the rope
instead of lifting up on it.
One of the effects of the
pulley is that there are
essentially two ropes coming
from it - one to the weight and
one to us. This will cause it’s
attachment to the ceiling to have
twice the force on it!
P
W = mg
P = Tension = W
Pulleys
Consider the single pulley in this diagram:
In this case, the single pulley essentially adds
a second rope to the weight. We still have
to pull up, but now the weight is split
between the two ropes, so we only have to
pull with
P
half the weight!
T
P=T, 2T = W; so P=W/2
T
W
Pulleys
We now consider putting two pulleys together
as in the diagram below:
The left pulley essentially adds a second rope
to the weight, reducing the tension in the
rope by half. The right pulley simply
redirects the tension
so the pull is down
instead of up.
P
Pulleys
Consider this combination of pulleys:
The rightmost pulley just redirects the tension
so we pull down.
The other four pulleys combine
to provide five ropes to
hold the weight, thus
reducing the tension in
the rope by a factor of 5.
P
Another Example of Newton’s 2nd Law:
Car going around a turn
Consider first a car making a right turn on a level
road. To make the turn, the car must go in a circle
(for a 90o turn, the car must go in a circle for 1/4
of the complete circle). This means that there will
be an acceleration towards the center of the circle,
which is to the right for a right turn.
From the circular motion relations, we know:
a = w2r, and v=wr; or a = v2/r. That says that we
have a bigger acceleration the faster we go and the
shorter the radius (sharper the turn).
Car going around a turn
What are the forces that cause this circular
acceleration?
Gravity (weight) acts down
Contact Force acts up (perpendicular to the surface)
Friction - which way
does it act (left or right)?
Fc
a=v2/r
W=mg
Car is heading into the screen.
Car going around a turn
Without friction the car will NOT make the
turn, it will continue straight - into the left
ditch! Therefore, friction by the road must
push on the car to the right. (Friction by
the car on the road will be opposite - to the
left.)
The fastest the car can go around
the turn without sliding is
when the friction is
maximum: Ff = Fc .
Fc
Ff  Fc
a=v2/r
W=mg
Car going around a turn
We now apply Newton’s Second Law - in
rectangular components:
Fx = Ff = ma = mv2/r
To make the car go around the turn fastest, we need
the maximum force of friction: Ff = Fc
Fy = Fc - W = 0
which says Fc = mg, so
mg = mv2/r, or vmax = [gr]
Fc
Ff = Fc
a=v2/r
W=mg
Car going around a turn
vmax = [gr]
Note that the maximum speed (without slipping)
around a turn depends on the coefficient of
friction, the amount of gravity (not usually under
our control), and the sharpness of the turn (radius).
If we go at a slower speed around the turn, friction
will be less than the maximum: Ff < Fc.
There is one other thing we can do to go faster
around the turn - bank the road! How does this
work?
Banked turn
By banking the road, we have not added any
forces, but we have changed the directions
of both the contact force and the friction
force!
Have we changed the direction of the
Fc
acceleration? No - the car
is still travelling in a
a=v2/r
horizontal circle.
Ff

W=mg
Banked turn
Since the acceleration is still in the x direction, we
will again use x and y components (rather than //
and )
Fx = Fc sin() + Ff cos() = mv2/r
Fy = Fc cos() - Ff sin() - mg = 0
If we are looking for the max
speed, we will need the max
friction: Ff = Fc .
This gives 3 equations for
3 unknowns: Ff, Fc, and v.
Fc
a=v2/r

W=mg
Ff
Banked turn
Fx = Fc sin() + Ff cos() = mv2/r
Fy = Fc cos() - Ff sin() - mg = 0
Ff = Fc . Using the third equation, we can
eliminate Ff in the first two:
Fc sin() + Fc cos() = mv2/r
Fc cos() - Fc sin() - mg = 0
We can now use the second equation to find Fc:
Fc = mg / [cos() -  sin()], and use this in the
first equation to get:
v = [gr {sin()+ cos()} / {cos() -  sin()} ]1/2
Banked Turn
v = [g r {sin()+ cos()} / {cos() -  sin()} ]1/2
Notice that the mass cancels out. This means
that the mass of the car does not matter!
(Big heavy trucks slip on slippery streets just like small
cars. When going fast, big heavy trucks flip over rather
than slide off the road; little cars don’t flip over like big
trucks. But flipping over is not the same as slipping!
We’ll look at flipping in Part 4 of the course.)
Note also that when  = 0, the above expression
reduces to the one we had for a flat road:
vmax = [gr]1/2 .
Banked Turn
v = [g r {sin()+ cos()} / {cos() -  sin()} ]1/2
Note that as  sin() approaches cos(), the
denominator approaches zero, so the maximum
speed approaches infinity!
What force really supports such large speeds (and so
large accelerations)? As the angle increases, the
contact force begins to act more and more to cause
the acceleration. And as the contact force
increases, so does friction.
Actually, there is a limit on the maximum speed
because there is a limit to the contact force.
Banked turn - Minimum Speed
Is there a minimum speed for going around a
banked turn? Consider the case where the
coefficient of friction is small and the angle
of bank is large. In that case the car, if
going too slow, will tend to
slide down (to the right) so
friction should act to the left.
Fc
Can you get an equation for the Ff
a=v2/r
minimum speed necessary?
What changes in what we did for

max speed?
W=mg
Multiple Objects:
Newton’s 2nd and 3rd Laws
Consider a dish on a tablecloth on a table. If
we pull on the tablecloth (P), can we keep
the dish on the table or will it be pulled off?
Note that while we pull (P) to the left on the
tablecloth, the pull by the tablecloth on us
is to the right (not shown).
P
Tablecloth
Dish
Table
Multiple Objects:
Newton’s 2nd and 3rd Laws
Consider a dish on a tablecloth on a table. If
we pull on the tablecloth (P), can we keep
the dish on the table or will it be pulled off?
We must be very careful to consider what
forces are acting on and what forces are
acting by each object in the problem.
On the diagrams in the next slides, all forces
on the tablecloth will be colored in blue.
All forces on the dish will be colored in
orange.
Multiple Objects:
Let’s look first at the weight. We recognize that
both the tablecloth and dish will have a mass, mc
and md, and so both will also have a weight:
Wc = mcg and Wd = mdg , both directed down.
The earth is doing the pulling, so by Newton’s 3rd
Law, the tablecloth and dish will also pull on the
Earth in the opposite direction (up) with the same
magnitude! However, the mass of the earth will
make the acceleration of the earth negligibly
small.
Dish
P
Tablecloth
mcg
m dg
Table
Multiple Objects:
We now recognize that the dish is pushing on
the tablecloth (due to the dish’s weight).
This is actually a contact force (Fcd) down
on the tablecloth by the dish.
But by Newton’s 3rd Law, there must be an
equal and opposite force (Fcd) up on the
dish by the tablecloth.
F
cd
Dish
P
Tablecloth
mcg
mdg
Fcd
Table
Multiple Objects:
We now recognize that the table is pushing on
the tablecloth. This is actually a contact
force (Fct) up on the tablecloth by the
table.
But by Newton’s 3rd Law, there must be an equal
and opposite force (down) on the table: Fct (not
shown). However, we will assume the table is solid
and will not fall down.
F
cd
P
Tablecloth
Dish
Fct
mdg
mcg
Fcd
Table
Multiple Objects:
We now recognize that there is friction between the
tablecloth and the dish. The dish is resisting the
pull to the left, so the direction of the force (Ff-cd)
on the tablecloth will be to the right.
But by Newton’s 3rd Law, there must be an equal
and opposite force on the dish: Ff-cd. Note that
this will be directed to the left (same direction as
P) trying to move the dish off the table.
P
Tablecloth
Fcd
Ff-cd
Dish
Fct
Ff-cd
mdg
Fcd
mcg
Table
Multiple Objects:
We now recognize that there is friction
between the tablecloth and the table. The
table is resisting the pull, so the direction of
the force (Ff-ct) on the tablecloth will be to
the right.
But by Newton’s 3rd Law, there must be an equal
and opposite force by the tablecloth on the table
(not shown). But since the table is bolted down, we
assume it will not move.
Fcd
Ff-cd
P
Tablecloth
Ff-cd
Dish
Fct Ff-ct
mdg
Fcd
mcg
Table
Multiple Objects:
Now we can use Newton’s 2nd Law:
DISH:
Fx = Ff-cd = mdad
Fy = Fcd - mdg = 0
Tablecloth: Fx = P - Ff-cd - Ff-ct = mcac
Fy = Fct - Fcd - mcg = 0
P
Tablecloth
Fcd
Ff-cd
Ff-ct
Dish
F
ct
Ff-cd
mdg
Fcd
mcg
Table
Multiple Objects:
DISH: Fx = Ff-cd = mdad, or ad = Ff-cd / md
Fy = Fcd - mdg = 0, or Fcd = mdg
Tablecloth: Fx = P - Ff-cd - Ff-ct = mcac, or
P = mcac + Ff-cd + Ff-ct
Fy = Fct - Fcd - mcg = 0, or
Fct = Fcd+mcg = (md+mc)g
Ff-cd = cdFcd = cd mdg, or
ad = cd mdg / md = cd g
Ff-ct = ctFct = ct (md+mc)g
Multiple Objects:
ad = cd g
This says that there will always be a constant
acceleration on the dish, but this will last
only as long as the tablecloth is in contact
with the dish. To minimize the speed, we
need to minimize the time the acceleration
lasts. It will also help to reduce the speed of
the dish if we reduce the friction between
the dish and the tablecloth.
Multiple Objects:
P = mcac + Ff-cd + Ff-ct
= mcac + cd mdg + ct (md+mc)g
This says that the Pull we need will depend on
the masses of the dish and tablecloth, on
both coefficients of friction, and upon the
amount of acceleration of the tablecloth we
want. We would like to have as large an
acceleration as possible in order to keep the
time down.