Ch0123 - Chapter 23 Statics 23 Statics It bears repeating:...

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Chapter 23 Statics 155 23 Statics It bears repeating: Make sure that any force that enters the torque equilibrium equation is multiplied by a moment arm, and that any pure torque (such as τ o in the solution of example 23-2 on page 151) that enters the torque equilibrium equation is NOT multiplied by a moment arm. For any rigid body, at any instant in time, Newton’s 2 nd Law for translational motion = F h h m 1 a and Newton’s 2 nd Law for Rotational motion = τ a h I 1 a both apply. In this chapter we focus on rigid bodies that are in equilibrium. This topic, the study of objects in equilibrium, is referred to as statics . Being in equilibrium means that the acceleration and the angular acceleration of the rigid body in question are both zero. When 0 = a h , Newton’s 2 nd Law for translational motion boils down to 0 = F h (23-1) and when 0 = a a , Newton’s 2 nd Law for Rotational motion becomes 0 = τ h (23-2) These two vector equations are called the equilibrium equations. They are also known as the equilibrium conditions. In that each of the vectors has three components, the two vector equations actually represent a set of six scalar equations: = 0 x F = 0 y F = 0 z F = 0 x < = 0 y < = 0 z <
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Chapter 23 Statics 156 In many cases, all the forces lie in one and the same plane, and if there are any torques aside from the torques resulting from the forces, those torques are about an axis perpendicular to that plane. If we define the plane in which the forces lie to be the x-y plane, then for such cases, the set of six scalar equations reduces to a set of 3 scalar equations (in that the other 3 are trivial 0=0 identities): = 0 x F (23-3) = 0 y F (23-4) = 0 z < τ
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Ch0123 - Chapter 23 Statics 23 Statics It bears repeating:...

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