Download Laws of Motion

Force and
Motion
PHYSICS HONORS
Lecture Notes
Laws of Motion (Chapter 4) 2048
Forces
Types
Range
Size
Gravitational
Unlimited
100
Electromagnetic
Unlimited
106
Weak Nuclear
 10-12 m
Strong Nuclear
 10-15 m
1020
1035
Laws of Motion (Chapter 4) 2048.ppt
Newton’s First Law
(Law of Inertia)
If no external force acts,
an object maintains a constant velocity.
Laws of Motion (Chapter 4) 2048.ppt
Mass
Gravitational mass
mg
Inertial mass
mi
mg = m i
Laws of Motion (Chapter 4) 2048.ppt
Newton’s Second Law
 Fx  ma x
 F  ma
 Fy  ma y
 Fz  ma z
Units of Force
System
SI
British
Mass
kg
slug
Acceleration
Force
m/s2
N = kg m/s2
ft/s2
lb = slug ft/s2
Laws of Motion (Chapter 4) 2048.ppt
Force of Gravity and Weight
T2
m
FN
Fg
Fg = mg
Weight= mg
Laws of Motion (Chapter 4) 2048.ppt
Newton’s Third Law
F1  - F2
F2
F1
Action/Reaction Forces
Applications of Newton’s Laws
Find Tensions T1 and T2
 Fy  0
q
mg  T1 sin q 
mg
20( 9.8)
T1 

sin q 
sin 60 
 Fx  0
T1 sin(60)
 226 N
T1
T2
T1 cos(60)
mg
T2  T1 cos q 
m
 mg 
T2  
 cosq 
 sin q  
T2 
mg
tanq 

20(9.8)
tan60 
 113 N
q = 60o
m = 20 kg
Laws of Motn (Chapter 4) 2048.ppt
Applications of Newton’s Laws
Another method
q
T1
mg
T1
q
T2
T2
mg
mg
tanq  
T2
mg
 T2 
tanq 
mg
sin q  
T1
mg
 T1 
sin q 
m
Laws of Motion (Chapter 4) 2048.ppt
Applications of Newton’s Laws
vo= 0
F
Dt = 5 s
F = 20 N
m = 5 kg
v= ?
m
Dv
a
Dt
F  ma
 Dv 
F  m 
 Dt 
FDt
v
m
Find velocity of
block after Dt
 v - vo 
 m

 Dt 
205 

5
 20 m/s
Laws of Motion (Chapter 4) 2048.ppt
Applications of Newton’s Laws
F = 10 N
m = 5 kg
vo= 20 m/s
F
v= 0
m
d
v 2  v o2 - 2ad
F  ma
Find distance
block moves

F
2
0  v o - 2 d
m
520 2
mv o2

d
210 
2F
F
a
m
 100 m
Laws of Motion (Chapter 4) 2053.ppt
Forces on m1
N
 Fy  0  N  m 1g
 Fx  m 1a
T
T  m 1a
m1
T
m1g
m2
m2g
Forces on m2
 F  ma
m 2 g - T  m 2a
T  m 2 g - m 2a
m 1a  m 2 g - m 2 a
a
m 2g
m1  m 2
Laws of Motion (Chapter 4) 2053.ppt
T
m2
Mass 1
Mass 2
 F  m 1a
 F  m 2a
T - m 1 g  m 1a
m 2 g - T  m 2a
T  m 1a  m 1 g
T  m 2 g - m 2a
T
m1
m1g
m 1a  m 1 g  m 2 g - m 2 a
m2g

m 2 - m1 g
a
m1  m 2
Laws of Motion (Chapter 4) 2053.ppt
Frictionless incline
 Fy  may
y
 Fy  0
N
 Fx  m a
N  mg cos q 
mg sin q   m a
mg sin(q)
mg cos(q)
q
a  g sin q 
x
mg
q
Laws of Motion (Chapter 4) 2053.ppt
m1
T
T
m2
m1g sin(q1)
q1
m1g
m2g
m2g sin(q2)
q2
 F 2  m 2a
 F 1  m 1a
T - m 1g sin q1   m 1a
m 2 g sin q 2  - T  m 2 a
T  m 1g sin q1   m 1a
T  m 2 g sin q 2   m 2 a
m 1g sin q1   m 1a  m 2 g sin q 2   m 2 a

m 2 sinq 2  - m1 sinq1 g
a
m1  m 2
Laws of Motion (Chapter 4) 2053.ppt
Friction:
Force of Static Friction
v0
N
F
fs
m
mg
F  0
fs   s N
fs  F
f s(max)   s N
Laws of Motion (Chapter 4) 2053.ppt
Friction:
Force of Kinetic Friction
fk
v
N F
m
mg
fk   k N
Laws of Motion (Chapter 4) 2053.ppt
Table 5.2
Coefficients
Of Friction
s
Steel on steel
Aluminum on steel
Copper on steel
Rubber on concrete
Wood on wood
Glass on glass
Waxed wood on wet snow
Waxed wood on dry snow
Metal on metal (lubricated)
Ice on ice
Teflon on Teflon
0.74
0.61
0.53
1.0
0.25-0.5
0.94
0.14
-----0.15
0.1
0.04
k
0.57
0.47
0.36
0.8
0.2
0.4
0.1
0.04
0.06
0.03
0.04
Laws of Motion (Chapter 4) 2053.ppt
 Fy  0
mg sin q  - f  m a
N
N  mg cos q 
f  k N
f   k mg cos q 
 Fx  m a
y
mg sin q  -  k mg cos q   m a
a  g sin q  -  k cos q 
a
fk
mg sin(q)
mg cos(q)
q
x
mg
q
Acceleration on a rough incline
Laws of Motion (Chapter 4) 2053.ppt
 Fy  0 
Rough surface
N
T
fk
T
m1 g
m2 g
 F2  m 2a
m 2 g - T  m 2a
T  m 2 g - m 2a
N  m 1g
 F x  m 1a
T - f k  m 1a
T -  k m 1 g  m 1a
T  m 1a   k m 1 g
fk   k N
f k   k m 1g
m 2 g - m 2 a  m 1a   k m 1 g
m 2 g -  k m 1g  m 1  m 2 a
a
m 2 -  k m1 g
m1  m 2
Laws of Motion (Chapter 4) 2053.ppt
Three mass system - find acceleration
m
F
2m
3m
 F  m sa
F  m  2m  3m a  6m a
F
a
6m
Laws of Motion (Chapter 4) 2053.ppt
Three mass system - find T2
m
2m
T2
F
3m
F
a
6m
 F  m 3a
F - T2  3m  a
T2  F - 3ma
 F 
T2  F - 3m

 6m 
F
 T2 
2
Laws of Motion (Chapter 4) 2053.ppt
Three mass system - find T1
m
T1
T2
2m
F
3m
F
a
6m
F
T2 
2
 F  m 2a
T2 - T1  2m  a
T1  T2 - 2ma
T1 
F
 F 
- 2m

2
 6m 
F
 T1 
6
Laws of Motion (Chapter 4) 2053.ppt
Pulling a block with constant speed
N
k
f
F
q
m
y
N
q
f
x
mg
mg
The normal force
F
 Fy  0
N  F sin q  - mg  0
N  mg - F sin q 
Laws of Motion (Chapter 4) 2053.ppt
Pulling a block with constant speed
N
k
f
F
q
m
y
N
q
f
x
mg
mg
The frictional force
F
 Fx  0
F cos q  - f  0
f  F cos q 
Laws of Motion (Chapter 4) 2053.ppt
Pulling a block with constant speed
N
k
f
q
m
y
F
F
q
f
x
mg
mg
The coefficient of friction
N
f  k N
f
k 
N
Laws of Motion (Chapter 4) 2053.ppt
m = 200 kg
s = 0.500
vo= 20 m/s
Crate on truck
Truck’s maximum
deceleration
 F  ma
f   s mg  ma
a  sg
a  0 .500 9 .8 
 4.9 m/s 2
Laws of Motion (Chapter 4) 2053.ppt
mg cosq
Frictionless
A block is released from rest
N
at the top of an incline.
Find the final speed and the
time to side to the bottom.
q
 F  ma
mg q
mg sin q  ma
a  g sin q
v 2  v o2  2ax
v 2  2ad
v 2  2g sin q d
v  2gd sin q
at 2
x  vot 
2
at 2 g sin q t 2
d

2
2
2d
t
g sin q
Laws of Motion (Chapter 4) 2053.ppt
f = kN = kmg cosq
N
f
mg cosq
With friction
q
mg q
v 2  v o2  2ax
v 2  2ad
v 2  2g sin q -  k cos q d
v  2 gd sin q -  k cos q 
A block is released from rest
at the top of an incline.
Find the final speed and the
time to side to the bottom.
 F  ma
mg sin q -  k mg cos q  ma
a  g sin q -  k cos q 
at 2
x  vot 
2
at 2 g sin q -  k cos q t 2
d

2
2
2d
t
g sin q -  k cos q 
Laws of Motion (Chapter 4) 2053.ppt
Force and
Motion
END
Laws of Motion (Chapter 4) 2053.ppt