Download Newton`s Second Law I

Newton’s Second
Law I
Review: Newton’s First Law
IF the object experiences NO net
external force….
Resting objects remain at rest.
 Moving objects move at a constant
velocity.

The ubiquitous block on an
inclined plane.
Fblock-on-book
Ffriction
Fgravity
Newton’s Second Law
The acceleration of an object is:

Directly proportional to the net
external force acting on it, and…

Inversely proportional to its mass
In other words…
Fnet = m×a
Let’s try a different incline:
Fnormal
Ffriction
Ice
Fgravity
Example 1:
Suppose you want to push a 8 kg
block up a frictionless inclined plane
of 30 degrees. What would be the
acceleration of the block, if a force of
50 Newtons were exerted parallel to
the plane?
Example 1 (cont.)


Step 1 – Draw a picture
Step 2 – Force-body diagram
Y
Fnormal

Fapp.
Fg
X
Example 1 (cont.)
3 – What do we know?
 Mass = 8 kg
 Fapp.= 50 N.
 θ = 30˚
Y
X
 g = 9.8m/s2
 Step

Fapp.
Fg
Example 1 (cont.)

Step 4 – Apply Newton’s Second Law, in a
single direction.
F
F
x
 m a
x
 Fapp.  mg sin  m  a
Y

Fapp.
Fg
X
Example 1 (cont.)
 Step
5 – Plug ‘n Chug
50 N  8kg  9.8
m
 sin 30  8kg  a

s2
a  135
.

Fapp.
Fg
m
s2
Y
X
Example 2
On October 29, 1998,
space shuttle Discovery
lifted off with a total
mass of 2,000,000 kg
http://spaceflight.nasa.gov/gallery/images/s
huttle/sts-103/html/sts103s005.html
The Problem:
What would be the
acceleration if the thrust
force (Fthrust) were
33,000,000 Newtons (an
accurate assumption)?
http://trc.dfrc.nasa.gov/Gallery/Gra
phics/STS/Small/STS_launch.gif
Example 2 (cont.)
Y
X
Fg
Ft

Step 1 – Picture

Step 2 – Forcebody diagram
Example 2 (cont.)
Y
X
Fg
Ft

Step 3 – What do
we know?
Mass = 2,000,000kg

Fthrust = 33MN

g = 9.8m/s2

Example 2 (cont.)
Y
X

Step 4 – apply
Newton’s second law
in one direction (Y)
F
Y
Fg
 m a
Ft  mg  m  a
Ft
Example 2 (cont.)
 Step 5 – Plug ‘n Chug
But…..in
reality, the acceleration
33,000,000
N  2,000
kg stages
9.8 m s  2of
,000launch
,000kg  a
during
the,000
first
2
was closer to 14m/s
m
2
a  6.7
s2
Why the difference?
Fg
Ft
a=14m/s2
θ
Ft
Ft Ft F
g Fgsin
 mma a
Fg∙sinθ
Launch video