Transcript Slide 1 - Particle and Astroparticle Physics

Lecture Outlines
Chapter 6
Physics, 3rd Edition
James S. Walker
© 2007 Pearson Prentice Hall
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Chapter 6
Applications of Newton’s
Laws
Units of Chapter 6
• Frictional Forces
• Strings and Springs
• Translational Equilibrium
• Connected Objects
• Circular Motion
6-1 Frictional Forces
Friction has its basis in surfaces that are not
completely smooth:
6-1 Frictional Forces
Kinetic friction: the friction experienced by
surfaces sliding against one another
The static frictional force depends on the normal
force:
(6-1)
The constant
kinetic friction.
is called the coefficient of
6-1 Frictional Forces
6-1 Frictional Forces
The kinetic frictional force is also independent
of the relative speed of the surfaces, and of their
area of contact.
Example 6-1
Pass the Salt – Please
Example 6-2 (p.145) (x=3,0 m, θ=23º, μk= 0,26, t =?)
Making a Big Splash
6-1 Frictional Forces
The static frictional force keeps an object from
starting to move when a force is applied. The
static frictional force has a maximum value, but
may take on any value from zero to the maximum,
depending on what
is needed to keep
the sum of forces
zero.
6-1 Frictional Forces
(6-2)
where
(6-3)
The static frictional force is also independent
of the area of contact and the relative speed of
the surfaces.
Example 6-3 (p.147) (m=95,0 kg, θ = 23,2 º, μs=?)
Slightly Tilted
Conceptual Checkpoint 6-1
Friction for Rolling Tires
Figure 6-4
Stopping distance with and without ABS
6-2 Strings and Springs
When you pull on a string or rope, it becomes
taut. We say that there is tension in the string.
6-2 Strings and Springs
The tension in a real rope will vary along its
length, due to the weight of the rope.
Here, we will assume that
all ropes, strings, wires,
etc. are massless unless
otherwise stated.
6-2 Strings and Springs
An ideal pulley is one that simply changes the
direction of the tension:
Example 6-4 (p.151) (T=165 N, m=?)
A Bad Break: Setting a Broken Leg with Traction
Conceptual Checkpoint 6-2
Compare the Readings on the Scales
6-2 Strings and Springs
Hooke’s law for springs states that the
force increases with the amount the
spring is stretched or compressed:
[Konstanten k kallas fjäderkonstanten och
vi kommer bara att behandla ideala fjädrar,
det vill säga masslösa fjädrar som lyder
Hookes lag]
Figure 6-8
Spring forces
Active Example 6-2 (p.153)
(F=0,22N, x=3,5 mm. a) k = ? b) Om x = 4,0 mm, F =? )
Nasal Strips
6-3 Translational Equilibrium
When an object is in translational equilibrium,
the net force on it is zero:
(6-5)
This allows the calculation of unknown forces.
6-3 Translational Equilibrium
Conceptual Checkpoint 6-3
Comparing Tensions
Conceptual Checkpoint 6-3
Comparing Tensions
Example 6-5 (p.156) (m=6,20 kg, θ=40,0º, T1=?, T2=?)
Suspended Vegetation
Figure 6-16
Conceptual Exercise 1
Active Example 6-3 (p.157) (m = 1,94 kg, θ = 3,50º, T = ?)
The Forces in a Low-Tech Laundry
Chapter 6 Opener
Newton’s Laws
6-4 Connected Objects
When forces are exerted on connected objects,
their accelerations are the same.
If there are two objects connected by a string,
and we know the force and the masses, we can
find the acceleration and the tension:
6-4 Connected Objects
We treat each box as a separate system:
6-4 Connected Objects
F –T = m1a1 = m1a
T
= m 2 a 2 = m2 a
(6-7)
F = m1a + m2a = (m1+m2)a
a = F/(m1+m2)
(6-8)
T = m2 F/(m1+m2)
(6-9)
6-4 Connected Objects
If there is a pulley, it is easiest to have the
coordinate system follow the string (NO FRICTION)
Conceptual Checkpoint 6-4 Tension in the string
(Attention, the box slides without friction!)
T
= m1 a
W2 - T
= m2 a
W2
= m 2g
W2
= (m1+m2)a
a
= m2g/(m1+m2)
T
= m1m2g/(m1+m2)
Example 6-7 (p.161)
Atwood’s Machine
Example 6-7 Atwood’s Machine 6-4
T - m1g
= m1 a
m2g - T
= m2 a
Eliminate T by adding the two equations
g(m2- m1)
= a(m2+m1)
a
= (m2-m1)g/(m2+m1)
If m1=3,1 kg, m2 = 4,4 kg blir a = 1,7 m/s2
Figure 6-12
Swinging a ball in a circle
6-5 Circular Motion
An object moving in a circle must have a force
acting on it; otherwise it would move in a straight
line.
The direction of the
force is towards the
center of the circle.
Table 6-2
sin θ / θ for values of θ Approaching Zero
6-5 Circular Motion
Some algebra gives us the magnitude of the
acceleration, and therefore the force, required
to keep an object of mass m moving in a circle
of radius r.
The magnitude of the force is given by:
(6-15)
6-5 Circular Motion
This force may be provided by the tension in a
string, the normal force, or friction, among others.
Example 6-8 (p.163) Rounding a Corner
(m= 1200 kg, r = 45 m, μs = 0,72, vmax = ?)
Example 6-8 Rounding a Corner
(fs = μsN)
fs
= max
N - mg
= 0
μ sN
= mvmax2/r
vmax
= (0,82•9,81•45)1/2 = 19 m/s
(when μs is maximum)
Photo 6-7
Banking
Example 6-9 Bank on It (p.165)
(m= 900 kg, v = 20,5 m/s, r = 85 m, θ = ?)
Example 6-9 Bank on it
N sinθ
= max = maCP= mv2/r
N cosθ - W = 0
Division gives
tanθ = mv2/mgr = v2/gr = (20,5)2/(9,81•85,0) = 0,504
θ = arctan(0,504) = 26,7 º
Active Example 6-4
Find the Normal Force
Active example 6-4 Find the Normal Force
(v = 17,0 m/s, m = 80,0 kg, r = 65,0 m)
N - mg
= may
ay = v2/r
N = mg + mv2/r = m(9,81 + 4,47) = 1140 N
Figure 6-14
Simplified top view of a centrifuge in operation
Exercise 6-1 (p.167)
(Find the magnitude of the acceleration when
v = 89,3 m/s, r = 8,50 cm)
acp = v2/r = (89,3)2/(0,0850) = 93800 m/s2 = 9560 g
6-5 Circular Motion
An object may be changing its speed as it
moves in a circle; in that case, there is a
tangential acceleration as well:
Summary of Chapter 6
• Friction is due to microscopic roughness.
• Kinetic friction:
• Static friction:
• Tension: the force transmitted through a
string.
• Force exerted by an ideal spring:
Summary of Chapter 6
• An object is in translational equilibrium if the
net force acting on it is zero.
• Connected objects have the same acceleration.
• The force required to move an object of mass m
in a circle of radius r is: