Physics 211
Lecture 5
Today’s Concepts:
a) Free Body Diagrams
b) Force due to strings
c) Force due to springs
d) Force due to gravity
Classical Mechanics
Free Body Diagrams
1)Choose system (object or group)
2)Draw forces acting on object
-Gravity
-Things touching (contact)
T
m
mg
FREE BODY DIAGRAMS! What can I consider a force?
Free Body Diagrams
Free Body Diagrams
Free Body Diagrams
Tension Force
Direction: Parallel to rope, always pulling.
Magnitude: As much or little as needed for Fnet=ma
T
m
mg
Checkpoint
BOX IN THE ELEVATOR
A box of mass m is hung with a string
from the ceiling of an elevator that is
accelerating upward. Which of the
following best describes the tension T
in the string:
A) T < mg
B) T = mg
C) T > mg
Fnet = mass*acceleration.
Because the elevator is
accelerating upward, there is a
net force directed upward as
well. This means the force
pulling the elevator up (tension)
is greater than it is down
(gravity).
Since the elevator is accelerating upward, the
tension on the box has to be greater than the
gravitational force acting on the box, T - mg = ma
a
Tension Clicker Question
A block weighing 4 lbs is hung from a rope attached to a scale. The
scale is then attached to a wall and reads 4 lbs. What will the scale
read when it is instead attached to another block weighing 4 lbs?
?
m
A) 0 lbs.
m
B) 4 lbs.
m
C) 8 lbs.
Normal Force
Normal force is the force perpendicular to the surface that
prevents the objects from passing through each other.
1) Direction: Perpendicular to surface and out
2) Magnitude: As much or little required keep objects separate.
"The normal force is simply what it has to be to do what it does.“
Oh God, what doe this mean??
Checkpoint
A block sits at rest on a horizontal frictionless surface. Which of the
following sketches most closely resembles the correct free body
diagram for all forces acting on the block? Each arrow represents a
force.
A
B
If an object is sitting at rest
on a frictionless surface, then
the only forces acting upon it
are the gravitational force and
the normal force
C
D
Spring Force
Direction: Parallel to spring: Push or pull
Magnitude:F=-kx
Springs Example
How to actually solve for something like the spring problem would be nice
A 5 kg mass is suspended from a
spring with spring constant 25 N/m.
How far is the spring stretched from
its equilibrium position?
F
 ma
kx  Mg  0
x 
Mg
5* 9.8

k
25
A) 0.5 m
B) 2 m
C) 5 m
Checkpoint
A box of mass m is hung by a spring
from the ceiling of an elevator.
When the elevator is at rest the
length of the spring is L = 1 m. If the
elevator accelerates upward the
length of the spring will be:
A) L = 1 m
B) L < 1 m
C) L > 1 m
a
L
The spring must exert a
greater force on the box
because T>mg so its length will
increase proportionally.
m
Student Comment/Question
I feel really comfortable with these
concepts. There is an interesting example
that we encountered last year in high school
physics with a person standing on a scale in
an elevator. The question was when does the
personal weigh the most (i.e. when does the
scale read the heaviest)? I was wondering
about the explanation for that question.
V
Clicker Question
You are traveling on an elevator
up the Sears tower. As you near
the top floor and are slowing
down, your acceleration
A) is upward
B) is downward
C) is zero
V
Clicker Question
You are traveling on an
elevator up the Sears tower,
and you are standing on a
bathroom scale.
As you near the top floor and
are slowing down, the scale
reads
A) More than your usual
weight
B) Less than your usual
weight
C) Your usual weight
Clicker Question
A cart with mass m2 is connected to a mass m1 using a
string that passes over a frictionless pulley, as shown
below. The cart is held motionless.
m2
The tension in the string is
A) m1g
B) m2g
C) 0
m1
Clicker Question
A cart with mass m2 is connected to a mass m1 using a string
that passes over a frictionless pulley, as shown below.
Initially, the cart is held motionless, but is then released and
starts to accelerate.
a
m2
After the cart is released, the tension in the string is
A) = m1g
B) > m1g
C) < m1g
a
m1
Universal Gravitation
Direction: Toward other object (down near earth surface)
Magnitude:F=GMm/r2
(mg near earth surface)
F G
Mm
r2
Newtons relation of the earths gravity to the gravitational
forces between two objects was difficult to follow.
Near surface of earth r=Rearth
F G
Mearth m
2
Rearth
 6.67  10 11 m3 / kg s2 
 9.8 m / s2  m
5.97  1024  kg 
 6.4  10 m
6
2
m
Most of it was really easy to
understand.... the only problem
are the gravity examples.. we
should do more of those....
USE ALGEBRA!

Mm
FGravity  G 2 rˆ
r

Mm
| FGravity | G
((2.2  1)r ) 2
G*mmars/(3.2*rmars)^2


Fnet  ma
Mm
v2
G 2 m
r
r
v  GM / r
d
t
v
2r
t
GM / r
3
r
t  2
GM
Careful, check units. Asks for time in
hours, not seconds!
v  GM / r
r3
t  2
GM
3
1
r
2
t1
GM

3
t0
r0
2
GM
3/ 2
t1  r1 
  
t0  r0 
2/3
 t1 
r1  r0  
 t0 