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 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 = dv/dt
which is a differential equation with v(t) being
the solution.
Air Resistance and Gravity
-mg + bv2 = 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 = 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.
FAR = bv2 = 0.03 Nt-s2/m2 * [-9.8 m/s]2 = 2.88 Nt.
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 201
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.
Terminal Velocity
As we can see from the spreadsheet, the velocity
does not continue to increase (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 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.
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
W=mg
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
P
W=mg
Ramps
From the perpendicular equation, we see that
Fc = mg cos(), so the parallel equations,
with a// = 0 (just balancing weight and
friction) gives: P = mg sin() +  mg cos()
Fc
Ff =Fc
P
W=mg
Ramps
P = mg sin() +  mg 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.
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
acceleration? No - the car
is still travelling in a
Fc
horizontal circle.
a=v2/r

W=mg
Ff
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.
Can you get an equation for the
Fc
minimum speed necessary? Ff
a=v2/r
What changes in what we did for
max speed?

W=mg
Example of Torque
Consider two people carrying a board with a
weight on it – see the diagram below. The
weight is 100 lb, and is located 7 feet away
from the left end. The board is 8 feet long
and we neglect the weight of the board.
How much force does each person have to
exert?
FL = ?
FR = ?
W = 100 lb
7 feet
8 feet
Example of Torque
We recognize this as a statics problem, and a
one dimensional problem (in y). This gives
us one relation: F = 0. We identify the
force the left person exerts, FL, the force the
right person exerts, FR, and the Weight.
F = +FL + FR – 100 lb = 0 .
FL = ?
FR = ?
W = 100 lb
7 feet
8 feet
Example of Torque
F = +FL + FR – 100 lb = 0 .
This is one equation in two unknowns. There
are lots of ways of dividing up the weight
among the left and right persons. What
does determine how much each has to
exert?
FL = ?
FR = ?
W = 100 lb
7 feet
8 feet
Example of Torque
F = +FL + FR – 100 lb = 0 .
What is important is where the weight is placed!
To get our second equation, we recognize this as a
torque situation: t = 0 where each of the three
forces has a distance (moment arm). But where do
we measure each moment arm from – since there
is no actual rotation about any point?
FL = ?
FR = ?
W = 100 lb
7 feet
8 feet
Example of Torque
F = +FL + FR – 100 lb = 0 .
Since there is no obvious point about which the board
rotates, we are free to choose any point.
Since the diagram indicates the distances measured
from the left end, we can initially choose that point.
Also for definiteness sake, let’s choose a counterclockwise (CCW) rotation as being positive.
FL = ?
FR = ?
W = 100 lb
7 feet
8 feet
Example of Torque
F = +FL + FR – 100 lb = 0 .
From the choices of the previous slide, we see that xL = 0 ft,
xR = 8 ft, and xW = 7 ft, and the torque due to FL is zero,
the torque due to FR is CCW and so is positive, and the
torque due to W is CW and so is negative:
t = {FL * 0 ft} + {FR * 8 ft} - (100 lb * 7 ft) = 0.
FL = ?
FR = ?
W = 100 lb
7 feet
8 feet
Example of Torque
F = +FL + FR – 100 lb = 0 .
t = {FL * 0 ft} + {FR * 8 ft} - (100 lb * 7 ft) = 0.
We now have two equations for two unknowns, and
we can solve the problem.
From the torque equation we can solve for FR:
FR = (100 lb * 7 ft) / 8 ft = 87.5 lb. And from
the force equation: FL = 100 lb – 87.5 lb = 12.5 lb.
FL = ?
FR = ?
W = 100 lb
7 feet
8 feet
Example of Torque
FR = (100 lb * 7 ft) / 8 ft = 87.5 lb, and
FL = 100 lb – 87.5 lb = 12.5 lb.
Note that the further the weight is from the person,
the lighter is the necessary force.
Also note that by choosing the measuring point at the
left force location, we effectively eliminated the FL
unknown from the torque equation making it easier
to solve.
FL = ?
FR = ?
W = 100 lb
7 feet
8 feet