1. © 2005 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. Lecture PowerPoints Chapter 9 Physics: Principles with Applications, 6 th edition Giancoli

2. Chapter 9 Static Equilibrium; Elasticity and Fracture

3. Units of Chapter 9 The Conditions for Equilibrium Solving Statics Problems Applications to Muscles and Joints Stability and Balance Elasticity; Stress and Strain Fracture Spanning a Space: Arches and Domes

4. 9-1 The Conditions for Equilibrium An object with forces acting on it, but that is not moving, is said to be in equilibrium.

5. 9-1 The Conditions for Equilibrium The first condition for equilibrium is that the forces along each coordinate axis add to zero.

6. 9-1 The Conditions for Equilibrium The second condition of equilibrium is that there be no torque around any axis; the choice of axis is arbitrary.

7. 9-2 Solving Statics Problems 1. Choose one object at a time, and make a free- body diagram 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.

8. 9-2 Solving Statics Problems The previous technique may not fully solve all statics problems, but it is a good starting point.

9. Solution Need to look at the forces in the beam. Label them The weight of the beam acts at its center of gravity Pick an axis for the torque that is convenient Use the torque equation to solve for F B

10. 5-5 Solving Statics Problems  = - (10.0m) (1500kg) g - (15.0m) (15,000kg) (g)+ (20.0m) F B =0 Solve for F B = 118,000 N Use sum of forces = 0 to solve for F A  F = F A +F B -(1500kg)g-(15,000kg)g=0 F A = -118,000+14,700+147,000 = 43,700 N

11. 9-2 Solving Statics Problems If there is a cable or cord in the problem, it can support forces only along its length. Forces perpendicular to that would cause it to bend.

12. 9-4 Stability and Balance 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.

13. 9-4 Stability and Balance If, however, the forces tend to move it away from its equilibrium point, it is said to be in unstable equilibrium.

14. 9-4 Stability and Balance An object in stable equilibrium may become unstable if it is tipped so that its center of gravity is outside the pivot point. Of course, it will be stable again once it lands!

15. 9-4 Stability and Balance 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.

16. 9-5 Elasticity; Stress and Strain Hooke’s law: the change in length is proportional to the applied force. (9-3)

17. 9-5 Elasticity; Stress and Strain This proportionality holds until the force reaches the proportional limit. Beyond that, the object will still return to its original shape up to the elastic limit. Beyond the elastic limit, the material is permanently deformed, and it breaks at the breaking point.

18. 9-5 Elasticity; Stress and Strain The change in length of a stretched object depends not only on the applied force, but also on its length and cross-sectional area, and the material from which it is made. The material factor is called Young’s modulus, and it has been measured for many materials. The Young’s modulus is then the stress divided by the strain.

19. 9-5 Elasticity; Stress and Strain In tensile stress, forces tend to stretch the object.

20. 9-5 Elasticity; Stress and Strain Compressional stress is exactly the opposite of tensional stress. These columns are under compression.

21. 9-5 Elasticity; Stress and Strain Shear stress tends to deform an object:

22. 9-6 Fracture If the stress is too great, the object will fracture. The ultimate strengths of materials under tensile stress, compressional stress, and shear stress have been measured. When designing a structure, it is a good idea to keep anticipated stresses less than 1/3 to 1/10 of the ultimate strength.

23. 9-6 Fracture A horizontal beam will be under both tensile and compressive stress due to its own weight.

24. 9-6 Fracture What went wrong here? These are the remains of an elevated walkway in a Kansas City hotel that collapsed on a crowded evening, killing more than 100 people.

25. 9-6 Fracture Here is the original design of the walkway. The central supports were to be 14 meters long. During installation, it was decided that the long supports were too difficult to install; the walkways were installed this way instead.

26. 9-6 Fracture The change does not appear major until you look at the forces on the bolts: The net force on the pin in the original design is mg, upwards. When modified, the net force on both pins together is still mg, but the top pin has a force of 2mg on it – enough to cause it to fail.

27. 9-7 Spanning a Space: Arches and Domes The Romans developed the semicircular arch about 2000 years ago. This allowed wider spans to be built than could be done with stone or brick slabs.

28. 9-7 Spanning a Space: Arches and Domes The stones or bricks in a round arch are mainly under compression, which tends to strengthen the structure.

29. 9-7 Spanning a Space: Arches and Domes Unfortunately, the horizontal forces required for a semicircular arch can become quite large – this is why many Gothic cathedrals have “flying buttresses” to keep them from collapsing.

30. 9-7 Spanning a Space: Arches and Domes A dome is similar to an arch, but spans a two-dimensional space.

31. Summary of Chapter 9 An object at rest is in equilibrium; the study of such objects is called statics. In order for an object to be in equilibrium, there must be no net force on it along any coordinate, and there must be no net torque around any axis. An object in static equilibrium can be either in stable, unstable, or neutral equilibrium.

32. Summary of Chapter 9 Materials can be under compression, tension, or shear stress. If the force is too great, the material will exceed its elastic limit; if the force continues to increase the material will fracture.

33. 9-3 Applications to Muscles and Joints These same principles can be used to understand forces within the body.

34. Problem How much force must the biceps muscle exert when a 5.0 kg mass is held in the hand (a) with the arm horizontal and (b) when the arm is at 45 degree angle as in the lower picture? Assume that the mass of the forearm and hand together is 2.0 kg and their CG is as shown.

35. B) The lever arm, as calculated about the joint is reduced by sin 45 degrees for all three forces. Our torque equation will look the same as the one above except that each term will have it lever arm reduced by the same factor, which will cancel out. So the same result is obtained, F M = 400 N

36. 9-3 Applications to Muscles and Joints The angle at which this man’s back is bent places an enormous force on the disks at the base of his spine, as the lever arm for F M is so small.