Physics 135A - PowerPoint PPT Presentation

1 / 22
About This Presentation
Title:

Physics 135A

Description:

10/3/09. 1. Physics 135A. Physics 135A. Lecture 15: Statics. Dr. Z. Gong. 10/3/09. 2. Physics 135A ... Statics problem solving ... – PowerPoint PPT presentation

Number of Views:51
Avg rating:3.0/5.0
Slides: 23
Provided by: physi1
Learn more at: http://physics.usc.edu
Category:
Tags: 135a | physics | statics

less

Transcript and Presenter's Notes

Title: Physics 135A


1

Physics 135A
  • Physics 135A
  • Lecture 15 Statics
  • Dr. Z. Gong

2

Physics 135A
Administrative Information
  • Homework Set 7 due this Thursday

3


Physics 135A
Review -1
  • Moment of Inertia
  • The angular acceleration of an object is
    determined by the net torque acting on it.
  • The Newtons 2nd law of motion for rotation
  • where St is the net torque, and a is the angular
    acceleration

4


Physics 135A
Review -2
  • The kinetic energy of an object with both the
    translational and rotational motion can be
    represented as
  • The mechanic energy of a system is conserved if
    theres only conservative force involved.

5



Physics 135A
Physics 135A
Review -3
  • If the net torque acting on a rotating system is
    zero, the total angular momentum of the system is
    conserved.
  • The direction of the angular velocity ? and
    angular momentum L of a body rotating along the
    axis of rotation follows the right hand rule.

6



Physics 135A
Physics 135A
Equilibrium state
  • A system is in equilibrium when the overall
    status of the system remains the same after an
    extended period of time.
  • If a system is in the equilibrium state, it does
    not necessarily mean theres nothing happening in
    that system. One example is gas in a container.
    From outside it is in equilibrium, but internally
    the gas molecular are moving around all the time.

7

Physics 135A
Conditions for equilibrium
  • When a system is in equilibrium, it should have
    no linear acceleration. The net force acting on
    the system is zero.
  • For the same reason, it should have no angular
    acceleration. The net torque acting on the system
    w.r.t. any axis is zero.

8

Physics 135A
Statics problem solving
  • Draw the free body diagram of one object at a
    time, including all the forces at the point of
    action with correct direction.
  • Setup a coordinate axis and find the axis of
    rotation. Define the positive direction of torque
    (either way is fine, as long as you dont change
    it in the middle)
  • Write down the net force on the x- and y-
    direction, as well as the net torque.

9

Physics 135A
Example -1
  • A traffic light hangs from a structure as shown
    below. The uniform pole AB is 7.5m long and 8.0kg
    in mass. The mass of the traffic light is 12.0kg.
    Determine the tension in the horizontal massless
    cable CD and the vertical and horizontal
    components of the forces exerted by the pivot A
    on the pole.

37o
3.8m
10

Physics 135A
Example -1
First, draw all the forces on the pole. From the
1st condition of equilibrium, the net force on
the horizontal and vertical direction are
zero. However, both FT , FN and ? are
unknown. We need to apply the 2nd condition of
equilibrium.
FT
37o
3.8m
FN
m2g
?
m1g
11

Physics 135A
Example -1
Choose the pivot as the rotational axis, the net
torque w.r.t. the pivot is zero. So we
get Substitute into the previous equations we
get
FT
37o
3.8m
FN
m2g
?
m1g
12

Physics 135A
Example -2
A group of 10 students with mass 70kg each join
an adventure journey. During the trip they have
to go across an old bridge of 10m long and 100kg
in mass that can sustain at most 3000N at each
end of the bridge. If the students walk with even
space between them, what is the minimum space so
that the bridge will not collapse into the river?
13

Physics 135A
Example -2
FN2
FN1
mg
mg
mg
Mg
Lets suppose the distance between each student
is l. The critical point will be when the 1st
student is at the other end of the bridge. The
bridge is in equilibrium state.
14

Physics 135A
Example -2
FN2
FN1
B
A
mg
mg
mg
Mg
If FN1 FN2 then, from the 1st equilibrium
condition
15

Physics 135A
Example -2
FN2
FN1
B
A
mg
mg
mg
Mg
If FN1 ?FN2 then, from the 2st equilibrium
condition w.r.t. point A as the axis of
rotation
16

Physics 135A
Example -2
FN2
FN1
B
A
mg
mg
mg
Mg
If FN1 ?FN2 then, from the 2st equilibrium
condition w.r.t. point B as the axis of
rotation
17

Physics 135A
Application to muscles and joints
  • Two kinds of muscles to make bones move. Flexor
    muscle those that moves the bone inward
    Extensor muscle those that moves the body
    outward.
  • The overall method to solve muscle and joint
    problem is the same as the previous problems. If
    it is a static state then both the net force and
    net torque are zero.

18

Physics 135A
Stability
  • The equilibrium state is not always the stable
    state.
  • Stable equilibrium the system returns to its
    original state after perturbation
  • Unstable equilibrium the system leaves its
    original state under small perturbation
  • Quasi-stable equilibrium the system returns to
    its original under small perturbation, but will
    leave under finite perturbation
  • Neutral equilibrium the system remains stable
    at the new position.

19

Physics 135A
Stability
Nonstable equilibrium
Quasi-stable equilibrium
Stable equilibrium
F
Ffr
Neutral equilibrium
20

Physics 135A
Example -3
A cube of side L rests on a rough floor. It is
subjected to a steady horizontal pull F exerted a
distance h above the floor. As F increase, the
block will either slide or tip over. What should
the coefficient of static friction be so that the
cube will slide but not tip? What should the
coefficient of static friction be so that the
cube will tip?
F
h
21

Physics 135A
Example -3
FN
The net force acting on the block is zero, and
the net torque w.r.t point A is also zero.
F
Ffr
x
h
mg
A
22

Physics 135A
Example -3
FN
F
Ffr
x
h
When x is within the block the normal force will
still exist to provide necessary static friction.
The overall object is balanced. When xlt0 the
normal force will be outside the body, which is
impossible for this case. The block hence starts
to tip.
mg
A
Write a Comment
User Comments (0)
About PowerShow.com