Physics I
Class 04
Newton’s Second Law
for More Complex Cases
Rev. 09-Jan-04 GB
04-1
Newton’s Second Law A Review
Newton’s Second Law:
 
 F  Fnet

 Fnet

 m a or a 
m
In a complex situation, we will need to apply this to more than one
object and/or in more than one dimension.
When we do this, we will get a set of linear equations and we will solve
them for the unknown quantities.
04-2
Using Newton’s Second Law to
Solve Complex Problems
1.
2.
3.
4.
5.
6.
Identify all forces acting on the object.
Today: Gravity, normal, and ropes/strings.
Choose a coordinate system.
If you know the direction of acceleration, one
coordinate axis should be in that direction.
Draw a “Free-Body Diagram.”
We will use two dimensions today.
Express the force vectors in components.
We will use trigonometry today.
Use Newton’s Second Law to write one
equation for each direction considered.
Solve the equations.
04-3
A Common Example:
Atwood’s Machine
a
W = mg
X
W = Mg
T
T
m
w = mg
a
X
M
W = Mg
04-4
Solution to Atwood’s Machine
T  mg  ma
Mg  T  Ma
Add these to eliminate T .
Mg  mg  Ma  ma
( M  m) g  ( M  m)a
M  m

a
g
 M  m
04-5
Inclined Plane
04-6
Coordinate Systems and
Free-Body Diagrams
Y
X
a
X
Use trigonometry to determine X & Y components
of forces not aligned with coordinate system.
04-7
Solving for Acceleration
For mass 1:
T  m1g sin( )  m1 a
X:
Y:
N  m1g cos( )  0
For mass 2:
m 2g  T  m 2 a
X:
To solve for a, add the two X equations:
T  m1g sin( )  m 2 g  T  m1 a  m 2 a
m  m1 sin( )
a 2
g
m1  m 2
What would it mean if we found a < 0 after plugging in the values?
If  = 0, does the “inclined” plane resemble something in class?
04-8
Class #4
Take-Away Concepts
1.
2.
3.
4.
Keep using the six-step process for doing Newton’s Second Law
problems for complex cases, it will help you keep things straight.
For each dimension and each object, you will get one equation. You
may or may not need to know the forces in the “normal” direction.
You should have the same number of unknowns as equations.
The easiest way to solve is usually to add the two equations resulting
from opposite ends of a rope or string, or opposite sides of a contact
surface where two objects push on each other.
04-9
Class #4
Problem of the Day
1. Ike and Mike are having a game of tug-of-war, where each pulls on the end of a rope to
see which one is pulled forward into a mud pit between them. To pull on the rope
effectively, they must each exert a horizontal force on the ground through the soles of their
shoes. Ike is exerting a horizontal force on the ground of 500 N (magnitude) while trying
to pull Mike in the –X direction . Mike is exerting a horizontal force on the ground of 514
N (magnitude) while he tries to pull Ike in the +X direction. They each have mass = 70
kg. Assume that the mass of the rope is negligible and that its tension, T, is not near its
breaking point. Find Ike’s acceleration, Mike’s acceleration, and the tension of the rope.
X
Ike
Mike
04-10
Answer to Problem 1 for Class #4
X
a
F1
Ike
T
T
Ike and Mike have the same
acceleration because they
are connected by a rope.
Mike
F2
T – F1 = m a
F2 – T = m a
a = (F2–F1)/(m+m) = (514–500)/(70+70) = 0.1 m/s2.
T = m a + F1 = 70 * 0.1 + 500 = 507 N
04-11
Activity #4 - Forces and Motion in
Coupled Systems
Objectives of the Activity:
1.
2.
Making detailed theoretical predictions and comparing
with measured data.
Understanding forces and motion in coupled systems.
04-12