Transcript Thomson Template - University of North Texas

Physics 1710 Chapter 5: Laws of Motion—III
1′ Lecture
• Newton’s Laws of Motion are:
 Acceleration (or deceleration) occurs if and
only if there is a net external force.
 a = F/m [Note this is a vector eqn.]
 The force exerted by a first object on a
second is always equal and opposite the
the force exerted by the second on the
first. F12 = - F21
Physics 1710 Chapter 5: Laws of Motion—III
1′ Lecture –continued (30″ more)
• Weight is the force of gravity equal to
the mass of the object.
g times
• g =9.80 N/kg
• The force of friction is opposed to the motion of a
body and proportional to the normal force.
• Free body diagrams are sketches of all the
forces acting on a body.
Physics 1710 Chapter 5: Laws of Motion—III
Laws of Motion
1st Law:
An acceleration is caused by a net external
force.
REVIEW
Physics 1710 Chapter 5: Laws of Motion—III
2nd Law of Motion:
a= F /m
Note the vector nature of the equation:
ax= Fx /m
ay= Fy /m
REVIEW
Physics 1710 Chapter 5: Laws of Motion—III
Third Law of Motion:
If two objects interact,
the force of the 1st on the 2nd is equal and opposite
to the force of the 2nd acting on the 1st body.
F12 = - F21
F12
#2
#1
F21
Physics 1710 Chapter 5: Laws of Motion—III
3rd Law in Action—Tug of War
#2
F21
?
It’s the Law!
F12 = - F21
#1
F12
Physics 1710 Chapter 5: Laws of Motion—III
Third Law of Motion, Implications :
Baron von Munchenhausen
Draw “free body diagram”
Arm force =
- Reaction force
Lou Geels
As Baron von Munchenhausen
Net force = 0
F12 = - F21
Physics 1710 Chapter 5: Laws of Motion—III
1. The tree is not pushing on the elephant.
2. The tree is pushing back, but the elephant is
pushing harder.
3. The tree is pushing harder on the elephant that
she is pushing on the tree.
4. The tree and elephant push equally against each
other.
But how can you have acceleration in this case?
The acceleration of object #2 is
a2 = F12 /m2
Not 0 = ( F12 +F21 )/(m1 + m2)
Physics 1710 Chapter 5: Laws of Motion—III
Third Law of Motion, Implications :
Tug of war
Draw “free body diagram”
Physics 1710 Chapter 5: Laws of Motion—III
Third Law of Motion, Implications :
Tug of war :“free body diagram”
Friction: m
-μ (mg)
Floor reaction:
mg
Mg
Rope tension
F12
F21
-mg
Resultant:
Weight
Fm = F12 – μ(mg)
FM = 0
M
Friction:
μ (Mg)
-Mg
Physics 1710 Chapter 5: Laws of Motion—III
Third Law of Motion, Implications :
• Why does mud fly backward from the
tire when an automobile accelerates in
mud?
• For each “action” and equal but
“opposite action” must happen.
Physics 1710 Chapter 5: Laws of Motion—III
2nd and 3rd Laws of Motion: Implications
Why does air have “resistance?”
Fdrag
Fdrag
- gm
- gm
Physics 1710 Chapter 5: Laws of Motion—III
Think!
Peer Instruction Time
No Talking!
Confer!
Physics 1710 Chapter 5: Laws of Motion—III
2nd and 3rd Laws of Motion: Implications
Why does air have “resistance?”
Inertia of the air in the way
Fdrag
- gm
Fdrag
- gm
Δmair
= ρAv Δt
aair = v/ Δt
Fair, max = Δmair aair
= (ρAv Δt)(v/ Δt) = ρAv 2
Fair
= ½ D ρAv 2 = - Fdrag
Physics 1710 Chapter 5: Laws of Motion—III
• Why do objects stop moving?
• Friction is a force that resists motion.
•Friction is proportional to the “normal” force
(ie perpendicular force) and opposed to the
direction of the applied force.
Physics 1710 Chapter 5: Laws of Motion—III
Empirical Observation:
Ff = - μN,
in opposite direction of applied force.
μ is the coefficient of friction, which
depends upon the nature of the two surfaces in contact
and the state of motion.
Physics 1710 Chapter 5: Laws of Motion—III
Friction is caused by “welding” of the
two surfaces.
Lubrication changes the nature of the
contact and/or the surface, atomically.
Physics 1710 Chapter 5: Laws of Motion—III
Friction is caused by “welding” of the two
surfaces.
Welds
Physics 1710 Chapter 5: Laws of Motion—III
Effect of Normal Force—more contact.
N
N
More contact
points
Ff ,max = μN
Physics 1710 Chapter 5: Laws of Motion—III
Friction is caused by “welding” of the two
surfaces.
Lubrication changes the nature of the
contact and/or the surface, atomically.
Physics 1710 Chapter 5: Laws of Motion—III
Rolling Friction:
Adhesion
Ff = μN
N Normal
Force
Physics 1710 Chapter 5: Laws of Motion—III
Rolling Friction: Lubrication
No Adhesion
Ff = μN
N Normal
Force
Physics 1710 Chapter 5: Laws of Motion—III
Think!
Peer Instruction Time
No Talking!
Confer!
Physics 1710 Chapter 5: Laws of Motion—III
Friction, Application:
• For constant velocity F = -Ff = -(- μ N)
• N = m g = (5.2 kg)(9.8 N/kg) = 51. N
• F = (0.30) (51. N) = 15. N
Physics 1710 Chapter 5: Laws of Motion—III
Third Law of Motion, Implications :
• A father (100 kg) and his daughter
(50 kg) are ice skating. The push off
from each other. Who pushes harder?
• What is there relative acceleration?
Physics 1710 Chapter 5: Laws of Motion—III
Third Law of Motion, Implications :
• A father (100 kg) and his daughter
(50 kg) are ice skating. The push off from each other.
Who pushes harder?
M
FDad
m
Fkid
Physics 1710 Chapter 5: Laws of Motion—III
Third Law of Motion, Implications :
• A father (100 kg) and his daughter
(50 kg) are ice skating. The push off from each other.
Who pushes harder?
3rd Law of Motion says they
push the same and opposite.
M
-F
aM =-F/M
m
F
am =F/m
Physics 1710 Chapter 5: Laws of Motion—III
Summary:
• Newton’s Laws of Motion are:
(1) Acceleration (or deceleration) occurs if and only
if there is a net external force.
(2) a = F/m [Note this is a vector eqn.]
(3) The force exerted by a first object on a second is
always equal and opposite the the force exerted by
the second on the first. F12 = - F21
Physics 1710 Chapter 5: Laws of Motion—III
Summary (cont’d.) :
• Weight is the force of gravity equal to g times the
mass of the object.
• g =9.80 N/kg
• The force of friction is opposed to the motion of a
body and proportional to the normal force.
• Free body diagrams are sketches of all the forces
acting on a body.