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Unformatted text preview: t is in seconds . Required: The integral of the force from t = 2s to t = 5s Solution: ( 29 ( 29 ( 29 s m kg dt t t dt s s t s s t / * 4 / cos 20 4 / sin 20 5 2 5 2 = = = k i F p p Again, the extra factor of s comes from the integral and that t is in s. ( 29 ( 29 s m kg dt t dt t s s t s s t / * 4 / cos 20 4 / sin 20 5 2 5 2 = = = p p k i ( 29 ( 29 ( 29 ( 29 [ ] s m kg t t t t / * 4 / sin / 4 20 4 / cos / 4 20 5 2 5 2 = == p p p p k i ( 29 ( 29 ( 29 ( 29 [ ] s m kg / * 1 2 / 2 / 4 20 2 / 2 / 4 20= p p k i ( 29 ( 29 [ ] s m kg / * 1 2 / 2 2 / 2 / 80 k i + + = p 1.18 Done in class ( 29 ( 29 ( 29 lb lb s t L k i k i += += = p p 2 5 . 2 2 / 2 2 / 2 5 3 & v...
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This note was uploaded on 04/18/2008 for the course TAM 212 taught by Professor Keane during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 Keane

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