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

Ladder stability

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

Department of Physics and Applied Physics95.141, Fall 2013, Lecture 22

Thank youSee you on Wednesday