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Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
Lecture 24
Chapter 12
Static Equilibrium
Physics I
Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
Today we are going to discuss:
Chapter 12:
Static Equilibrium: Section 12.8
IN THIS CHAPTER, you will discuss static equilibrium of an object
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
Bicycle wheel/turntable as a demo of Angular Momentum Conservation
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
Bicycle wheel/Ang. Mom. Conservation
xL
wzL
pzL
Initial situation Final situation
finz
inz LL
Angular Momentum Conservation (z comp):
pz
wz
finz LLL 0 w
zpz LL wwpp II
wp
wp I
I )( counterclockwiseclockwise
0inzL w - wheel
p - person
ConcepTest Spinning Bicycle WheelYou are holding a spinningbicycle wheel while standing on astationary turntable. If yousuddenly flip the wheel over sothat it is spinning in the oppositedirection, the turntable will:
The total angular momentum of the system is L upward, and it is conserved. So if the wheel has−L downward, you and the table must have +2L upward.
A) remain stationary
B) start to spin in the same direction as before flipping
C) start to spin in the same direction as after flipping
finz
inz LL
tablemeLLL zLL tableme 2
L
L2L
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
The 1st Condition for Equilibriumprevents translational motion
0F
0 xF 0 yF 0 zF
amF
N. 2nd law describes translational motion
He doesn’t want to have any sliding of a ladder, i.e. 0a
Force equilibrium
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
The 2nd Condition for Equilibriumprevents rotational motion
0
I
Rotational N. 2nd law describes rotational motion:v
He doesn’t want to have any rotation of a ladder, i.e. 0
There must be no net torque around any axis (the choice of axis is arbitrary).
0 x 0 y 0 z
Torque equilibrium
ConcepTest Static equilibriumConsider a light rod subject to the two forces of equal magnitude as shown in figure. Choose the correct statement with regard to this situation:
(A) The object is in force equilibrium but not torque equilibrium.(B) The object is in torque equilibrium but not force equilibrium(C) The object is in both force equilibrium and torque equilibrium(D) The object is in neither force equilibrium nor torque equilibrium
F
extforce equilibrium
torque equilibrium
Here, the 1st condition is satisfied but the 2nd isn’t, so there will be rotation.So, to have static situation, both conditions must be satisfied.
origin
Fr
1
2
0
FF
0 X21
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
Reduce # of Equilibrium Equations
For simplicity, we will restrict the applications to situations in which all the forces lie in the xy plane.
1st condition:
2nd condition:
0 xF 0 yF 0 zF
0 x 0 y 0 z
0)1 xF 0)2 yF 0)3 z
There are three resulting equations, which we will use
Fr
Fr ,
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
Axis of rotation for the 3rd equation
Does it matter which axis you choose for calculating torques?0 z
0F
Any axis of rotation works.It is up to you which one to choose
If an object is in a force equilibrium and the net torqueis zero about one axis, then the net torque must be zero about any other axis
We should be smart to choose a rotation axis to simplify problems
So. The choice of an axis is arbitrary
EH
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
Any axis of rotation works for the 3rd equation (proof)(Read only if you want)
The choice of an axis is arbitrary 0 z
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
Concurrent/Nonconcurrent forces
1F 2F
3F
Concurrent forces:when the lines of action of the forces intersect at a common point, there will be no rotation. So
1F 2F
3F
0 xF 0 yF0 z 0 xF 0 yF0 z
Nonconcurrent forces:when the lines of action of the forces do not intersect at a common point, there will be rotation. So
Automatically satisfied
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
Recall how to calculate torque (shortcut)
Fr
F
r
• Draw a line of force • Find the perpend. distance (r┴) from
the axis of rotation to that line.• Magnitude of torque is r┴F• Torque is positive if it tries to rotate an
object CCW
sinrF
Line of force (action)
r Fr
Lecture 20
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
Example
0 z
0 xF
A 3.0-m-long ladder leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.40. What is the minimum angle the ladder can make with the floor without slipping?
Ladder stability (12.58)
0 yF
The forces are nonconcurrent, so we need all equilibrium
conditions
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
Find the tension in the two wires supporting the traffic light
Example Traffic light
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
1. Choose one object at a time, and make a free-body diagram by showing all the forces on it and where they act.
2. Choose a coordinate system and resolve forces into components.
3. Write equilibrium equations for the forces.
4. Choose any axis perpendicular to the plane of the forces and write the torqueequilibrium equation. A clever choice here can simplify the problem enormously.
5. Solve.
Solving Statics Problems
Department of Physics and Applied PhysicsPHYS.1410 Lecture 24 Danylov
Thank youSee you on the Final Exam