Perplexing Rope

Fix a coordinate system with the origin at the point where the string meets the block. At some point on the string, let $\theta$ be the angle that the tangent to the string makes with the horizontal, and let $s$ the arc-length of the string from the origin. In other words, $(s,\theta)$ are the intrinsic coordinates of the string. Considering the small length of string between $\theta$ and $\theta+\delta\theta$ , and resolving forces horizontally and vertically, we have $(T + \delta T)\cos(\theta+\delta\theta) - T\cos\theta \; \approx \; 0 \hspace{2cm} (T + \delta T)\sin(\theta+\delta\theta) - T\sin\theta \; \approx \; \tfrac{mg}{\ell}\delta s$ Linearising, and letting $\delta\theta \to0$ , we deduce that $\frac{dT}{d\theta} \; = \; T\tan\theta \hspace{2cm} \frac{mg}{\ell}\,\frac{ds}{d\theta} \; = \; \frac{dT}{d\theta}\sin\theta + T\cos\theta$ and hence $T \; = \; A\sec\theta \hspace{1cm} s \; = \; c\tan\theta \hspace{1cm} \mathrm{where} \hspace{1cm} c \; = \; \frac{A\ell}{mg}$ But then $\sec\theta \; = \; \frac{ds}{dx} \; = \; c\sec^2\theta \frac{d\theta}{dx} \; \Rightarrow \; \frac{dx}{d\theta} \; = \; c\sec\theta \; \Rightarrow \; x \; = \; c\ln(\sec\theta + \tan\theta)$ and hence $\sec\theta = \cosh\big(\tfrac{x}{c}\big)$ and $\tan\theta = \sinh\big(\tfrac{x}{c}\big)$ , so that $s = c\sinh\big(\tfrac{x}{c}\big)$ . But then $c\frac{dy}{dx} = c\tan\theta = s = c\sinh\big(\tfrac{x}{c}\big)$ , so that $y \; = \; c\Big(\cosh\big(\tfrac{x}{c}\big) - 1 \Big)$ But this implies that $(y+c)^2 - s^2 = c^2$ , so that $s^2 - y^2 \; = \; 2cy$ and, in addition, $T = A\cosh\big(\tfrac{x}{c}\big) = A\big(1 + \tfrac{y}{c}\big)$ , so that $T \; = \; A + \frac{mgy}{\ell}$

We have $m=3$ , $h=2$ , $\ell=5$ and $g=10$ . Thus $c=\tfrac14(5^2-2^2) = \tfrac{21}{4}$ , so that $A = \tfrac{mgc}{\ell} = \tfrac{63}{2}$ . The maximum tension in the string is $T = A + \tfrac{mhg}{\ell} = \tfrac{87}{2}$ , making $\cos\alpha = \tfrac{A}{T} = \tfrac{21}{29}$ . The tension at the bottom of the string must equal the frictional force on the block, and so $\beta = A = \tfrac{63}{2}$ .

In the new configuration we must have $A = \mu Mg = 40$ , and so $c = \tfrac{A\ell}{mg} = \tfrac{20}{3}$ . Then $(h+c)^2 = c^2 + \ell^2$ , and hence $h = \tfrac53$ , and so $\gamma = \tfrac53$ .

Thus $5\cos\alpha + \beta + 4\gamma = \tfrac{7271}{174}$ . Taking the integer part of this, we obtain the answer $\boxed{41}$ .

$\textbf{Hint and Answers:-}$

$\text{Tension in the rope varies with height }y \text{ above lower end as } T(y)=T(0)+mgy/(\ell) \cdots \cdots (1)$ .

$\text{Using this with Free body diagram of rope and block,}$ $\cos \alpha =\dfrac{\ell^2-h^2}{\ell^2+h^2}$ $\text{friction}=T(0)=\dfrac{mg( \ell^2-h^2)}{2\ell h}$ $H=\gamma=\dfrac{- \mu M \ell + \sqrt{(\mu M)^2+m^2}}{m}$ $\text{Consider the free body diagram of a small element of the rope.}$

Free Body Diagram Resolving forces along (parallel to) the element, $T(y+dy)-T(y)= (dm)g \sin \theta = \lambda g dy \implies \dfrac{dT}{dy}= \lambda g$ $\displaystyle \int_{T(0)}^{T(y)} dT =\lambda g \displaystyle \int_{0}^{y} dy \implies (1).$ where $\lambda \equiv \dfrac{dm}{d\ell}$ is the linear mass density.

Now for the whole rope in stasis, $T \sin \alpha = mg, T \cos \alpha = T(0)$

This can be used to solve for $T(0) \text{ and } \cos \alpha$ .

$\text{For the third part, set } T(0)=\mu M g \text{ and express } T(0) \text{ as a function of height } H \text{ of the upper end using equation (1)}.$