lec04-NL-hwkprob5-20

# lec04-NL-hwkprob5-20 - 7.5 N m a 0.798 2(1.3 3.2 4.9 kg s F...

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m 1 m 2 m 3 F 12 F 32 F=(m 1 +m 2 +m 3 )a + 75 0 798 1 3 3 2 4 9 2 . N m a. ( . . . ) kg s   1 2 3 F a (m m m ) F 23 F 21 Find F 21 and F 23 I first find “a”

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m 1 m 2 m 3 F 12 F 32 F=(m 1 +m 2 +m 3 )a + 75 0 798 1 3 3 2 4 9 2 . N m a. ( . . . ) kg s   1 2 3 F a (m m m ) F 23 F 21 32 4 9 798 2 F . (. ) kg m/s 32 3 F m a
m 1 m 2 m 3 F 12 F 32 F=(m 1 +m 2 +m 3 )a + 75 0 798 1 3 3 2 4 9 2 . N m a. ( . . . ) kg s   1 2 3 F a (m m m ) 12 1 F F m a  F 23 F 21 32 4 9 798 2 F . (. ) kg m/s 2 12 1 7 5 1 3 798 6 46 F F m a . N . (. )kg m / s . N 32 3 F m a sign – direction included

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m 1 m 2 m 3 F 12 F 32 + 75 0 798 1 3 3 2 4 9 2 . N m a. ( . . . ) kg s   F 23 F 21 check 21 23 2 F F m a  2 6 46
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lec04-NL-hwkprob5-20 - 7.5 N m a 0.798 2(1.3 3.2 4.9 kg s F...

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