HR12 - Chapter 12 Equilibrium and Elasticity In this...

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Chapter 12 Equilibrium and Elasticity In this chapter we will define equilibrium and find the conditions needed so that an object is at equilibrium We will then apply these conditions to a variety of practical engineering problems of static equilibrium We will also examine how a “rigid” body can be deformed by an external force. In this section we will introduce the following concepts: Stress and strain Young’s modulus (in connection with tension and compression) Shear modulus (in connection with shearing) Bulk modulus (in connection to hydraulic stress) (12-1)
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We say that an object is in equilibrium when the following two conditions are satisfied: The linear momentum of the center of mass is constant The angular momentum about the cent P L Equilibrium 1. 2. G G er of mass or any other point is a constant Our concern in this chapter is with situations in which 0 and 0 That is we are interested in objects that are not moving in any way (this include PL == GG s translational as well as rotational motion) in the reference frame from which we observe them. Such objects are said to be in In chapter 8 we differentiated between stable and uns static equilibrium table static equilibrium If a body that in equilibrium is displaced slightly from this position the forces on it may return it to its old position. In this case we say that the equilibrium is . stable If the body does not return to its old position then the equilibrium is unstable (12-2)
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An example of is shown in the figures In fig.a we balance a domino with the domino's center of mass vertically above the supporting edge. The torque of the gravitational force g F G unstable equilibrium about the supporting edge is zero because the line of action of passes through the edge. g F G Thus the domino is in equilibrium. Even a slight force on the domino ends the equilibrium. As the line of action of moves to one side of the supporting edge (see fig.b) the torque due to i g g F F G G s non-zero and the domino rotates in the clockwise direction away from its equilibrium position of fig.a. The domino in fig.a is in a position of unstable equilibrium. The domino is fig.c is not quite as unstable. To topple the domino the applied force would have to rotate it through and beyond the position of fig.a. A flick of the finger against the domino can topple it. (12-3)
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In chapter 9 we calculated the rate of change for the linear momentum of the center of mass of an object. If an object is in translational equilibrium then net dP F dt = The Conditions of equilibrium G G constant and thus 0 In chapter 11 we analyzed rotational motion and saw that Newton's second law takes the form: For an object in rotational equilibrium we hav 0 e: net net dP P dt dL dt F τ == = = G G G G G constant 0 The two requirements for a body to be i 0 n equilibrium are: net L dL dt = →= = G G G 1. The vector sum of all the external forces on the body must be zero 2. The vector sum of all the external torques that act on the body measured about any point must be zero (12-4) net F = G 0 net = G
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This note was uploaded on 10/31/2010 for the course EE 423 taught by Professor Mitin during the Spring '10 term at SUNY Buffalo.

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HR12 - Chapter 12 Equilibrium and Elasticity In this...

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