Transcript Document

The Calculus
connection
When Newton first wrote the relationship we
call the second law, it was not it the form
we use now. Instead, he wrote
dp
F=
dt
dp dmv
dv
F= 
m
 ma
dt
dt
dt
More than one force can act on an object at once. For
example, two people could push on a book at the
same time. One person could push toward the left
and the other could push toward the right. In this
case the two forces would act against each other.
What is the NET force?
2 N towards the left
If the book had a mass of ½ kg, what would be its
acceleration?
a = Fnet / m
a = 2 / .5
a = 4 m/s2 (left)
What if the opposing
forces were equal?
What is the net Force?
What is the
acceleration?
What if the forces were
in the same direction?
What is the net force?
If the book had a mass
of 2 kg, what is its
acceleration?
a = Fnet / m
a = 14 / 2 = 7 m/s2
• What are the unknown forces for the given
net force?
We usually label forces as
negative or positive.
Forces upward are positive.
Forces downward are
negative.
Forces to the right are
positive.
Forces to the left are
negative
+
+
-
What is the net Force?
- 6 N + 18 N = + 12 N
What is the acceleration?
a = Fnet / m
12 / 3 = 4 m/s2
6N
3 kg
18N
What is the net Force?
- 6 N – 15 N + 18 N = - 3 N
What is the acceleration?
a = Fnet / m
a = - 3 N / 3 kg = - 1 m/s2
6N
15N
3 kg
18N
Weight
Weight, Wt. is the gravitational force on an
object
Weight = mass x gravity
W = mg
Since weight is a force, it is measured in
Newtons, N
Remember, “g” on Earth is 9.8 m/s2(10 m/s2)
What is the weight of a 42 kg child on Earth?
W = mg
m = 42 kg
W = 42 kg x 10 m/s2
W = 420 N
What is the mass of a 15000 N car?
W = mg
m=W/g
m = 15000 N / 10 m/s2
m = 1500 kg
Remember, one Newton is not a
very big force (about the same
as a ¼ pound).
So, your weight in Newtons is
MUCH bigger than your weight
in pounds!
In fact, you would have to multiply
your weight in pounds by 4.45
to get your weight in Newtons.
How much do you weigh in
Newton’s?
Even if you weigh
550 Newtons,
You still wouldn’t
be much of a
Sumo Wrestler!
(that’s only around 120 lbs)
A car weighs 1680 N on Earth. What is its
mass?
W = mg
m=W/g
m = 1680 N / 10 m/s2
m = 168.0 kg
What is the net force on the 50 kg
parachutist?
-100 N
What is his acceleration?
a = Fnet / m
a = -100 / 50
a = -2 m/s2
What is the net Force?
Hold on, there’s another force not
drawn!
The gravitational force of weight!
Wt = mg, (g = 10 m/s2)
Wt = 2 kg x 10 m/s2 = 20 N
Draw the weight vector also!
Now, what is the net Force?
Net force = +19 N – 5 N – 20 N =
Net Force = - 6 N
What is the acceleration?
a = Fnet / m =
a=-6/2
a = -3 m/s2 It will accelerate downward.
19 N
2 kg
5N
mg = 20 N
Tension
Tension, T, is the force
that cables, ropes,
and strings pull with.
A child pulls up on a string
that is holding 2 fish of total
mass 5 kg. If he is
providing a tension of 60 N,
what is the net force on the
fish?
Fnet = + T – W
Fnet = 60 N – 50 N
Fnet = 10 N
What is the acceleration of the
fish?
a = Fnet / m
a = 10 / 5
a = 2 m/s2
, “sigma” is a Greek letter that is used
to signify “the sum of”
Quite often, in Newton’s 2nd Law, we
write  F = ma instead of Fnet = ma
A child pulls a 5 kg bucket out of well with a
rope. If the bucket accelerates upward at
1.2 m/s2, what is the tension in the rope?
m = 5 kg
a = 1.2 m/s2
T
T=?
 F = ma
T – mg = ma
mg
T = ma + mg
T = 5 x 1.2 + 5 x 9.8
T = 55 N
Three kids pull on a 2
kg cat. One child
pulls left with a force
of 20 N. One child
pulls right with a force
of 50 N. If the cat
accelerates to the left
at 3 m/s2, what was
the force exerted by
the third child, F3?
F1 = 20 N
F2 = 50 N
F = ma
-F1 + F2 + F3 = ma
F3 = ma + F1 – F2
F3 = 2(-3) + 20 – 50
F3 = -36 N
The “Normal” Force, N
When an object is pressed against a
surface, the surface pushes back. (That’s
Newton’s 3rd Law)
This “push back” from the surface is
called the Normal Force, N
The word “normal” in math terminology
means “perpendicular”
The surface pushes back in a direction that
is perpendicular to the surface.
Normal force, N
Weight = mg
If the box is not accelerating,
then the Normal force must be
balanced by the Weight
But… what if you were accelerating up or down?
The Normal force would NOT be equal to your
weight if you’re accelerating up or down. And…
your weight seems to change!
Apparent Weight
When you ride an elevator, you “feel”
heavier or lighter than you actually weigh
because of the acceleration of the
elevator.
Your “apparent weight” is found by taking
your REAL weight, mg, and adding the
term ma, where “a” is your acceleration
Apparent weight = mg + ma
A 50 kg woman steps in an elevator that
accelerates upward at 1.5 m/s2.
What is her REAL weight?
mg = 500 N
What is her APPARENT weight?
Fnet = ma
N – mg = ma
Her APPARENT weight is what she feels
like she weighs, given by the
Normal force, N.
N = ma + mg
N = 50 x 1.5 + 500
Apparent weight = 575 N
N
mg
Do falling objects REALLY
accelerate toward the Earth
at 9.8 m/s2?
No, because of air resistance.
Air resistance is a force that
pushes up on an object as it
falls.
The faster you fall…
The greater the air resistance.
Eventually, the air resistance
pushing up on you is just as
large as your weight that is
pulling down on you!!
The faster the man falls, the more air resistance
pushes up on him. Eventually, there will be just as
much air resistance pushing up on him as his weight
pulling him down. What will be the NET force acting
then?
What will be his acceleration?
Once the air resistance pushing up is as large as
the weight pushing down, the NET force acting on
you is ZERO!
If the net force is zero, what is your acceleration?
ZERO!
This doesn’t mean you stop in mid air. But it does
mean that you stop accelerating!
You still continue to fall towards the Earth, but you
don’t pick up any more velocity- you continue to
fall towards the Earth at the same velocity.
This speed is called your “terminal velocity”
You will reach your terminal velocity when
Air resistance = your weight
Which one will have a faster terminal velocity?
You don’t reach terminal velocity until the air
resistance grows to as large a force as your
weight.
The more massive skydiver will have a faster terminal
velocity and hit the ground at a faster speed
The “Spring Force”
If an object is attached to a
spring and then pulled or
pushed, the spring will exert
a force that is proportional
to the displacement of the
object but opposite in
direction.
Fs = -kx
Where k is called the “spring
constant” or the “force
constant” and describes
how stiff the spring is.
The units for k are N/m
Friction, f
• A force that always opposes motion
• Depends on two things: the roughness of
the surfaces and how hard they are
pressed together.
f = mN
m , mu- the “coefficient of friction” tells how
rough the surfaces are.
N, the Normal force tells how hard the
surfaces are pressed together
Example: How large is the frictional force
between 2 surfaces if the coefficient of
friction is 0.2 and the Normal force is 80 N?
f = mN
f = 0.2 x 80
f = 16 N
There are two kinds of friction:
“static friction” (not moving) must be
overcome to initiate motion.
“kinetic friction” must be overcome while
an object is moving
Static friction > Kinetic friction
You pull on a box with an applied force of 30 N. The coefficient of friction is 0.4. If the
mass of the box is 2 kg, what is its acceleration?
1. Draw the box and all FOUR forces acting on it.
2. Write what you know and don’t know.
3. Write the equations, Fnet = ma and f = mN
4. Calculate the Normal force and the friction force.
5. Calculate the value of the net Force and then the acceleration.
FA = 30 N
m = 2 kg
m = 0.4
a=?
Fnet = ma
f = mN
f = mN
N = mg = 2 x 10 = 20N
f = mN = 0.4(20) = 8N
Horizon: Fnet = FA - f
30N – 8N = 22N
a = Fnet / m
a = 11 m/s2
Normal force, N
FA= 30 N
Weight = mg
FA = 25N
m = 2 kg
q = 43
m = 0.4
a=?
Fx = ma
f = mN
 Fx = - mN + FA cos q = ma
Fy = N – mg - Fsin q = ma = 0
N = mg + Fsin q
 36.65 N
Normal force, N
(-0.4(36.65) + 25cos 43 ) / 2= a
a = 1.81 m/s2
f
q
FA
Weight = mg
FA = 15 N
m = 2 kg
q = 43
m = 0.4
a=?
Fx = ma
f = mN
 Fx = - mN + FA cos q = ma
Fy = N – mg + Fsin q = ma = 0
N = mg – Fsin q
 9.77 N
Normal force, N
FA
(-0.4(9.77) + 15cos 43) / 2 = a
a = 3.53 m/s2
f
q
Weight = mg
If the box is moving at
constant speed, there is no
acceleration,
Therefore the net force must
be zero…
so the horizontal forces must
cancel each other.
f
mN = Fcos q
Normal force, N
FA
q
If you push hard enough to
just get the box moving, the
acceleration is zero in that
Weight = mg
case also, but the friction is
static, not kinetic.
Inclines: With friction, case 1: at
rest or sliding down
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector, both perpendicular and parallel to the
incline. The components help form a right triangle. Label the angle.
Write Newton’s Second Law for the
box.
 F = ma
What is friction?
N
f
- mgsin q = ma
mmgcos q - mgsinq = ma
f
mgcos q - gsinq = a
q
f
= mN
What is N?
N = mgcos q
f = mmgcos q
mg
q
Inclines: With friction, case 2:
pushed downward
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector, both perpendicular and parallel to the
incline. The components help form a right triangle. Label the angle.
Write Newton’s Second Law for the
box.
 F = ma
What is friction?
N
f
- mgsin q – FA = ma
mmgcos q - mgsinq – FA = ma
f
FA
q
mg
q
f
= mN
f = mmgcos q
Inclines: With friction, case 3:
pushed upward
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector, both perpendicular and parallel to the
incline. The components help form a right triangle. Label the angle.
Write Newton’s Second Law for the
box.
 F = ma
N
FA - mgsin q – f = ma
FA - mgsinq - mmgcos q = ma
FA
What is friction?
f
= mN
f = mmgcos q
q
mg
f
q
Friction along an incline
An object placed along an incline will
eventually slide down if the incline is
elevated high enough. The angle at which
it slides depends on how rough the incline
surface is. To find the angle where it
Since it doesn’t move:
slides:
mgsinq = mmgcosq
Therefore:
Tan qmax = mmax, the coefficient of static friction
Or
The angle qmax = tan -1 mmax
q