TAM 212 Spring, 2004, Exam 1A Name: solutions1. RmvFsmFmaFFnormalnormalnormalnormalhorizontal/sin/sin22=θ⇒ρ=θ⇒==∑∑&(1)mgFmgFmaFnormalnormalVerticalVertical=θ⇒=−θ⇒=∑cos0cos(2) ( ) ( )⇒21()RgvRgv/tan/tan212−=θ⇒=θ2. Cable / tractor problem: The total length of the cable is constant and given by ()()lyhconstL+−=+2., i.e.()()222.xhyhconstL++−=+(1) The velocity of bale B (up) is given byy&, so we should differentiate (1) w.r.t. time: 2220xhxxyL++−==&&&Therefore, the vertical velocity of the bale is 2221xhxxy+=&&(3) Now we have to write (3) in terms of time, so we have to determine xand x&. We are given aA= c, which is the same as cx=&&. Integration w.r.t. time givesDctx+=&. But the tractor starts from rest, so D=0. Integration one more time gives Etcx+=22. 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: 4214223tchcty+=&. 3.Vertical bead problem: xxmaNFF=−=∑θsin. ax=0, so θsinFN=. (1)dtdvmmamgNFFyky==−−=∑µθcos(2) where vis the vertical velocity of the collar. We know that θ= kt(problem statement) and from (1) that
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