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Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
Lecture 22
Chapter 12
Static Equilibrium
12.02.2013Physics I
Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI
Lecture Capture: http://echo360.uml.edu/danylov2013/physics1fall.html
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
Chapter 12
The Conditions for Equilibrium Solving Statics Problems Stability and Balance
Outline
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
Exam III Info
Exam III Wed Dec. 4, 9:00-9:50am, OH 150.Exam III covers Chapters 9-11
Same format as Exam IIPrior Examples of Exam III posted
Ch. 9: Linear Momentum (no section 10)Ch. 10: Rotational Motion (no section 10)
Ch. 11: Angular Momentum; General Rotation (no sections 7-9)
Exam Review Session TBA, Ball 210
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
There are three branches of Mechanics:
Kinematics Motion Forces Dynamics Motion Forces Statics Motion Forces
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
An object with forces acting on it, but with zero net force, is said to be in equilibrium.
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
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
The 2nd Condition for Equilibriumprevents rotational motion
0 ext
Iext
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 external torque around any axis (the choice of axis is arbitrary).
0 x 0 y 0 z
ConcepTest 1 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
0 FFF
Iext
force equilibrium
torque equilibrium X
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.
1r 2r
1
2
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
Reduce # of Equilibrium EquationsFor 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
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
Axis of rotation for the 3rd equation
Does it matter which axis you choose for calculating torques?NO. The choice of an axis is arbitrary
0 z
0F
Any axis of rotation works
If an object is in translational equilibrium and the net torque is 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
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
Any axis of rotation works for the 3rd
equation (proof)The choice of an axis is arbitrary
0 z
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
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
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
Example:traffic light
Find the tension in the two wires supporting the traffic light
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
Torque due to gravityHere, we will often have objects in which there is a torque exerted by gravity. There is a simple rule how to get gravitational torque:
gMRCM
gMW
CMR
CM
For extended objects, gravitational torque acts as if all mass were concentrated at the center of mass
sinMgRCMR
gM
As if the whole mass were here
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
Recall how to calculate torqueFr
1F
1r
• Draw a line of force • Find the perpend. distance (r┴) from
the axis of rotation to that line.• Magnitude of torque is r┴F• Use the right-hand rule to find its
direction
sinrFLine of force (action)
2F
1r
2r
2r
111 Fr Direction: out of the board
222 Fr Direction: into the board
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
Ladder stabilityA uniform ladder of mass m and length l leans at an angle θagainst a frictionless wall. If the coefficient of static friction between the ladder and the ground is μs, determine a formula for the minimum angle at which the ladder will not slip.
wF
gxF
gyF
CM
gmW
The forces are nonconcurrent, so we need all equilibrium conditions
0 xF0 z 0 yF
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
Stability and Balance
Assume that we found equilibrium of a system 0 xF 0 z0 yF
What is going to happen with the system if we disturb it slightly?Is it stable or unstable?
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
If the forces on an object are such that they tend to return it to its equilibrium position, it is said to be in stable equilibrium.
Stable equilibrium
Equilibrium position
Disturbed systemNet force returns the ball back to its equilibrium
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
Unstable equilibrium
If, however, the forces tend to move it away from its equilibrium point, it is said to be in unstable equilibrium.
0 xF
0 z
0 yF
0 xF
0 z
0 yF
r
gm
Equilibrium position Disturbed system
Torque turns the pencil away from equilibrium
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
An object in stable equilibrium may become unstable if it is tipped so that its center of mass (CM) is outside its base of support. Of course, it will be stable again once it lands!
Border between a stable/unstable equilibrium
outside its base of supportinside its base of support
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
People carrying heavy loads automatically adjust their posture so their center of mass is over their feet. This can lead to injury if the contortion is too great.
Stability and Balance
ConcepTest 2 Tipping Over I
1 2 3
A) all
B) 1 only
C) 2 only
D) 3 only
E) 2 and 3
A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a blue dotin each case. In which case(s) does the box tip over?
The torque due to gravity acts like
all the mass of an object is
concentrated at the CM. Consider
the bottom right corner of the box
to be a pivot point. If the box
can rotate such that the CM is
lowered, it will!!
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22
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 torque equilibrium equation. A clever choice here can simplify the problem enormously.
5. Solve.
Solving Statics Problems
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