Download Chapter 5 Applying Newton`s Laws

Chapter 5
Applying Newton’s Laws
5.1
Using Newton’s First Law
First Law. A body acted on by no net force, i.e.
!
⃗i = 0
F
i
has a constant velocity (which may be zero) and zero acceleration.
Example 5.1. A gymnast with mass mG = 50 kg suspends herself from
the lower end of a hanging rope of negligible mass. The upper end of the rope
is attached to the gymnasium ceiling. (a) What is the gymnasts’s weight? (b)
What force (magnitude and direction) does the rope exert on her? (c) What
is the tension at the top of the rope?
Step 1: Choose coordinate system. Let’s choose x-axis to point upward.
Step 2: Draw free-body diagrams.
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CHAPTER 5. APPLYING NEWTON’S LAWS
50
Step 3: Apply Newton’s Laws.
(a) What is the gymnasts’s weight? Although weight is a force (due to
gravity near the surface of Earth) and thus a vector, one usually speaks of
weight as a magnitude of force vector given in general by Eq. (4.16). For the
gymnast it is
w = 9.8 m/s2 · 50 kg = 490 N.
(5.1)
If we choose a coordinate system such that x-axis point upward, then
⃗ earth on gymnast = (−490 N) î.
w
(5.2)
(b) What force (magnitude and direction) does the rope exert on her?
Draw a free-body diagram for the gymnast. There are only two forces action
⃗ rope on gymnast and due to
⃗ earth on gymnast and tension T
the gymnast: weight w
first law they must balance each other
⃗ rope on gymnast = 0
⃗ earth on gymnast + T
w
(5.3)
⃗ rope on gymnast = (+490 N) î.
T
(5.4)
Trope on gymnast = 490 N
(5.5)
or using Eq. (5.2)
whose magnitude is
and the direction is pointing upward.
(c) What is the tension at the top of the rope? Draw a free-body diagram
for the rope. There are again only two forces action on the rope and due to
first law they must also balance each other
⃗ gymnast on rope + T
⃗ ceiling on rope = 0.
T
(5.6)
We can now use the third law and Eq. (5.4) to find
⃗ gymnast on rope = −T
⃗ rope on gymnast = (−490 N) î.
T
(5.7)
By combining (5.6) and (5.7) we get
⃗ ceiling on rope = (+490 N) î
T
(5.8)
Tceiling on rope = +490 N.
(5.9)
whose magnitude is
Example 5.2. Find the tension at each end of rope in Example 5.1 if
the weight of the rope is 120 N.
CHAPTER 5. APPLYING NEWTON’S LAWS
51
Step 1: Coordinate system. Was already chosen in Example 5.1.
Step 2: Draw a free-body diagram for the gymnast and the rope.
Step 3: Apply Newton’s Laws. Nothing changed for the gymnast’s diagram and thus we can still follow the same logic and conclude that
⃗ gymnast on rope = (−490 N) î.
T
(5.10)
What is however different for the rope diagram is that there is yet another
force acting on the rope - the weight - and thus the first law implies
⃗ gymnast on rope + T
⃗ ceiling on rope + w
⃗ earth on rope = 0.
T
(5.11)
⃗ earth on rope = (−120 N) î.
w
(5.12)
where
By combining together Eqs. (5.10), (5.11) and (5.12) we get
⃗ ceiling on rope = −T
⃗ gymnast on rope −⃗
T
wearth on rope = − (−490 N) î−(−120 N) î = (+610 N) î.
(5.13)
Example 5.4. A car of weight w rests on a slanted ramp attached to a
trailer. (See figure below. Angle α is given. ) Only a cable running from the
trailer to the car prevents the car from rolling off the ramp. (The car brakes
are off and its transmission is neutral.) Find the tension in the car and the
force that the ramp exerts on the car’s tires.
Step 1: Coordinate system. To proceed we must choose a 2D coordinate
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CHAPTER 5. APPLYING NEWTON’S LAWS
system with x-axis pointing to the right and y-axis pointing up (note that
this is intentionally different from the choice made in your textbook).
Step 2: Draw a free-body diagram for the car.
Step 3: Apply Newton’s laws. From the first law we have
⃗ + ⃗n = 0.
⃗ +T
w
(5.14)
Then Eq. (5.14) can be written in components as
(0, w) + (T cos α, T sin α) + (−n sin α, n cos α) = 0
(5.15)
or for each component separately
T cos α − n sin α = 0
w + T sin α + n cos α = 0.
(5.16)
We now have two equations with two unknowns which can be easily solved
T = n
w+n
sin α
cos α
sin2 α
+ n cos α = 0.
cos α
(5.17)
and thus
n = w cos α
T = w sin α.
(5.18)
CHAPTER 5. APPLYING NEWTON’S LAWS
5.2
53
Using Newton’s Second Law
Second Law. If a net external force acts on a body, the body accelerates.
The direction of acceleration is the same as the direction of the net force.
The mass of the body times the acceleration vector of the body equals to the
net force vector, i.e.
!
⃗ i = m⃗a
F
(5.19)
i
Example 5.6. An iceboat is at rest on a frictionless horizontal surface.
Due to the blowing wind, 4.0 s after the iceboat is released, it is moving to the
right at 6.0 m/s. What constant horizontal force FW does the wind exert on
the iceboat? The combined mass of iceboat and rider is 200 kg.‘
Step 1: Coordinate system: We can choose a 1D coordinate system with
x-axis pointing in the direction of wind and origin at the place where the
boat was released.
Step 2: Initial Conditions are
t0 = 0 s
x0 = 0 m
v0x = 0 m/s
(5.20)
Step 3: Final conditions are
v(4.0 s) = 6.0 m/s
(5.21)
CHAPTER 5. APPLYING NEWTON’S LAWS
54
and from equation of motions with constant acceleration we get
vx (t) = v0x + ax t
vx (4.0 s) = 0 m/s + ax · 4.0 s.
(5.22)
By equating Eqs. (5.21) and (5.22) we arrive at the expression for acceleration due to wind
ax = 1.5 m/s2
(5.23)
but since the mass of the boat is
m = 200 kg
(5.24)
"
#
⃗ W = (200 kg) 1.5 m/s2 î = (300 N) î.
F
(5.25)
the second law implies
Example 5.10. A toboggan loaded with physics students (total weight w)
slides down a snow-covered hill that slopes at a constant angle α. The toboggan is well waxed, so there is virtually no friction. (a) what is its acceleration? (b) What is normal force?
Step 1: Coordinate system. This time let’s work with a coordinate suggested in the textbook: x-axis pointing in the direction of the slope and
y-axis normal to the slope. Exact location of origin is not important.
Step 2: Free-body diagram. Draw a free-body diagram for the toboggan.
Step 3: Apply Newton’s Laws. From the second law
⃗ + ⃗n = m⃗a
w
(5.26)
CHAPTER 5. APPLYING NEWTON’S LAWS
55
where mass can be obtained from the weight
m=
w
.
g
(5.27)
The second law can also be written in components as
w
(w sin α, −w cos α) + (0, n) = ( ax , 0)
g
(5.28)
or as two separate equations
w
ax
g
−w cos α + n = 0.
w sin α =
(5.29)
Then the acceleration is
⃗a = g sin αî
(5.30)
⃗n = w cos αĵ.
(5.31)
and normal force is
5.3
Friction forces
Friction coefficients. There two types of contact forces between macroscopic objects. One is the normal force which is perpendicular to the contact
surface and another one is parallel to the contact surface:
⃗n − normal force always perpendicular to the contact surface
⃗f − friction force is always parallel to the contact surface .
Both forces arise due to microscopic (electromagnetic) interaction between
molecules, but we shall only study their macroscopic properties.
If there is a motion along the surface of contact, then these two forces are
related to each other by the so-called coefficient of kinetic friction
µk =
fk
n
(5.32)
or
fk = µk n.
(5.33)
(Note that the friction coefficients are dimensionless, i.e. have no units.)
If there is no motion along the surface of contact, then the friction force
is bounded from above
fs ≤ (fs )max = µs n.
(5.34)
CHAPTER 5. APPLYING NEWTON’S LAWS
56
It turns out that
µs > µk .
(5.35)
and so a larger force T must be applied to an object at rest for it to start
moving, i.e. fsmax , but then a smaller force will be sufficient to continue
motion, i.e. fk , since
fk < fsmax .
(5.36)
Here is a plot of frictional force as a function of applied force
The exact values of both coefficients depend on the materials in constant,
e.g.
Material
µs
µk
steel on steel
0.74 0.57
ice on steel
0.03 0.015
dry rubber on dry concrete 1.0
0.8
wet rubber on wet concrete 0.3 0.25
There is also rolling friction which is typically much smaller. For steel
wheels on steel rails it is ∼ 0.002 − 0.003.
Example 5.13. You want to move a 500 N crate across a level floor. To
start the crate moving, you have to pull with a 230 N horizontal force. Once
the crate starts to move, you can keep it moving at constant velocity with
only 200 N force. What are the coefficient of static and kinetic friction?
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CHAPTER 5. APPLYING NEWTON’S LAWS
Step 1: Choose coordinate system.
Step 2: Draw free-body diagrams.
Step 3: Apply Newton’s Laws. For the static case the First Law implies
⃗ + (f⃗s )
⃗n + w
⃗ +T
max = 0
(5.37)
nĵ − w ĵ + T î − (fs )max î = 0
(5.38)
or
and thus
n = w
T = (fs )max .
(5.39)
By combining with Eq. (5.34) we get
µs =
(fs )max
T
230N
= =
= 0.46.
n
n
500N
(5.40)
For the kinetic case the First Law implies
⃗ + f⃗k = 0
⃗n + w
⃗ +T
(5.41)
nĵ − w ĵ + T î − fk î = 0
(5.42)
or
and thus
n = w
T = fk
(5.43)
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CHAPTER 5. APPLYING NEWTON’S LAWS
By combining with Eq. (5.32) we get
µk =
fk
T
200N
= =
= 0.40.
n
n
500N
(5.44)
Example 5.16. A toboggan loaded with physics students (from Example
5.10) slides down a snow-covered hill. The wax has worn off , so there is a
nonzero coefficient of kinetic friction µk . The slope has just the right angle
α to make the toboggan slide with constant velocity. Find the angle in terms
of w and µk .
Step 1: Coordinate system. Let’s choose x-axis to point in the direction
of the slope and y-axis normal to the slope. Exact location of origin is (once
again) not important.
Step 2: Draw a free body diagram.
Step 3: Apply Newton’s Laws. For the kinetic case the First Law implies
⃗n + w
⃗ + f⃗k = 0
(5.45)
nĵ + (w sin α) î − (w cos α) ĵ − fk î = 0
(5.46)
or
and thus
n = w cos α
w sin α = fk
(5.47)
By combining with Eq. (5.32) we get
µk =
fk
w sin α
=
= tan α
n
w cos α
(5.48)
CHAPTER 5. APPLYING NEWTON’S LAWS
59
and so
α = arctan µk .
(5.49)
Fluid (air) resistance. The (magnitude of) force of fluid resistance
depends on the velocity,
$
kv
for "small" velocities
f=
(5.50)
2
Dv for "large" velocities
where the coefficients depend on many factors: type of fluid, shape of object,
etc. We can apply the second law (in the vertical direction) for a falling
object to get
mg − (kv) = ma
(5.51)
for small velocities. This acceleration will keep accelerating the object until
the (so-called terminal) velocity is
vt =
mg
k
(5.52)
in which case according to Eq. (5.51) it will start moving with constant
velocity, i.e. a = 0.
Similarly in the regimes where the large velocities approximation of resistance force of Eq. (5.50) is valid, the Second Law implies
mg − (Dv 2) = ma.
(5.53)
and thus the terminal velocity is
vt =
%
mg
.
D
(5.54)
Example 5.18. For a human body falling through air in a spread-eagle
position, the numerical value of the constant D in Eq. (5.6) is about 0.25
kg/m. Find the terminal speed for a 50 kg skydiver.
Step 1: Coordinate system. Let’s choose the x-axis to point upward.
Step 2: Draw a free body diagram.
CHAPTER 5. APPLYING NEWTON’S LAWS
60
Step 3: Apply Newton’s laws. We have already used the First Law to derive
Eq. (5.54) and thus we can just use it to get the terminal velocity
&
%
mg
(50 kg)(9.8 m/s2 )
=
= 44 m/s.
(5.55)
vt =
D
0.25 kg/m
5.4
Dynamics of circular motion
We have already looked at a circular motion with constant speed in Section
3.4 and derived the following relations
a⊥ =
v2
R
(5.56)
CHAPTER 5. APPLYING NEWTON’S LAWS
and
a⊥ =
4π 2 R
T2
61
(5.57)
where
a⊥
v
R
T
−
−
−
−
magnitude of acceleration
constant speed
radius of circular path
period of motion.
(5.58)
An object in such a motion experiences a constant (in magnitude) acceleration and thus according to Second Law the (magnitude) of net force must
be
v2
4π 2 R
Fnet = ma⊥ = m = m 2 .
(5.59)
R
T
Example 5.19. A sled with a mass of 25.0 kg rests on a horizontal sheet
of essentially frictionless ice. It is attached by a 5.00 m rope to a post set in
the ice. Once given a push, the sled revolves uniformly in a circle around the
post . If the sled makes five complete revolutions every minute, find the force
F exerted on it by the rope.
Step 1: Coordinate system. Let’s choose a moving coordinate system with
perpendicular direction towards center and parallel direction in the direction
of motion of the sled.
Step 2: Draw a free body diagram.
CHAPTER 5. APPLYING NEWTON’S LAWS
62
Step 3: Apply Newton’s Laws. From the Second Law we got Eq. (5.59) which
implies that
4π 2 R
Fnet = m
T
where
60 s
= 12 s
5
m = 25.0 kg
R = 5.00 m.
T =
and thus
Fnet = 34.3 N.