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# hwsol03 - ME 358 Homework Solution 3 Problem 1 For the...

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Unformatted text preview: ME 358 Homework Solution 3 Problem 1 For the linkage shown, the velocity of point B is v B = 40 m/s to the left, the distance between points A and B is R BA = 400 mm and the angles are ψ = 15 o and φ = -45 o as indicated in the figure. Using the analytical method discussed in class, find the acceleration of point A and the angular acceleration of link 3. Solution ( ) ( ) ( ) ( ) 2 4 2 3 3 4 2 4 2 3 3 2 3 3 2 4 2 4 4 3 3 4 2 4 2 4 3 3 4 3 3 2 3 3 3 3 4 2 4 2 4 2 3 3 2 2 4 4 3 3 2 2 4 4 3 2 4 3 AB 2 B 4 A AB B A sin sin r sin sin cos cos sin r sin cos r cos r sin sin r sin sin cos cos sin sin r cos cos r r sin r cos r r r sin sin cos cos sin r sin r sin r cos r cos r cos r θ − θ θ − θ = θ θ − θ θ − θ θ − θ θ − = θ − θ θ − θ = θ θ − θ θ − θ θ θ θ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ θ θ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ θ θ − θ θ − θ = θ − θ θ = θ − θ = − = = = + = r r r r R r R r R R R R ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) mm 7 . 772 15 sin 150 sin mm 400 sin sin r r mm 8 . 1092 15 sin 15 150 sin mm 400 sin sin r r 15 150 30 180 180 mm 400 R r o o o o 2 4 2 3 3 4 o o o o 2 4 4 3 3 2 o 4 o o o o 3 o 2 BA 3 = − − = θ − θ θ − θ = = − − = θ − θ θ − θ = = ψ = θ = − = φ − = θ = θ = = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) k ω j i r v r r r 3 3 4 4 4 4 4 A 3 4 3 2 4 2 3 4 3 4 3 2 4 2 4 2 3 3 4 3 3 4 2 2 4 2 2 4 3 3 4 3 2 2 3 4 3 4 3 3 2 3 2 3 2 3 3 4 3 3 4 3 3 2 2 3 3 2 2 4 2 2 2 2 3 4 3 3 4 3 3 4 2 2 3 3 3 4 4 2 2 3 3 3 4 4 4 2 3 B 2 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 2 3 4 sin r cos r cos r sin r sin sin cos cos r sin cos cos sin r cos r sin sin r cos sin r sin cos r cos cos cos r sin sin cos cos r sin sin cos cos r r cos r sin sin r cos cos r sin r sin r cos r r sin r cos r r cos r sin sin r cos sin r cos r sin r cos r sin r cos r r v r cos r sin r cos r sin r cos r sin r sin r cos r sin r cos r sin r cos r θ = θ + θ = = θ − θ θ − θ = θ θ + θ θ θ θ − θ θ = θ − θ θ θ θ θ θ θ = θ θ − θ θ − θ = θ θ + θ θ θ θ + θ θ = θ − θ θ θ θ − θ θ θ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ θ θ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ θ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ θ − θ θ θ θ = θ θ − θ θ = θ θ + θ = θ = θ = = θ θ + θ = θ θ + θ − θ θ + θ θ θ − θ = θ θ − θ − θ θ − θ = − & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) k k ω j i j i j i r v s rad 6 . 36 s rad 6 . 36 150 15 cos m 4 . 15 sin s m 40 cos r sin r s m 7 . 12 s m 3 . 47 15 sin s m . 49 15 cos s m . 49 sin r cos r s m . 49 150 15 cos 150 cos s m 40 cos cos r r s m 40 r 3 3 o o o o 3 4 3 2 4 2 3 4 4 4 4 4 A o o o o 3 4 3 2 2 4 2 = θ = = − − − = θ − θ θ − θ = θ − − = − − = θ + θ = = − = − − − = θ − θ θ − θ = − = & & & & & & & & & o o ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )...
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## This note was uploaded on 09/14/2008 for the course ME 358 taught by Professor Smith during the Spring '08 term at Stevens.

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hwsol03 - ME 358 Homework Solution 3 Problem 1 For the...

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