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ME 242 Lecture 11 02.13

# ME 242 Lecture 11 02.13 - ME 242 Dynamics Today's Key...

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1 ME 242 Dynamics February 13, 2009 Eric Wang Today’s Key Concepts Chapter 3: Kinetics of a Particle Newton’s 2 nd law: In Path Coordinates F = d dt mv ( ) Path Coordinates Drop the sign with the understanding that we always mean the sum F = ma Vector form F t e t + F n e n + F z k = m ( a t e t + a n e n + a z k ) Scalar form F t = m ˙ v F n = m v 2 r c F z = m ˙ ˙ ˙ z Example (prob. 3.3.2) Given: m=300 kg • r c =80 m v(0)=0 m/s • a t =1.1 m/s 2 (constant) Find: at t=4s F tires Always start with FBD=IRD • FBD • IRD e n e t k e t Write Eqns based on the Diagrams

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2 Count the Unknowns m = 300 kg a t = 1.1 m s 2 r c = 80 m v (0) = 0 Substitute in the Givens Total force on tires: m = 300 kg a t = 1.1 m s 2 r c = 80 m v = 4.4 m s Example (prob. 3.3.12) Given: r=10 m m=120 kg ω =0.1 rad/s θ =45° Find: Total force between seat and Ferris wheel (constant) Always start with FBD=IRD • FBD • IRD R 1 ma n ma t R 2 e n e t mg θ Write Eqns based on the Diagrams F t = R 1 mg cos θ = ma t = m ˙ v F n = R 2 + mg sin θ = ma n = m v 2 r c = mr c ω 2 v = r c ω F t = ma t = m ˙ v F n = ma
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ME 242 Lecture 11 02.13 - ME 242 Dynamics Today's Key...

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