PHY 113 A General Physics I 9 9

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PHY 113 A General Physics I 9 -9: 50 AM MWF Olin 101 Plan

PHY 113 A General Physics I 9 -9: 50 AM MWF Olin 101 Plan for Lecture 20: Chapter 12 – Static equilibrium 1. Balancing forces and torques; stability 2. Center of gravity 3. Note: May not have time to cover elastic properties of materials 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 1

10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 2

10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 2

Comment on exam question #4 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20

Comment on exam question #4 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 3

10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 4

10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 4

From Webassign #17 (11. 34) 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20

From Webassign #17 (11. 34) 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 5

Conditions for stable equilibrium 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 6

Conditions for stable equilibrium 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 6

Stability of “rigid bodies” N mi g 10/17/2012 PHY 113 A Fall 2012 --

Stability of “rigid bodies” N mi g 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 7

Center-of-mass Torque on an extended object due to gravity (near surface of the earth)

Center-of-mass Torque on an extended object due to gravity (near surface of the earth) is the same as the torque on a point mass M located at the center of mass. ri 10/17/2012 mi r. CM PHY 113 A Fall 2012 -- Lecture 20 8

Notion of equilibrium: Notion of stability: q F=ma r T- mg cos q =

Notion of equilibrium: Notion of stability: q F=ma r T- mg cos q = 0 -mg sin q = -maq T t=I a r mg sin q = mr 2 a = mraq mg(-j) 10/17/2012 Example of stable equilibrium. PHY 113 A Fall 2012 -- Lecture 20 9

Unstable equilibrium: Support above CM: Support below CM: q r T mg(-j) 10/17/2012 PHY

Unstable equilibrium: Support above CM: Support below CM: q r T mg(-j) 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 10

Analysis of stability: 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 11

Analysis of stability: 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 11

* X * 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 12

* X * 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 12

10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 13

10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 13

RCM Fg 1 * X * mg 10/17/2012 PHY 113 A Fall 2012 --

RCM Fg 1 * X * mg 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 14

iclicker question: F 2 F 1 L/3 Mg mg L Consider the above drawing

iclicker question: F 2 F 1 L/3 Mg mg L Consider the above drawing of the two supports for a uniform plank which has a total weight Mg and has a weight mg at its end. What can you say about F 1 and F 2? (a) F 1 and F 2 are both up as shown. (b) F 1 is up but F 2 is down. (c) F 1 is down but F 2 is up. 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 15

F 1 F 2 * X * L/3 Mg mg L 10/17/2012 PHY 113

F 1 F 2 * X * L/3 Mg mg L 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 16

iclicker question: The fact that we found F 1<0 means: A. We set up

iclicker question: The fact that we found F 1<0 means: A. We set up the problem incorrectly B. The analysis is correct, but the direction of F 1 is opposite to the arrow C. Physics makes no sense iclicker question: What would happen if we analyzed this problem by placing the pivot point at F 1 ? : A. The answer would be the same. B. The answer would be different. C. Physics makes no sense 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 17

10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 18

10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 18

Fwall Mg = 120 N mg = 98 N mg T < 110 N

Fwall Mg = 120 N mg = 98 N mg T < 110 N Mg N T 10/17/2012 PHY 113 A Fall 2012 -- Lecture 20 19