Transcript (or free-body) diagrams

Objectives: The student will be
able to:
• Draw an accurate free body diagram
locating each of the forces acting on an
object or a system of objects.
• Use free body diagrams and Newton's
laws of motion to solve word problems.
• Use the methods of vector algebra to
determine the net force acting on an
object.
The Statue of Liberty has a mass of 225,000
kg. How much does she weigh?
Write the formula:
W = m
*
g
Substitute known values:
W = (225,000 kg) * 9.80 m/s²
Present solution with units: W = 2205000 N
The Statue of Liberty weighs 2205000 Newtons,
which is 495,704 pounds!
If she is exerting a 495,704 lb force down, the
how much does Liberty Island have to push up
on her to maintain static equilibrium?
495,704 lb, according to Newton’s 3rd law
Note: 1 N = 0.22 lbs
A scalar is simply a number, a magnitude alone.
A force is usually shown as a vector, which includes both
magnitude and a direction.
Force (or free-body) diagrams show the relative magnitude
and direction of all forces acting upon an object. The object
must be isolated and “free” of its surroundings.
This is a free-body diagram of the Statue
of Liberty. She is represented by a simple
box. The forces acting on her are labeled
with a magnitude and the arrow shows
direction. Notice the surrounding objects
are stripped away and the forces acting on
the object are shown.
495,704 lb
495,704 lb
W here represents the force of the weight
of the statue.
W =495,704 lb
N is the normal force, which represents
the force Liberty Island is pushing back
up on the statue.
The island has a great resistance to
compression. The ground is exerting a
force upward on the statue perpendicular,
or normal, to the surface.
N = 495,704 lb
(Positive y-direction)
+y
Think of the diagram on an XY
plane.
W =495,704 lb
If “up” is assumed to be the positive
direction, then N is positive and W is
negative.
N = 495,704 lb
+x
(Positive x-direction)
The first line of this calculation reads,
“The sum of the Forces in the positive y direction is
W + N” (  is the Greek symbol for “sum” )
+
(Positive y-direction)
+y
W =495,704 lb
Fy = W + N
Fy = (-495704 lb) + (+495704 lb )
Fy = 0
The sum of the forces in the y is zero.
N = 495,704 lb
+x
(Positive x-direction)
The forces acting on the object cancel each other out.
•We know F = m * a, where “a” is acceleration.
•If a = 0, then F = m * 0 = 0.
•When  F = 0, the object is not accelerating.
•We we can then say that the forces acting on the
object cancel each other out and it is in a state of
static equilibrium.
Create a free body diagram (FBD) for each of the following
situations. Draw a FBD of the gorilla:
N
W
Sitting Gorilla
Free Body Diagram of the Sitting
Gorilla (The box represents the
gorilla, W = weight of the gorilla,
N = Normal force)
Create a free body diagram (FBD) for each of the following
situations. Draw a FBD of the gorilla:
W
N
This is also an acceptable
diagram.
Sitting Gorilla
Draw a FBD of the wooden swing:
T1
T2
W
Parrot on wooden
swing hung by ropes
Free Body Diagram of the wooden
swing (The box represents the wooden
swing, W = weight of the swing and the
parrot, T represents the ropes that are
in tension supporting the weight)
Draw a FBD of bucket the bungee jumper leaped from:
Bungee jumping
from crane
Draw a FBD of bucket the bungee jumper leaped from:
T
W
Bungee jumping
from crane
Free Body Diagram of the bucket (T
represents the tensile force of the
cable the bucket is suspended from,
and W is the weight of the diver and
the bucket)
Draw a FBD of the ring at point C:
A
B
C
D
Traffic Light
supported by cables
Draw a FBD of the ring at point C:
A
B
C
TCA
TCB
D
TCD
Traffic Light
supported by cables
Free Body Diagram of the ring at
point C (T represents the force of the
cables that are in tension acting on
the ring)
Draw a FBD of the traffic light:
A
B
C
D
Traffic Light
supported by cables
Draw a FBD of the traffic light:
A
B
TCD
C
D
W
Traffic Light
supported by cables
Free Body Diagram of the traffic light
(TCD represents the force of the
cables acting on the light and W is
the weight acting on the light)
Draw a FBD of the pin at point A:
A
C
B
E
D
Pin-Connected Pratt Through
Truss Bridge
Draw a FBD of the pin at point A:
TAB
A
TAC T TAE
AD
B
Free Body Diagram of pin A
C
E
D
Pin-Connected Pratt Through
Truss Bridge
(If you consider the third dimension, then
there is an additional force acting on point A
into the paper: The force of the beam that
connects the front of the bridge to the back of
the bridge.)
4-7 Solving Problems with Newton’s Laws
–
Free-Body Diagrams (work out problem)
1. Draw a sketch.
2. For one object, draw a free-body
diagram, showing all the forces acting
on the object. Make the magnitudes
and directions as accurate as you
can. Label each force. If there are
multiple objects, draw a separate
diagram for each one.
3. Resolve vectors into components.
4. Apply Newton’s second law to each
component.
5. Solve.
Example 4-11 (Solve)
4-7 Solving Problems with Newton’s Laws
–
Free-Body Diagrams
(Solve Example 4-12)
When a cord or rope pulls on
an object, it is said to be under
tension, and the force it exerts
is called a tension force.
An Atwood’s machine is two masses connected by a
strong light string that are hung over an ideal pulley
(light and frictionless). The masses have identical
velocity and acceleration magnitudes at every instant.
If we define up on the left and down on the right as
positive directions, then the masses have identical
velocities and accelerations period. This simplifies the
analysis a lot.
Atwood Machine
Would this move?
100 kg 100 kg
Atwood Machine
Would this move?
Which way?
FT
FT
Forces?
100 kg
200 kg
Fg = w
Fg = w
Let’s break this up and analyze each mass…
The most natural choice that makes up on the left and
down on the right positive directions is to focus on the
pulley. Up on the left and down on the right are the
same to the pulley – both are clockwise rotations. This
the coordinate system to use for this problem.
cw
Let’s look at the clockwise forces that act on the
system of two masses and the affect they have on
their acceleration.
cw
Fnet ,cw
acw 
(m1  m2 )
m2 g  m1 g
acw 
(m1  m2 )
m1 g
m2 g
Atwood Device
Assume m1 < m2 and that
the clockwise direction is
+.
T
T
m1
m1g
m2
m2g
If the rope & pulley have
negligible mass, and if the
pulley is frictionless, then
T is the same throughout
the rope.
If the rope doesn’t stretch,
a is the same for both
masses.
Atwood Analysis
Remember, clockwise has been defined as
+.
2nd Law on m1: T - m1g = m1a
2nd Law on m2: m2g - T = m2 a
T
Add equations:
T
m1
m1g
m2
m2g – m1g = m1a + m2
a
(The T’s cancel out.)
Solve for a:
a=
m2g
m2 – m1
m +m
g
Atwood as a system
Treated as a system (rope & both
masses), tension is internal and
the T’s cancel out (one clockwise, one counterclockwise).
T
T
m1
m1g
m2
m2g
Fnet = (total mass)  a implies
(force in + direction) (force in - direction)
= m2g - m1g = (m1 + m2) a.
Solving for a gives the same
result. Then, knowing a, T can
be found by substitution.
Atwood: Unit Check
m2 – m1
a=m +m g
1
2
units:
kg - kg
kg + kg
m
m
= 2
2
s
s
Whenever you derive a formula you should check
to see if it gives the appropriate units. If not, you
screwed up. If so, it doesn’t prove you’re right,
but it’s a good way to check for errors.
Remember, you can multiply or divide scalar
quantities with different units, but you can only add
or subtract quantities with the same units!
Atwood: Checking Extremes
Besides units, you should
m2 – m1
also check a formula to see
a=m +m g
if what happens in extreme
1
2
& special cases makes sense.
m2 >> m1 : In this case, m1 is negligible compared
to m2. If we let m1 = 0 in the formula, we get
m1
a = (m2 / m2 )g = g, which makes sense, since with
only one mass, we have freefall.
m2 << m1 : This time m2 is negligible compared to
m2
m1, and if we let m2 = 0 in the formula, we get
a = (-m1 / m1 )g = -g, which is freefall in the negative
(counterclockwise) direction.
m2 = m1 : In this case we find a = 0 / (2m1)g = 0, which is
what we would expect considering the device is balanced.
Note: The masses in the last case can still move but only
with constant velocity!
Do sample Atwood Machine
Problem
An Atwood’s machine has m1 = 2 kg, m2 = 3 kg, hung
from an ideal pulley. What is the acceleration of the
masses? Calculate the tension in the string attached to
each mass.
Modified Atwood Machine and Lab for 2nd Law
Will need to solve for acceleration, draw FBD, and
solve for tension.
Do example.
Discuss theoretical aspect of the lab.
Homework for Chapter 4
• Problems # 19, 23, 25, 26, 31