chapter 12 static equilibrium. equilibrium we already introduced the concept of equilibrium in...
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Equilibrium Static equilibrium: Stable equilibrium: the body returns to the state of static equilibrium after having been displaced from that state. Unstable equilibrium: the state of equilibrium is lost after a small force displaces the bodyTRANSCRIPT
![Page 1: Chapter 12 Static Equilibrium. Equilibrium We already introduced the concept of equilibrium in Chapter 7: dU(x)/dx = 0 More general definition of equilibrium:](https://reader036.vdocuments.site/reader036/viewer/2022062306/5a4d1b7c7f8b9ab0599b9654/html5/thumbnails/1.jpg)
Chapter 12
Static Equilibrium
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Equilibrium
• We already introduced the concept of equilibrium in Chapter 7: dU(x)/dx = 0
• More general definition of equilibrium:
• Static equilibrium:
• Stable equilibrium: the body returns to the state of static equilibrium after having been displaced from that state. Unstable equilibrium: the state of equilibrium is lost after a small force displaces the body
constLconstP
;
0;0 LP
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Equilibrium
• Static equilibrium:
• Stable equilibrium: the body returns to the state of static equilibrium after having been displaced from that state. Unstable equilibrium: the state of equilibrium is lost after a small force displaces the body
0;0 LP
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Equilibrium
• Stable equilibrium:
• Unstable equilibrium:
02
2
dx
Ud
02
2
dx
Ud
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Chapter 12Problem 25
A particle’s potential energy as a function of position is given by U = 2x3 – 2x2 – 7x + 10,with x in meters and U in joules. Find the positions of any stable and unstable equilibria.
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Center of mass: stable equilibrium
• We consider the torque created by the gravity force (applied to the CM) and its direction relative to the possible point(s) of rotation
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Center of mass: stable equilibrium
• We consider the torque created by the gravity force (applied to the CM) and its direction relative to the possible point(s) of rotation
![Page 8: Chapter 12 Static Equilibrium. Equilibrium We already introduced the concept of equilibrium in Chapter 7: dU(x)/dx = 0 More general definition of equilibrium:](https://reader036.vdocuments.site/reader036/viewer/2022062306/5a4d1b7c7f8b9ab0599b9654/html5/thumbnails/8.jpg)
Center of mass: stable equilibrium
• We consider the torque created by the gravity force (applied to the CM) and its direction relative to the possible point(s) of rotation
![Page 9: Chapter 12 Static Equilibrium. Equilibrium We already introduced the concept of equilibrium in Chapter 7: dU(x)/dx = 0 More general definition of equilibrium:](https://reader036.vdocuments.site/reader036/viewer/2022062306/5a4d1b7c7f8b9ab0599b9654/html5/thumbnails/9.jpg)
Center of mass: stable equilibrium
• We consider the torque created by the gravity force (applied to the CM) and its direction relative to the possible point(s) of rotation
![Page 10: Chapter 12 Static Equilibrium. Equilibrium We already introduced the concept of equilibrium in Chapter 7: dU(x)/dx = 0 More general definition of equilibrium:](https://reader036.vdocuments.site/reader036/viewer/2022062306/5a4d1b7c7f8b9ab0599b9654/html5/thumbnails/10.jpg)
The requirements of equilibrium
• For an object to be in equilibrium, we should have two requirements met
• Balance of forces: the vector sum of all the external forces that act on the body is zero
• Balance of torques: the vector sum of all the external torques that act on the body, measured about any possible point, is zero
0 ; netFdtPdconstP
0 ; netdtLdconstL
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Equilibrium: 2D case
• If an object can move only in 2D (xy plane) then the equilibrium requirements are simplified:
• Balance of forces: only the x- and y-components are considered
• Balance of torques: only the z-component is considered (the only one perpendicular to the xy plane)
0;0 ,, ynetxnet FF
0, znet
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Examples of static equilibrium
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Examples of static equilibrium
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Chapter 12Problem 38
A crane in a marble quarry is mounted on the quarry’s rock walls and is supporting a 2500-kg marble slab as shown in the figure. The center of mass of the 830-kg boom is located one-third of the way from the pivot end of its 15-m length, as shown. Find the tension in the horizontal cable that supports the boom.
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Indeterminate structures
• Indeterminate systems cannot be solved by a simple application of the equilibrium conditions
• In reality, physical objects are not absolutely rigid bodies
• Concept of elasticity is employed
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Questions?
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Answers to the even-numbered problems
Chapter 12
Problem 24:
(a) 47 m from the origin (b) unstable
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Answers to the even-numbered problems
Chapter 12
Problem 36:
1.2 m
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Answers to the even-numbered problems
Chapter 12
Problem 54:
L/6 of the bottom book can overhangthe desk