Exam1_solns - TAM 212 Spring, 2004, Exam 1A 1. Name:...

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TAM 212 Spring, 2004, Exam 1A Name: solutions 1. R mv F s m F ma F F normal normal normal normal horizontal / sin / sin 2 2 = θ ρ = θ = = & (1) mg F mg F ma F normal normal Vertical Vertical = θ = θ = cos 0 cos (2) ()( ) 2 1 ( ) Rg v Rg v / tan / tan 2 1 2 = θ = θ 2. Cable / tractor problem: The total length of the cable is constant and given by () ( ) l y h const L + = + 2 . , i.e. ( ) 2 2 2 . x h y h const L + + = + (1) The velocity of bale B (up) is given by y & , so we should differentiate (1) w.r.t. time: 2 2 2 0 x h x x y L + + = = & & & Therefore, the vertical velocity of the bale is 2 2 2 1 x h x x y + = & & (3) Now we have to write (3) in terms of time, so we have to determine x and x & . We are given a A = c , which is the same as c x = & & . Integration w.r.t. time gives D ct x + = & . But the tractor starts from rest, so D =0. Integration one more time gives E t c x + = 2 2 . We can say that the tractor starts at x =0 when t =0, so E =0. If we plug these relations into (3), we get the vertical velocity of bale B as a function of time: 4 2 1 4 2 2 3 t c h ct y + = & . 3. Vertical bead problem: x x ma N F F = = θ sin . a x =0, so θ sin F N = . (1) dt dv m ma mg N F F y k y = = = µ θ cos (2) where v is the vertical velocity of the collar. We know that θ = kt (problem statement) and from (1) that
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This note was uploaded on 04/09/2008 for the course TAM 212 taught by Professor Keane during the Spring '08 term at University of Illinois at Urbana–Champaign.

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