ME 242 Lecture 21 03.25

# ME 242 Lecture 21 03.25 - ME 242 Dynamics Today's Key...

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1 ME 242 Dynamics March 25, 2009 Travis Fields Today’s Key Concepts Chapter 5: Multiparticle Systems Force Balance Example Angular Momentum Angular Impulse G is a special spot Recall: Force Balance • For the i th particle (constant mass) • System with n -particles j = 1 n ˜ F ji + F i = m i ˙ ˙ r i i = 1 n F i = m ˙ ˙ r G = ma G Example (prob. 5.1.7) Given: w A =w B= 120 lb, w C =110 lb, w D =150 lb Find: horizontal force on the platform v A = a A = v B = a B = 0 a C = a D = 6 i ft s 2 FBD=IRD A m A ˙ ˙ x A m A g N A B m B ˙ ˙ x B m B g N B F A F B FBD=IRD C m C ˙ ˙ x C m C g N C D m D ˙ ˙ x D m D g N D F C F C

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2 FBD=IRD for platform m P ˙ ˙ x P N D F C F D F A F B N C N A N B m P g N P Substitute the givens Angular Momentum H O = r × mv For any fixed point O • For a single particle: H O = i = 1 n r i × m i v i ( ) • For a system of particles: Angular Momentum H P = i = 1 n r i / P × m i v i ( ) P r i / P For a moving point P: ≠ ∑ i = 1 n r i / P × m i v i / P ( ) Cannot use the relative velocity
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ME 242 Lecture 21 03.25 - ME 242 Dynamics Today's Key...

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