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Question Number 23548 by ajfour last updated on 01/Nov/17

Commented by ajfour last updated on 01/Nov/17

Find v in terms of u,a, b, and l   at a moment when the collars  at distances a, and b from the  vertical rod of the stand.

$${Find}\:\boldsymbol{{v}}\:{in}\:{terms}\:{of}\:\boldsymbol{{u}},\boldsymbol{{a}},\:\boldsymbol{{b}},\:{and}\:\boldsymbol{{l}}\: \\ $$$${at}\:{a}\:{moment}\:{when}\:{the}\:{collars} \\ $$$${at}\:{distances}\:\boldsymbol{{a}},\:{and}\:\boldsymbol{{b}}\:{from}\:{the} \\ $$$${vertical}\:{rod}\:{of}\:{the}\:{stand}. \\ $$

Commented by mrW1 last updated on 01/Nov/17

let h=distance of the horizontal rods  L^2 =a^2 +b^2 +h^2   L, h = constant  differentiate w.r.t. time  0=2a(da/dt)+2b(db/dt)+0  0=2au+2b(−v)+0  ⇒v=(a/b)×u

$$\mathrm{let}\:\mathrm{h}=\mathrm{distance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{rods} \\ $$$$\mathrm{L}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{h}^{\mathrm{2}} \\ $$$$\mathrm{L},\:\mathrm{h}\:=\:\mathrm{constant} \\ $$$$\mathrm{differentiate}\:\mathrm{w}.\mathrm{r}.\mathrm{t}.\:\mathrm{time} \\ $$$$\mathrm{0}=\mathrm{2a}\frac{\mathrm{da}}{\mathrm{dt}}+\mathrm{2b}\frac{\mathrm{db}}{\mathrm{dt}}+\mathrm{0} \\ $$$$\mathrm{0}=\mathrm{2au}+\mathrm{2b}\left(−\mathrm{v}\right)+\mathrm{0} \\ $$$$\Rightarrow\mathrm{v}=\frac{\mathrm{a}}{\mathrm{b}}×\mathrm{u} \\ $$

Commented by ajfour last updated on 01/Nov/17

Excellent sir ! I like the property  too well.

$$\mathscr{E}{xcellent}\:{sir}\:!\:{I}\:{like}\:{the}\:{property} \\ $$$${too}\:{well}. \\ $$

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