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Question Number 107711 by ajfour last updated on 12/Aug/20

Commented by ajfour last updated on 12/Aug/20

If motion begins from rest when  blue block is at A and by the time it  reaches B string goes slack, then  find length AB.

$${If}\:{motion}\:{begins}\:{from}\:{rest}\:{when} \\ $$$${blue}\:{block}\:{is}\:{at}\:{A}\:{and}\:{by}\:{the}\:{time}\:{it} \\ $$$${reaches}\:{B}\:{string}\:{goes}\:{slack},\:{then} \\ $$$${find}\:{length}\:{AB}. \\ $$$$ \\ $$

Answered by mr W last updated on 12/Aug/20

Commented by mr W last updated on 12/Aug/20

T cos θ=ma_x   mg−T=ma=ma_x cos θ  ⇒((T cos θ)/(mg−T))=(1/(cos θ))  T(1+cos^2  θ)=mg  ⇒T=((mg)/(1+cos^2  θ))>0  the string can never become slack!

$${T}\:\mathrm{cos}\:\theta={ma}_{{x}} \\ $$$${mg}−{T}={ma}={ma}_{{x}} \mathrm{cos}\:\theta \\ $$$$\Rightarrow\frac{{T}\:\mathrm{cos}\:\theta}{{mg}−{T}}=\frac{\mathrm{1}}{\mathrm{cos}\:\theta} \\ $$$${T}\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\theta\right)={mg} \\ $$$$\Rightarrow{T}=\frac{{mg}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\theta}>\mathrm{0} \\ $$$${the}\:{string}\:{can}\:{never}\:{become}\:{slack}! \\ $$

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