61 - TAM 212 Spring, 2004 1.61 Given: The length of the...

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TAM 212 Spring, 2004 1.61 Given: The length of the string is 44 m . The system shown starts from rest in the position shown. At that instant, 2 / 2 s m x B = . (Also given is the information in the picture.) Required: The speed of A just before it hits the pulley, that is, after it has traveled 5m up. Solution: Let ( 29 t y A be the distance of the object A from the pulley P, so ( 29 t y A is the velocity of A at an arbitrary time t (positive down, given that ( 29 t y A is measured down from the pulley). Let ( 29 t x B be the distance of the object B from the wall, so ( 29 t x B is the velocity of B at an arbitrary time t (positive down, given that ( 29 t x B is measured to the right of the wall). Let ( 29 t s be the distance of the object B from the pulley P, so ( 29 t s is the speed at which this is changing, that is, the lengthening of the distance of the rope extended between P and B…at an arbitrary time t. We have two constraints to work with: (1) The length of the rope is constant: ( 29 ( 29 ( 29( 29 K t y t d t x m L A B + + + = = 44
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61 - TAM 212 Spring, 2004 1.61 Given: The length of the...

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