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Question Number 23489 by Tinkutara last updated on 31/Oct/17
A rectangular wire frame ABCD is in vertical plane is moving with a constant acceleration a into the plane. Direction of gravity is shown in figure. A collar can move on wire AC of length l. Coefficient of friction between wire and collar is μ. Find (i) The minimum acceleration a so that collar does not slip on wire. (ii) The time taken by collar to reach C if acceleration is half the value calculated in part (i)
$$\mathrm{A}\:\mathrm{rectangular}\:\mathrm{wire}\:\mathrm{frame}\:{ABCD}\:\mathrm{is}\:\mathrm{in} \\ $$$$\mathrm{vertical}\:\mathrm{plane}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\mathrm{acceleration}\:{a}\:\mathrm{into}\:\mathrm{the}\:\mathrm{plane}.\:\mathrm{Direction} \\ $$$$\mathrm{of}\:\mathrm{gravity}\:\mathrm{is}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{figure}.\:\mathrm{A}\:\mathrm{collar} \\ $$$$\mathrm{can}\:\mathrm{move}\:\mathrm{on}\:\mathrm{wire}\:{AC}\:\mathrm{of}\:\mathrm{length}\:{l}. \\ $$$$\mathrm{Coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{wire} \\ $$$$\mathrm{and}\:\mathrm{collar}\:\mathrm{is}\:\mu.\:\mathrm{Find} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{The}\:\mathrm{minimum}\:\mathrm{acceleration}\:{a}\:\mathrm{so}\:\mathrm{that} \\ $$$$\mathrm{collar}\:\mathrm{does}\:\mathrm{not}\:\mathrm{slip}\:\mathrm{on}\:\mathrm{wire}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{The}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{collar}\:\mathrm{to}\:\mathrm{reach}\:{C} \\ $$$$\mathrm{if}\:\mathrm{acceleration}\:\mathrm{is}\:\mathrm{half}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{calculated}\:\mathrm{in}\:\mathrm{part}\:\left(\mathrm{i}\right) \\ $$
Commented by Tinkutara last updated on 31/Oct/17
Commented by mrW1 last updated on 31/Oct/17
(i) μm((√(a^2 +g^2 cos^2 θ)))≥mg sin θ a^2 ≥(((sin^2 θ)/μ^2 )−cos^2 θ)g^2 a≥(g/μ)(√(sin^2 θ−μ^2 cos^2 θ)) (ii) with a=(g/(2μ))(√(sin^2 θ−μ^2 cos^2 θ)) ma_s =mg sin θ−μmg((√(((sin^2 θ−μ^2 cos^2 θ)/(4μ^2 ))+cos^2 θ))) a_s =(g/2)(2sin θ−(√(sin^2 θ+3μ^2 cos^2 θ))) t=(√((2L)/a_s ))=2(√(L/((2sin θ−(√(sin^2 θ+3μ^2 cos^2 θ)))g)))
$$\left(\mathrm{i}\right) \\ $$$$\mu\mathrm{m}\left(\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{g}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}\right)\geqslant\mathrm{mg}\:\mathrm{sin}\:\theta \\ $$$$\mathrm{a}^{\mathrm{2}} \geqslant\left(\frac{\mathrm{sin}^{\mathrm{2}} \:\theta}{\mu^{\mathrm{2}} }−\mathrm{cos}^{\mathrm{2}} \:\theta\right)\mathrm{g}^{\mathrm{2}} \\ $$$$\mathrm{a}\geqslant\frac{\mathrm{g}}{\mu}\sqrt{\mathrm{sin}^{\mathrm{2}} \:\theta−\mu^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$\left(\mathrm{ii}\right) \\ $$$$\mathrm{with}\:\mathrm{a}=\frac{\mathrm{g}}{\mathrm{2}\mu}\sqrt{\mathrm{sin}^{\mathrm{2}} \:\theta−\mu^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$\mathrm{ma}_{\mathrm{s}} =\mathrm{mg}\:\mathrm{sin}\:\theta−\mu\mathrm{mg}\left(\sqrt{\frac{\mathrm{sin}^{\mathrm{2}} \:\theta−\mu^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{4}\mu^{\mathrm{2}} }+\mathrm{cos}^{\mathrm{2}} \:\theta}\right) \\ $$$$\mathrm{a}_{\mathrm{s}} =\frac{\mathrm{g}}{\mathrm{2}}\left(\mathrm{2sin}\:\theta−\sqrt{\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{3}\mu^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}\right) \\ $$$$\mathrm{t}=\sqrt{\frac{\mathrm{2L}}{\mathrm{a}_{\mathrm{s}} }}=\mathrm{2}\sqrt{\frac{\mathrm{L}}{\left(\mathrm{2sin}\:\theta−\sqrt{\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{3}\mu^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}\right)\mathrm{g}}} \\ $$
Answered by ajfour last updated on 31/Oct/17
Commented by ajfour last updated on 31/Oct/17
N_z =ma N_x =mgcos θ N=m(√(a^2 +g^2 cos^2 θ)) (i) f−mgsin θ=0 ⇒ μN ≥ mgsin θ or a^2 +g^2 cos^2 θ=((g^2 sin^2 θ)/μ^2 ) a ≥ (g/μ)(√(sin^2 θ−μ^2 cos^2 θ)) (ii) l=(1/2)A_y t^2 mgsin θ−μN=mA_y N=m(√(a^2 /4+g^2 cos^2 θ)) A_y =gsin θ−μ(√((g^2 /(4μ^2 ))(sin^2 θ−μ^2 cos^2 θ)+g^2 cos^2 θ)) t=(√((2l)/A_y )) =(√((2l)/(gsin θ−(g/2)(√(sin^2 θ+3μ^2 cos^2 θ))))) .
$${N}_{{z}} ={ma} \\ $$$${N}_{{x}} ={mg}\mathrm{cos}\:\theta \\ $$$$\:{N}={m}\sqrt{{a}^{\mathrm{2}} +{g}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta} \\ $$$$\left({i}\right)\:\:\:\:\:\:{f}−{mg}\mathrm{sin}\:\theta=\mathrm{0} \\ $$$$\:\:\:\:\Rightarrow\:\mu{N}\:\geqslant\:{mg}\mathrm{sin}\:\theta \\ $$$${or}\:\:\:{a}^{\mathrm{2}} +{g}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta=\frac{{g}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mu^{\mathrm{2}} } \\ $$$$\:\:\:\:\:{a}\:\geqslant\:\frac{{g}}{\mu}\sqrt{\mathrm{sin}\:^{\mathrm{2}} \theta−\mu^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}\: \\ $$$$\left({ii}\right)\:\:{l}=\frac{\mathrm{1}}{\mathrm{2}}{A}_{{y}} {t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:{mg}\mathrm{sin}\:\theta−\mu{N}={mA}_{{y}} \\ $$$$\:\:\:{N}={m}\sqrt{{a}^{\mathrm{2}} /\mathrm{4}+{g}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}\: \\ $$$${A}_{{y}} ={g}\mathrm{sin}\:\theta−\mu\sqrt{\frac{{g}^{\mathrm{2}} }{\mathrm{4}\mu^{\mathrm{2}} }\left({s}\mathrm{in}\:^{\mathrm{2}} \theta−\mu^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta\right)+{g}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta} \\ $$$${t}=\sqrt{\frac{\mathrm{2}{l}}{{A}_{{y}} }} \\ $$$$\:=\sqrt{\frac{\mathrm{2}{l}}{{g}\mathrm{sin}\:\theta−\frac{{g}}{\mathrm{2}}\sqrt{\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{3}\mu^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}}}\:. \\ $$$$ \\ $$
Commented by Tinkutara last updated on 01/Nov/17
I got that μN≥mgsinθ. But only doubt is that N is of the form i^∧ +k^∧ . Then why it should be equated with mgsinθy^∧ ? Also, how you write 3^(rd) line of (ii) part?
$${I}\:{got}\:{that}\:\mu{N}\geqslant{mg}\mathrm{sin}\theta.\:{But}\:{only}\:{doubt} \\ $$$${is}\:{that}\:{N}\:{is}\:{of}\:{the}\:{form}\:\overset{\wedge} {{i}}+\overset{\wedge} {{k}}.\:{Then} \\ $$$${why}\:{it}\:{should}\:{be}\:{equated}\:{with}\:{mg}\mathrm{sin}\theta\overset{\wedge} {{y}}? \\ $$$${Also},\:{how}\:{you}\:{write}\:\mathrm{3}^{{rd}} \:{line}\:{of}\:\left({ii}\right)\:{part}? \\ $$

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