Transcript Document
Chapter 4
Forces and Newton’s Laws of
Motion
F=ma; gravity
0) Background
• Galileo
– inertia (horizontal motion)
– constant acceleration (vertical motion)
• Descartes & Huygens
– Conservation of momentum: mass x velocity = constant
• Kepler & Braha
– laws of planetary motion (kinematics only)
Question of the day:
Explain planetary motion
1) Newton’s first law: the law of inertia
A free object moves with constant velocity
a) Free object --> no forces acting on it
b) Constant velocity --> at rest or motion in a straight line with
constant speed
c) Natural state is motion with constant velocity
-
Aristotle: rest is natural state
Galileo: circular motion (orbits) is natural state
d) Inertial reference frames
-
A reference frame in which the law of inertia holds
- does not hold on a carousal, or an accelerating car
Requires ability to identify a free object: If no force acts on a body, a
reference frame in which it has no acceleration is an inertial frame.
1) Newton’s first law: the law of inertia
A free object moves with constant velocity
e) Velocity is relative
-
-
All frames moving at constant velocity with respect to an inertial frame
are also inertial frames
No local experiment can determine the state of uniform motion
Cannot define absolute rest: No preferred reference frame
- (Principle of Relativity)
2) Newton’s second law: F=ma
a) Mass
(i)
(ii)
quantity of matter (determined with a balance)
quantity that resists acceleration (inertial mass)
Define 1 kg as mass of a standard cylinder
Addition of masses (scalar): m = m1 + m2
-
in particular two identical masses have twice the mass, to satisfy quantity of
matter definition
(iii) Observe acceleration vs mass for a given force:
mass acceleration
1 kg
1 m/s2
2 kg
1/2 m/s2
3 kg
1/3 m/s2
mass is inversely proportional to acceleration
2) Newton’s second law: F=ma
b) Force
(i)
(ii)
push or pull
disturbs “natural” state: causes acceleration
2
Define 1 N (newton) as force required rto accelerate
1
kg
by
1
m/s
r
Addition of forces (vector): Fnet F1 F2
Identical forces in opposite
direction produce no acceleration
Two identical forces at 60º produce the same
acceleration as a third identical force at 0º
(cos(60º)=1/2)
Two identical parallel forces corresponds to twice the force.
2) Newton’s second law: F=ma
(iii) Observe acceleration vs. force for a given object
Force
1N
2N
3N
Acceleration
1 m/s2
2 m/s2
3 m/s2
Force is proportional to acceleration
(iv) Types of force:
- gravity
- electromagnetic
- weak nuclear
-strong nuclear
electroweak
2) Newton’s second law: F=ma
c) Second Law
Combine
1
F a for a given mass, and m for a given force
a
to give
F ma or F (const)ma
Define proportionality constant =1. Then,
r
F ma
For m = 1 kg, and a = 1 m/s2, F = 1 N by definition, and
F = ma gives F = 1 kg m /s2, so
1 N = 1 kg m/s2
2) Newton’s second law: F=ma
• F = ma can be used as the defining equation for force and
inertial mass, but only because of the physical observation that
force is proportional to acceleration (for a given mass), and
mass is inversely proportional to acceleration (for a given
force).
• Inertia is the tendency of an object not to accelerate
• Newton’s second law formally refers to the rate of change of
r
momentum:
(mv)
F
t
r
v
r
For constant mass,
Fm
ma
t
• Special case:
r
F 0 0 ma
r
a0
velocity is constant (1st law)
2) Newton’s second law: F=ma
d) Free-body diagrams
Replace object(s) by dot(s). Represent all forces from the dot.
Solve F=ma for each object
F2
FN
F1
F2
F1
m
mg
2) Newton’s second law: F=ma
d) Free-body diagrams
10 N
10 N
10 N
scale
m
?N
10 N
10 N
scale
m
m
2) Newton’s second law: F=ma
e) Components of force
r
F ma means
F
x
max &
sum of all forces
F
y
may
2) Newton’s second law: F=ma
e) Components of force
Example:
F1 = 15 N
m = 1300 kg
m
º
Find acceleration.
F2 = 17 N
y
F1
Fx F2 F1 cos 23 N
F
y
x
F2
F1 sin 14 N
ax
F
ay
x
m
Fy
m
0.018 m/s 2
0.011 m/s 2
3) Newton’s third law
For every action, there is an equal and opposite reaction
FAB
A
Conservation of momentum:
B
FBA
FAB = -FBA
r
(mB vB )
FAB
t
r
r
(mA vA )
FBA
t
r
r
r
r
(mB vB ) (mA vA )
FAB FBA
t
t
r
r
(mB vB mA vA )
0
t
4) Gravity
5) Normal Force
6) Friction
7) Tension and pulleys
• Tension: force exerted by rope or cable
– For an ideal (massless, inextensible) line, the same force is
exerted at both ends (in opposite directions)
– objects connected by a line (no slack) have the same
acceleration
• Pulley: changes direction of force
– For an ideal pulley (massless, frictionless) the magnitude of the
tension is the same on both sides
– magnitude of acceleration of connected objects is the same
7) Tension and pulleys
+a
T1
m1
• T 1 = T2 = T
• a1 = a 2 = a
• For the example,
a1y = -a2y
T2
m2
• Simplify problem, by choosing
sign for a sense of the motion
7) Tension and pulleys
T
+a
T1
m1g
m1
T2
m2
T
m2g
Equations of motion:
T m1g m1a
m 2 g T m2 a
Solve for a (eliminate T ):
Solve for T (eliminate a):
m2 m1
a
g
m m
2m1m2
T
g
m1 m2
1
2
7) Tension and pulleys
e.g. m1 = 5 kg; m2 = 10 kg
+a
T1
m1
T2
g
10 5
a
g
10 5
3
100kg
T
g 65N
15
m2
Solve for a (eliminate T ):
Solve for T (eliminate a):
m2 m1
a
g
m m
2m1m2
T
g
m1 m2
1
2
7) Tension and pulleys
+a
Acceleration can be determined by
considering external forces (tension is an
internal force holding objects together)
F
ext ()
m1
m1g
m a
m2 g m1g (m1 m2 )a
m2 m1
a
g
m m
m2
1
m2g
2
Example
If m1 = m2, and rope and pulley are
ideal, what happens when the monkey
climbs the rope?
T1
T2
m1g
m2g
T1
m1
T2
Quick Time™a nd a
TIFF ( Unco mpre ssed ) dec ompr esso r
ar e nee ded to see this pictur e.
m2
Since T1 = T2, any change in T2 to
cause the monkey to ascend, results in a
change in T1, causing the bananas to
ascend at the same rate.
Example
If m1 = m2, and rope and pulley are
ideal, what happens when the monkey
climbs the rope?
T1
T2
m1g
m2g
Since T1 = T2, any change in T2 to
cause the monkey to ascend, results in a
change in T1, causing the bananas to
ascend at the same rate.
Example
If m1 = m2, and rope and pulley are
ideal, what happens when the monkey
climbs the rope?
T1
T2
m1g
m2g
Since T1 = T2, any change in T2 to
cause the monkey to ascend, results in a
change in T1, causing the bananas to
ascend at the same rate.
Example
If m1 = m2, and rope and pulley are
ideal, what happens when the monkey
climbs the rope?
T1
T2
m1g
m2g
Since T1 = T2, any change in T2 to
cause the monkey to ascend, results in a
change in T1, causing the bananas to
ascend at the same rate.
Example
If m1 = m2, and rope and pulley are
ideal, what happens when the monkey
climbs the rope?
T1
T2
m1g
m2g
Since T1 = T2, any change in T2 to
cause the monkey to ascend, results in a
change in T1, causing the bananas to
ascend at the same rate.
8) Equilibrium applications
• Equilibrium means zero acceleration
• Balance forces in x and y directions
F
F
x
0
y
0
8) Equilibrium applications
Example: Find tension on leg (F)
Free body diagram for pulley:
T
Free body diagram for weight:
T
T=mg
F T1 cos 35º T2 cos 35º
mg
Since T T1 T2 ,
F 2T cos 35º 36 N
9) Non-equilibrium applications
• Non-equilibrium means non-zero acceleration
• Determine acceleration from 2nd law:
F
F
x
max
y
may
• Solve kinematic equations
Example: Apparent weight
Apparent weight (measured by scale) is
the normal force
At rest or moving with constant velocity
FN
W
FN W ma 0
FN W 700N
Example: Apparent weight
Apparent weight (measured by scale) is
the normal force
Accelerating up
FN
FN W may ma
FN W ma 1000N
W
FN W FN W
FN
a
g
1 g
W
m
W
If FN 2W , a g
For FN 1000N, and W 700N, a
3
g
7
Example: Apparent weight
Apparent weight (measured by scale) is
the normal force
Accelerating down
FN
FN W may ma
FN W ma 400N
W
W FN W FN
FN
a
g 1 g
m
W
W
If FN 0, a g
For FN 400N, and W 700N, a
3
g
7
Example: Apparent weight
Apparent weight (measured by scale) is
the normal force
Free fall
FN = 0
FN W may ma mg
FN W mg 0
W
weightlessness