Newton’s Three Laws
Newton’s First Law
Existence of inertial systems of reference
In inertial system of reference, any object remains at rest or continues its
motion along straight line with constant velocity
Newton’s Second Law
m-?


F  ma
 

  
F  Ftot   Fn  F1  F2  F3 ...
a~1/m
F-?
a~F
Newton’s Third Law
n


FAB   FBA
Note: these two forces act on different objects. Never add these forces!
Units: 1N  1kg  m / s 2
1dyne  1g  cm / s
(1 lb = 4.448 N)
2
Weight:


W  mg
Newton’s Third Law
For every force, or action force, there is an equal but opposite force,
or reaction force.

FBA


FAB   FBA
A
B

FAB
If object A exerts a force on object B (an “action”), then
object B exerts a force on body A (a “reaction”).
These two forces have the same magnitude but opposite direction.
Note: these two forces act on different objects.
Example (Snapped string): A small ball attached to the end of a string
moves in circles as shown below. If the string snaps, what will be the
trajectory of the ball?
Applying Newton’s First Law:
B
A
C
Example (Bowling on the Moon): An astronaut on Earth kicks a bowling
ball horizontally and hurts his foot. A year later, the same astronaut
kicks a bowling ball on the moon with the same force.
His foot hurts:
A. More
B. Less
The mass of the bowling ball (its inertia or reluctanceto-change-its-state-of-motion) is the same!
C. The same
Example: Is Ames a good Inertial Reference Frame?
Newton’s laws are true in Inertial Reference Frames (IRF).
In a non-inertial ref. frame, you can have an acceleration without having a force
→ we think there’s a force (we’re applying the 2nd law!) .
These are “fictitious” forces (the most popular one: the “centrifugal” force).
2
 2 
aAmes  R   R 

→
T 
where T  1 day = 8.64  10 4 s
2
REarth  6.4  106 m
aAmes ≈ 0.034 m/s2 << 9.8 m/s2
(pretty decent IRF)
Example
Given: FA , FC ,
a=0
y

FA

FC



x
?
FB ?
FC cos   FA cos   0
FA sin   FC sin   FB  0

FB
cos  
FA cos 
FC
FB  FA sin   FC sin 
FB  FA sin   FC
 FA
1  
 FC
2

 cos 2 

Example: A force F acting on a mass m1 results in an acceleration a1.
The same force acting on a mass m2 results in an acceleration a2 = 2a1.
a1
F
a2 = 2a1
F
m1
m2
If both masses are put together and the same force is applied to the
combination, what is the resulting acceleration?
a?
F
m1
A.
2/3 a1
B.
3/2 a1
C.
3/4 a1
m2
F
m1 
a1
F
F
m2 

a2 2a1
a1
F
F
2
a 


 a1
F
F
1
m1  m2
3

1
a1 2a1
2
Example: A skater is pushing a heavy box (m = 100 kg) across a sheet
of ice (horizontal and frictionless). He applies a horizontal force of 50 N
on the box. If the box starts at rest, what is its speed v after being
pushed over a distance d = 10 m?
a
F = 50 N
m = 100
kg
v0 = 0
d = 10 m
v?
target
v 2  v 02  2a x
F
v  v  2a x  0  2 x 
m
2
0
Fnet F
a 

m
m
2Fd
 3.2 m/s
m
F1
Example:
y
F2
θ1
Fnet
θ2
F1 - ?
x
F1 sin 1  F2 sin 2  0
F1  F2
sin 2
sin 1
F1 cos 1  F2 cos 2  ma
(5.0 kg)(1.1 m/s2 )
F2 

 2.2 N
sin 2
sin 40
 cos 40
 cos 2
tan20
tan 1
ma
F1  F2
sin 2
ma

sin 1

sin 1  sin 2
 cos 2 

sin 2  tan 1

(5.0 kg)(1.1 m/s2 )


 4.1 N
sin 1
sin20
 cos20
 cos 1
tan
40

tan 2
ma
θ1 = 20°
θ2 = 40°
mbox = 5.0 kg
a = 1.1 m/s2
F2 - ?
 sin 2

F2 
cos 1  cos 2   ma
 sin 1

Checks:
•Units
•Limits:
θ1 = θ2
θ1 = θ2 = 0
m → 0, ∞
a → 0, ∞
Example: Car brakes provide a force F for 5 s. During this time, the car
moves 25 m, but does not stop. If the same force would be applied for
10 s, how far would the car have traveled during this time?
1)
2)
3)
4)
5)
100 m
50 m < x < 100 m
50 m
25 m < x < 50 m
25 m
In the first 5 s, the car has still moved 25 m.
However, since the car is slowing down, in the
next 5 s, it must cover less distance. Therefore,
the total distance must be more than 25 m but less
than 50 m.
d1  25m
t 2  2t1
t1  5 s
at12
d1  v0 t1 
2
2
2
at 2
a2t1 
d 2  v0 t 2 
 v0 2t1 
 2d1  at12  2d1
2
2
t 2  10 s
d2  ?