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10/17/2012PHY 113 A Fall 2012 -- Lecture 201 PHY 113 A General Physics I 9-9:50 AM MWF Olin 101 Plan for Lecture 20: Chapter 12 – Static equilibrium 1.Balancing.

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Presentation on theme: "10/17/2012PHY 113 A Fall 2012 -- Lecture 201 PHY 113 A General Physics I 9-9:50 AM MWF Olin 101 Plan for Lecture 20: Chapter 12 – Static equilibrium 1.Balancing."— Presentation transcript:

1 10/17/2012PHY 113 A Fall Lecture 201 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

2 10/17/2012 PHY 113 A Fall Lecture 202

3 10/17/2012PHY 113 A Fall Lecture 203 Comment on exam question #4

4 10/17/2012PHY 113 A Fall Lecture 204

5 10/17/2012PHY 113 A Fall Lecture 205 From Webassign #17 (11.34)

6 10/17/2012PHY 113 A Fall Lecture 206 Conditions for stable equilibrium

7 10/17/2012PHY 113 A Fall Lecture 207 Stability of “rigid bodies” N migmig

8 10/17/2012PHY 113 A Fall Lecture 208 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. mimi riri r CM

9 10/17/2012PHY 113 A Fall Lecture 209 Notion of stability: mg(-j) r T  F=ma  T- mg cos   mg sin  ma   =I   r mg sin  = mr 2  mra  Notion of equilibrium: Example of stable equilibrium.

10 10/17/2012PHY 113 A Fall Lecture 2010 Unstable equilibrium: mg(-j) r T  Support above CM: Support below CM:

11 10/17/2012PHY 113 A Fall Lecture 2011 Analysis of stability:

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

13 10/17/2012PHY 113 A Fall Lecture 2013

14 10/17/2012PHY 113 A Fall Lecture 2014 * X F g1 mg R CM

15 10/17/2012PHY 113 A Fall Lecture 2015 iclicker question: F1F1 F2F2 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. L/3 L Mg mg

16 10/17/2012PHY 113 A Fall Lecture 2016 F1F1 F2F2 L/3 L Mg mg * X

17 10/17/2012PHY 113 A Fall Lecture 2017 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

18 10/17/2012PHY 113 A Fall Lecture 2018

19 10/17/2012PHY 113 A Fall Lecture 2019 mg Mg F wall N T Mg = 120 N mg = 98 N T < 110 N


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