The time to reach the top edge of the river is\[ t=\frac{37}{8}=4.625\text{ sec}\] The distance travelled in horizontal direction is therefore\[ x=\left ( 7\right ) \left ( 4.625\right ) =32.375\text{ ft}\]
Length of rope \(L\) is \[ L=2x_{B}+x_{A}\] Where \(x_{B}\) is distance from top to \(B\) and \(x_{A}\) is distance from top to \(A\). Taking derivatives gives\begin{align*} 0 & =2v_{B}+v_{A}\\ v_{B} & =-\frac{v_{A}}{2}=-\frac{-1.5}{2}=0.75\text{ m/s} \end{align*}
\begin{align*} \frac{d\theta }{dt} & =\left ( 1950\right ) \left ( \frac{2\pi }{\text{rotation}}\right ) \left ( \frac{\text{minute}}{60}\right ) \\ & =\left ( 1950\right ) \frac{2\pi }{60}\\ & =204.2035\text{ rad/sec} \end{align*}
But \begin{align*} L^{2} & =R^{2}+y_{c}^{2}-2\left ( R\right ) \left ( y_{c}\right ) \cos \left ( \theta \right ) \\ y_{c}^{2}-2Ry_{c}\cos \left ( \theta \right ) +\left ( R^{2}-L^{2}\right ) & =0 \end{align*}
Or\begin{align*} y_{c} & =\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}\\ & =R\cos \theta \pm \frac{1}{2}\sqrt{4R^{2}\cos ^{2}\theta -4\left ( R^{2}-L^{2}\right ) }\\ & =R\cos \theta \pm \frac{1}{2}\sqrt{4R^{2}\cos ^{2}\theta -4R^{2}+4L^{2}}\\ & =R\cos \theta \pm \frac{1}{2}\sqrt{4R^{2}\left ( \cos ^{2}\theta -1\right ) +4L^{2}}\\ & =R\cos \theta \pm \frac{1}{2}\sqrt{-4R^{2}\sin ^{2}\theta +4L^{2}}\\ & =R\cos \theta \pm \sqrt{L^{2}-R^{2}\sin ^{2}\theta } \end{align*}
At \(\theta =0\),\(\ y_{c}=R+L\), therefore we pick the plus sign\[ y_{c}=R\cos \theta +\sqrt{L^{2}-R^{2}\sin ^{2}\theta }\] Taking derivative with time\begin{align*} \dot{y}_{c} & =-R\dot{\theta }\sin \theta +\frac{1}{2}\frac{1}{\sqrt{L^{2}-R^{2}\sin ^{2}\left ( \theta \right ) }}\left ( -2R^{2}\sin \theta \left ( \dot{\theta }\cos \theta \right ) \right ) \\ & =-R\dot{\theta }\sin \theta -\frac{R^{2}\dot{\theta }\sin \theta \cos \theta }{\sqrt{L^{2}-R^{2}\sin ^{2}\left ( \theta \right ) }} \end{align*}
Plugging in values \(R=\frac{3.4}{12}\) ft, \(L=\frac{5.7}{12}\) ft and \(\theta =40^{0}\) and \(\dot{\theta }=204.2035\) gives\begin{align*} \dot{y}_{c} & =-\left ( \frac{3.4}{12}\right ) \left ( 204.2035\right ) \sin \left ( \frac{40}{180}\pi \right ) -\frac{\left ( \frac{3.4}{12}\right ) ^{2}\left ( 204.2035\right ) \sin \left ( \frac{40}{180}\pi \right ) \cos \left ( \frac{40}{180}\pi \right ) }{\sqrt{\left ( \frac{5.7}{12}\right ) ^{2}-\left ( \frac{3.4}{12}\right ) ^{2}\left ( \sin \left ( \frac{40}{180}\pi \right ) \right ) ^{2}}}\\ & =-55.59\text{ ft/sec} \end{align*}
Resolving forces in vertical direction\begin{equation} mg-2T=m\ddot{y} \tag{1} \end{equation} To find \(y\,\), since rope length is \[ L=2y+\sqrt{x_{A}^{2}+\left ( l-h\right ) ^{2}}\] Taking derivative gives\begin{align*} 0 & =2\dot{y}+\frac{x_{A}\dot{x}_{A}}{\sqrt{x_{A}^{2}+\left ( l-h\right ) ^{2}}}\\ \dot{y} & =\frac{-x_{A}\dot{x}_{A}}{\sqrt{x_{A}^{2}+\left ( l-h\right ) ^{2}}} \end{align*}
Taking another derivative\[ \ddot{y}=\frac{-\dot{x}_{A}^{2}}{2\sqrt{x_{A}^{2}+\left ( l-h\right ) ^{2}}}+\frac{x_{A}^{2}\dot{x}_{A}^{2}}{2\left ( x_{A}^{2}+\left ( l-h\right ) ^{2}\right ) ^{\frac{3}{2}}}\] But \(\dot{x}_{A}=v_{0}=3.2\) ft/sec. Hence\[ \ddot{y}=\frac{-v_{0}^{2}}{2\sqrt{x_{A}^{2}+\left ( l-h\right ) ^{2}}}+\frac{x_{A}^{2}v_{0}^{2}}{2\left ( x_{A}^{2}+\left ( l-h\right ) ^{2}\right ) ^{\frac{3}{2}}}\] When \(l=14.5,h=6.4,x_{A}=9.5\) then \begin{align*} \ddot{y} & =\frac{-\left ( 3.2\right ) ^{2}}{2\sqrt{9.5^{2}+\left ( 14.5-6.4\right ) ^{2}}}+\frac{\left ( 9.5\right ) ^{2}\left ( 3.2\right ) ^{2}}{2\left ( 9.5^{2}+\left ( 14.5-6.4\right ) ^{2}\right ) ^{\frac{3}{2}}}\\ & =-0.172\,64\text{ ft/sec}^{2} \end{align*}
From (1) we solve for tension\begin{align*} T & =\frac{mg-m\ddot{y}}{2}\\ & =\frac{m\left ( g-\ddot{y}\right ) }{2}\\ & =\frac{550\left ( 32.2-\left ( -0.172\,64\right ) \right ) }{2}\\ & =8902.476\text{ lb force} \end{align*}
Hence \begin{align*} T & =\frac{8902.476}{32.2}\\ & =276.474\text{ lb} \end{align*}
Resolving forces in the \(x\) direction gives equation of motion\begin{align*} m\ddot{x}+kx & =0\\ \ddot{x} & =-\frac{k}{m}x \end{align*}
Let \(\frac{d^{2}x}{dt^{2}}=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}\) and the above becomes\[ v\frac{dv}{dx}=-\frac{k}{m}x \] This is now separable\[ \int _{v_{i}}^{v_{stop}}vdv=-\int _{0}^{x_{i}}\frac{k}{m}xdx \] But \(v_{stop}=0\) and the above becomes\[ v_{i}^{2}=\frac{k}{m}x_{i}^{2}\] For \(v_{i}=4.3\) mph and \(x_{i}=0.21\) ft, we solve for \(k\) from the above\begin{align*} k & =\frac{mv_{i}^{2}}{x_{i}^{2}}=\frac{\left ( \frac{3340}{32.2}\right ) \left ( \left ( 4.3\right ) \left ( \frac{5280}{3600}\right ) \right ) ^{2}}{\left ( 0.21\right ) ^{2}}\\ & =93551.7\text{ lb/ft}\\ & =9.355\times 10^{-4}\text{ lb/ft} \end{align*}
Question, why had to divide by \(g\) in above to get correct answer? Problem said \(lb\) in statement?
Resolving forces in the \(x\) direction gives equation of motion\begin{align*} m\ddot{x}+\beta x^{3} & =0\\ \ddot{x} & =-\frac{\beta x^{3}}{m} \end{align*}
Let \(\frac{d^{2}x}{dt^{2}}=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}\) and the above becomes\[ v\frac{dv}{dx}=-\frac{\beta x^{3}}{m}\] This is now separable\[ \int _{v_{i}}^{v_{stop}}vdv=-\frac{\beta }{m}\int _{0}^{x_{stop}}x^{3}dx \] But \(v_{stop}=0\) and the above becomes\[ v_{i}^{2}=\frac{1}{2}\frac{\beta }{m}x_{stop}^{4}\] For \(v_{i}=3\) km/h and \(\beta =650\times 10^{6}\) and \(m=75000\) kg, we solve for \(x_{stop}\) from the above\begin{align*} x_{stop} & =\left ( \frac{2mv^{2}}{\beta }\right ) ^{\frac{1}{4}}\\ & =\left ( \frac{2\left ( 75000\right ) \left ( 3\left ( \frac{1000}{3600}\right ) \right ) }{650\times 10^{6}}\right ) ^{\frac{1}{4}}\\ & =0.11776\text{ m} \end{align*}