Transcript Document

Home work
Thesis
1. Hair tension and it’s applications
2. Frictions and their applications
3. Frictional reduction
4. The moon movements
5. Water moving by moon gravitation
6. Solar system movements
1. Net force and Newton's first law
2. Newton's second law
3. Newton's third law
4. Frictional forces
5. Gravitation
6. Circular motion
7. Centripetal force
8. Static equilibrium and
reference frames
Teflon is actually a substance called
polytetrafluoroethylene (PTFE),
which is considered to be the most
slippery substance that exists
Review about rotation
Learning check
Part 7
Centripetal Force
Centripetal Force:
Appear with CM
Fc= m.a=m.v2/R
Perpendiculer to v
Exercise: Calculating Centripetal
accelaration for the Moon?
Giving: Moon takes 27.3 days (2.36.106s) to orbit
the Earth with average distance of 3.84.108m
mass of the Moon = 7.36 × 10
22
kilograms
radius of the moon = 1 737.4 kilometers
Determine the centripetal
force acts to the
moon?
Determine the gravity from the moon’s surface?
Part 8
Static equilibrium and
reference frames
Static Equilibrium
Static Equilibrium :The object is either at rest
Dynamic Equilibrium: The object has its center of mass moving
with a constant velocity.
1-Translational Equilibrium:
F 0
Equilibrium in linear motion
However for object with size this is not sufficient. One
more condition is needed. What is it?

F
d
CM
d
-F
For rotational equilibrium:
the net torque acting on the
object about any axis must
be zero.
For an object to be in
static equilibrium, it should
not have linear or angular CM
speed.
  0
v 0  0
Conditions for Equilibrium
Suppose that forces acting on x-y plane, giving torque only along z-axis.
What is the conditions for equilibrium in this case?
Two vector equations turn to three equations.
F 0
  0
F  0 F
  0
x
y
0
z
What happens if there are many forces exerted on the object?
r5
O r’
If an object is in translational and rotational
static equilibrium, the net torque must be 0
about any arbitrary axis.
O’
Because the object is not moving, no matter what the
rotational axis is, there should not be a motion.
How did we solve equilibrium problems?
1. Identify all the forces and their directions and
locations
2. Draw a free-body diagram with forces indicated
on it
3. Write down vector force equation for each x and
y component with proper signs
4. Select a rotational axis for torque calculations 
Selecting the axis such that the torque of one of
the unknown forces become 0.
5. Write down torque equation with proper signs
6. Solve the equations for unknown quantities
Learning check
A uniform 40.0 N board supports a father and daughter
weighing 800 N and 350 N, respectively. If the support (or
fulcrum) is under the center of gravity of the board and the
father is 1.00 m from CoG, what is the magnitude of normal
force n exerted on the board by the support?
1m
x
Determine where the
child should sit to
balance the system.
n
F
D
MFg
MFg
MBg
Help
1m
F
MFg
x
n
D
MFg
MBg
Since there is no linear motion, this
system is in its translational equilibrium
F
F
x
0
y
 MBg  MF g  MDg n
0
Therefore the magnitude of the normal
n  40.0  800  350  1190N
force
The net torque about the fulcrum by the three forces is:
  M B g  0  M F g 1.00  M D g  x  0
Therefore to balance the system, the daughter must sit
x
MFg
800

1.00m 
1.00m  2.29m
MDg
350
Equilibrium Test
Determine the position of the child to balance the system for different
position of axis of rotation (half of X at the right side).
Rotational axis
1m
x
n
F
MFg
x/2
MBg
What is the conclusion of this exercise?
D
MFg
Rotational axis
1m
MFg

x
n
F
Solution
x/2
D
MFg
MBg
The net torque about the axis of
rotation by all the forces is
 M B g  x / 2  M F g  1.00  x / 2  n  x / 2  M D g  x / 2  0
n  MBg  MF g  MDg
Since the normal force is
The net torque can be rewritten

Therefore
x

 M B g  x / 2  M F g  1.00  x / 2
 M B g  M F g  M D g  x / 2  M D g  x / 2
 M F g 1.00  M D g  x  0
MFg
800
1.00m 
1.00m  2.29m
MDg
350
No matter where the
rotation axis is, net effect of
the torque is identical.
Mechanical Equilibrium
A uniform ladder of length l and weight mg = 50 N rests
against a smooth, vertical wall. If the coefficient of static
friction between the ladder and the ground is ms=0.40,
find the minimum angle qmin at which the ladder does not
slip.
P
l
FBD
q
n
Of
mg
P
n
Of
mg
Result
First the translational equilibrium, using
components
F  f P 0

F
x
 mg  n  0
n  mg  50 N
Thus, the normal force is
y
The maximum static friction force just before
slipping is, therefore, f max  m n  0.4  50N  20N  P
s
s
From the rotational equilibrium
l
 O  mg 2 cosq min  Pl sin q min  0
q min
 mg 
1  50 N 

 tan 1 

tan


  51
 40 N 
 2P 
Inertial Reference Frames:
• A Reference Frame is the place you measure from.
– It’s where you nail down your (x,y,z) axes!
• An Inertial Reference Frame (IRF) is one that is not
accelerating or having constant velocity.
– We will consider only IRFs in this course.
• Valid IRFs can have fixed velocities with respect to
each other.
– Remember that we can measure from different
vantage points.
Inertial Reference Frame (IRF) is different from
Non-Inertial Reference Frame (NIRF) that is
accelerating a0
Learning check
Choosing an IRF from these
a) a free fall
b) an UCM
c) a butterfly with 5m/s of velocity
d) a fly with 5m/s2 of acceleration
Home work
1-
2-
3-
4A person holds a 50.0N sphere in his hand. The forearm is horizontal. The
biceps muscle is attached 3.00 cm from the joint, and the sphere is 35.0cm
from the joint. Find the upward force exerted by the biceps on the forearm
and the downward force exerted by the upper arm on the forearm and
acting at the joint. Neglect the weight of forearm.
FB
d
O
FU
l
mg
5A uniform horizontal beam with a length of 8.00m and a weight of
200N is attached to a wall by a pin connection. Its far end is
supported by a cable that makes an angle of 53.0o with the
horizontal. If 600N person stands 2.00m from the wall, find the
tension in the cable, as well as the magnitude and direction of the
force exerted by the wall on the beam.
T
R
q
53.0o
200N
600N
2m
53.0o
FBD
Rsinq
Rcosq
8m
Tsin53
Tcos53
Result
R
q
53.0o
FBD
T
200N
600N
2m
53.0o
Rsinq
Rcosq
Tsin53
Tcos53
8m
First the translational equilibrium,
using components
 Fx  R cosq  T cos 53.0  0
F
y
 R sin q  T sin 53.0  600N  200N  0
From the rotational equilibrium
  T sin 53.0  8.00  600N  2.00  200N  4.00m  0

Using the
translational
equilibrium
T  313N
R cosq  T cos 53.0
R sin q  T sin 53.0  600 N  200 N
 800  313  sin 53.0 


q  tan 

71
.
7


 313 cos 53.0

1
And the magnitude of R is
T cos 53.0 313  cos 53.0
R

 582 N
cos 71.1
cosq
6-
???
7-
End of chapter 2
Thank you very much for your attention
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