Problems

A step-ladder has two sections \(AB\) and \(AC,\) each of length \(4a,\) smoothly hinged at \(A\) and connected by a light elastic rope \(DE,\) of natural length \(a/4\) and modulus \(W\), where \(D\) is on \(AB,\) \(E\) is on \(AC\) and \(AD=AE=a.\) The section \(AB,\) which contains the steps, is uniform and of weight \(W\) and the weight of \(AC\) is negligible. The step-ladder rests on a smooth horizontal floor and a man of weight \(4W\) carefully ascends it to stand on a rung distant \(\beta a\) from the end of the ladder resting on the floor. Find the height above the floor of the rung on which the man is standing when \(\beta\) is the maximum value at which equilibrium is possible.

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Solution:

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\begin{align*} N2(\uparrow): && 0 &= R_B+R_C - 5W \\ \Rightarrow && 5W &= R_B + R_C \\ \\ \overset{\curvearrowright}{A}: && 0 &= (R_B - R_C) \cdot 4a \cdot \cos \theta -W \cdot 2a \cdot \cos \theta - 4W \cdot (4 - \beta)a \cdot \cos \theta \\ \Rightarrow && R_B-R_C &= W \left ( \frac12 + (4-\beta)\right) \\ \Rightarrow && R_B &= \frac{W}2 \left ( 5+\frac12+(4-\beta)\right) = \frac{W}{2}\left(\frac{19}{2} - \beta\right) \\ && R_C &= \frac{W}{2} \left (5 - \frac12 - 4 +\beta \right) = \frac{W}{2} \left (\frac12 + \beta \right) \\ \\ \overset{\curvearrowright}{(A, AC)}: && 0 &= T \cdot a \cdot \sin \theta - R_C \cdot 4a \cdot \cos \theta \\ \Rightarrow && T &=4 \cot \theta \frac{W}{2} \left ( \frac12 + \beta\right) \\ &&&= 20W \cot \theta \\ \text{Hooke's Law}:&& T &= \frac{W(2a \cos \theta - \frac{a}{4})}{\frac{a}{4}} = W(8 \cos \theta - 1) \\ \Rightarrow && 8 \cos \theta -1 &= \cot \theta (2\beta+1)\\ \Rightarrow && 1+2\beta &=8\sin \theta-\tan \theta \\ \Rightarrow && \beta &= 4 \sin \theta - \frac12 \tan \theta - \frac12 \\ \Rightarrow && \frac{\d \beta}{\d \theta} &= 4 \cos \theta - \frac12 \sec^2 \theta \\ &&&= \frac{8\cos^3 \theta - 1}{\cos^2 \theta} \\ \Rightarrow && \cos \theta &= \frac12 \\ \Rightarrow && h &= \beta a \sin \theta \\ &&&= \left (4 \frac{\sqrt{3}}{2}-\frac12 \sqrt{3}-\frac12 \right) a \frac{\sqrt3}{2} \\ &&&= \left ( \frac{9-\sqrt{3}}{4}\right)a \end{align*}

In certain forms of Tennis two players \(A\) and \(B\) serve alternate games. Player \(A\) has probability \(p\low_{A}\) of winning a game in which she serves and \(p\low_{B}\) of winning a game in which player \(B\) serves. Player \(B\) has probability \(q\low_{B}=1-p\low_{B}\) of winning a game in which she serves and probability \(q\low_{A}=1-p\low_{A}\) of winning a game in which player \(A\) serves. In Shortened Tennis the first player to lead by 2 games wins the match. Find the probability \(P_{\text{short}}\) that \(A\) wins a Shortened Tennis match in which she serves first and show that it is the same as if \(B\) serves first. In Standard Tennis the first player to lead by 2 or more games after 4 or more games have been played wins the match. Show that the probability that the match is decided in 4 games is \[ p^{2}_Ap_{B}^{2}+q_{A}^{2}q_{B}^{2}+2(p\low_{A}p\low_{B}+q\low_{A}q\low_{B})(p\low_{A}q\low_{B}+q\low_{A}p\low_{B}). \] If \(p\low_{A}=p\low_{B}=p\) and \(q\low_{A}=q\low_{B}=q,\) find the probability \(P_{\text{stan}}\) that \(A\) wins a Standard Tennis match in which she serves first. Show that \[ P_{\text{stan}}-P_{\text{short}}=\frac{p^{2}q^{2}(p-q)}{p^{2}+q^{2}}. \]

During his performance a trapeze artist is supported by two identical ropes, either of which can bear his weight. Each rope is such that the time, in hours of performance, before it fails is exponentially distributed, independently of the other, with probability density function \(\lambda\exp(-\lambda t)\) for \(t\geqslant0\) (and 0 for \(t<0\)), for some \(\lambda>0.\) A particular rope has already been in use for \(t_{0}\) hours of performance. Find the distribution for the length of time the artist can continue to use it before it fails. Interpret and comment upon your result. Before going on tour the artist insists that the management purchase two new ropes of the above type. Show that the probability density function of the time until both ropes fail is \[ \mathrm{f}(t)=\begin{cases} 2\lambda\mathrm{e}^{-\lambda t}(1-\mathrm{e}^{-\lambda t}) & \text{ if }t\geqslant0,\\ 0 & \text{ otherwise.} \end{cases} \] If each performance lasts for \(h\) hours, find the probability that both ropes fail during the \(n\)th performance. Show that the probability that both ropes fail during the same performance is \(\tanh(\lambda h/2)\).

Three points, \(P,Q\) and \(R\), are independently randomly chosen on the perimeter of a circle. Prove that the probability that at least one of the angles of the triangle \(PQR\) will exceed \(k\pi\) is \(3(1-k)^{2}\) if \(\frac{1}{2}\leqslant k\leqslant1.\) Find the probability if \(\frac{1}{3}\leqslant k\leqslant\frac{1}{2}.\)