Stuff you asked about:
From what I learned in physics in the past, knowing how to correctly draw free body
diagrams is immensely helpful in solving these types of problems. Unfortunately, I find
that drawing free body diagrams is one of the most difficult skills to master. For this
reason, we should make sure that free body diagrams are understood completely.
The spring box in elevator problem was confusing. I'm also not too sure what the
universal gravitation force is used for.
Have trouble just listing what forces are acting on the object in question and then
decomposing them into the X and Y components (or when to do this.)
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
“I would like more free body diagram examples please.”
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
“Third Law Newton needs explaining. Why is gravity and
normal force not 3rd law?”
Free Body Diagrams
Free Body Diagrams
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.
Clicker Question
A block accelerates down a frictionless inclined plane. 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
C
- Gravity is down.
- Normal is perpendicular to surface.
D
In the pre-lecture question, I was asked to choose which free body
diagram matched a box sliding down a frictionless slope. As such,
shouldn't there be an arrow pointing downwards parallel to the slope?
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
“Tension acts upward on the box, while mg (weight) acts
downward on the box. Since the box accelerates upward,
T must be greater than mg.“
“Increases tension”
a
Watch out for subtle mistakes…
Many people got the right answer with
explanations like this one:
Because the elevator is accelerating
upwards the box is being pulled down
relative to the elevator, meaning that
there must be extra tension to keep the
box hanging.
Don’t view problem from accelerating
reference frame or you may get confused
T
m
mg
Just Remember: Net force causes acceleration
a
Aside: How a Spring Scale works
F
F
There is a spring in there
F
F
Dx
The dial just shows how much it has been stretched, Dx.
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.
Spring Force
Direction: Parallel to spring: Push or pull
Magnitude:F = kx
Springs Example
“can we go over springs more in depth in lecture?”
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
kDx  Mg  0
Dx 
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
The box is accelerating upward,
so the spring force must be
greater than the gravity force.
Thus, L is larger than normal
length.
a
L
m
Accelerating reference frames can confuse…
there will be more force pulling down on the
box causing it to stretch the spring
As in the previous problem, the acceleration
will induce a downwards force on the box.
This force will stretch the spring to a length
greater than what it was originally.
a
L
m
The same principle with the string, because
the there will be a 'reaction force'
downwards, the spring will stretch to be
longer than its original length.
Remember: Net force causes acceleration
Clicker Question
V
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
a
mg
N
Clicker Question
You are traveling on an elevator
up the Sears tower, and you are
standing on a bathroom scale
that measures N.
As you near the top floor and
are slowing down, the Normal
force is.
a
mg
A) More than your usual weight
B) Less than your usual weight
C) Your usual weight
N
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
Near surface of earth r = Rearth
Mearth m
F G 2
Rearth
 6.67  1011 m3 / kg s2 
5.97  1024  kg 
m / s2  m

9.8
m
2
6.4  10 m
6
r

Mm
FGravity  G 2 rˆ
r
Careful - r is measured from the center of Mars!
r


Fnet  ma
Mm
v2
G 2 m
r
r
v  GM / r
USE ALGEBRA!
r
d
t
v
2r
t
GM / r
3
r
t  2
GM
Careful, check units. Asks for time in
hours, not seconds!
r
v  GM / r