Problems

A uniform ladder of length \(l\) and mass \(m\) rests with one end in contact with a smooth ramp inclined at an angle of \(\pi/6\) to the vertical. The foot of the ladder rests, on horizontal ground, at a distance \(l/\sqrt{3}\) from the foot of the ramp, and the coefficient of friction between the ladder and the ground is \(\mu.\) The ladder is inclined at an angle \(\pi/6\) to the horizontal, in the vertical plane containing a line of greatest slope of the ramp. A labourer of mass \(m\) intends to climb slowly to the top of the ladder.

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1. Find the value of \(\mu\) if the ladder slips as soon as the labourer reaches the midpoint.
2. Find the minimum value of \(\mu\) which will ensure that the labourer can reach the top of the ladder.

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Solution:

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1. \begin{align*} \text{N2}(\uparrow): && R_1 + R_2\sin(\frac{\pi}{6})-2mg &= 0 \\ \text{N2}(\rightarrow): && R_2 \cos (\frac{\pi}{6})-F_r &= 0 \\ \overset{\curvearrowleft}{X}: && lmg \cos \tfrac{\pi}{6} - l R_2 \cos \tfrac{\pi}{6} &= 0 \\ \\ \Rightarrow && R_2 &= mg \\ \Rightarrow && R_1 &= 2mg - \frac12mg \\ &&&=\frac32mg \\ \Rightarrow && \frac{\sqrt{3}}2mg - \mu\frac32mg &= 0 \\ \Rightarrow && \mu &= \frac{1}{\sqrt{3}} \end{align*}
2. \begin{align*} \text{N2}(\uparrow): && R_1 + R_2\sin(\frac{\pi}{6})-2mg &= 0 \\ \overset{\curvearrowleft}{X}: && \frac12 lmg \cos \tfrac{\pi}{6}+xmg \cos \tfrac{\pi}{6} - l R_2 \cos \tfrac{\pi}{6} &= 0 \\ \\ \Rightarrow && R_2 &= mg(\frac{1}2+\frac{x}{l}) \\ \Rightarrow && R_1 &= 2mg - \frac12mg(\frac{1}2+\frac{x}{l}) \\ &&&=(\frac74 - \frac{x}{2l})mg \\ &&&\geq \frac{5}{4}mg\\ \text{N2}(\rightarrow): && R_2 \cos (\frac{\pi}{6})-\mu R_1& \leq 0 \\ \Rightarrow && \frac{\sqrt{3}}2mg - \mu\frac54mg &\leq 0 \\ \Rightarrow && \mu &\geq \frac{2\sqrt{3}}{5} \end{align*}

A smooth billiard ball moving on a smooth horizontal table strikes another identical ball which is at rest. The coefficient of restitution between the balls is \(e(<1)\). Show that after the collision the angle between the velocities of the balls is less than \(\frac{1}{2}\pi.\) Show also that the maximum angle of deflection of the first ball is \[ \sin^{-1}\left(\frac{1+e}{3-e}\right). \]

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Solution:

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Set up the coordinate frame so that the \(x\)-direction is the line of centres of the spheres. Then if the initial velocities are \(\displaystyle \binom{u_x}{u_y}\) and \(\displaystyle \binom{0}{0}\). Then the final velocities must be: \(\displaystyle \binom{v_{x1}}{u_y}\) and \(\displaystyle \binom{v_{x2}}{0}\) where \(mu_x = mv_{x1}+mv_{x2}\) by conservation of energy and \(\frac{v_{x1}-v_{x2}}{u_x} = -e\). \begin{align*} && \begin{cases} v_{x1}+v_{x2} &= u_x \\ v_{x1}-v_{x2} &= -eu_x \\ \end{cases} \\ \Rightarrow && 2v_{x1} &= (1-e)u_x \\ \Rightarrow && v_{x1} &= \frac{(1-e)}{2} u_x \\ && v_{x2} &= \frac{1+e}{2} u_x \end{align*} Notice that since \(0 < e < 1\) we must have \(v_{x1} > 0\) and so the ball on the left is still continuing in the positive direction, therefore the angle will be less than \(\frac12 \pi\). The angle the first ball is deflected through is the angle between: \(\displaystyle \binom{u_x}{u_y}\) and \(\displaystyle \binom{\frac{1-e}{2}u_x}{u_y}\). We can scale the velocities so \(u_y = 1\). So we are interested in the angle between \(\displaystyle \binom{x}{1}\) and \(\displaystyle \binom{\frac{1-e}{2}x}{1}\). To maximise \(\theta\) we can maximise \(\tan \theta\), so: \begin{align*} && \tan \theta &= \frac{\frac{2}{(1-e)x-\frac{1}{x}}}{1+\frac{2}{(1-e)x^2}} \\ &&&= \frac{2x-(1-e)x}{(1-e)x^2+2} \\ &&&= \frac{(1+e)x}{(1-e)x^2+2} \\ \\ \frac{\d}{\d t}: &&&= \frac{(1+e)((1-e)x^2+2)-2(1+e)(1-e)x^2}{\sim} \\ &&&= \frac{2(1+e)-(1+e)(1-e)x^2}{\sim}\\ \frac{\d}{\d t} = 0: &&0 &= 2(1+e)-(1+e)(1-e)x^2 \\ \Rightarrow && x &= \pm \sqrt{\frac{2}{1-e}} \\ \\ \Rightarrow && \tan \theta &= \frac{\pm(1+e)\sqrt{\frac{2}{1-e}}}{2+2} \\ &&&= \pm \frac{\sqrt{2}(1+e)}{4\sqrt{1-e}} \\ \Rightarrow && \cot^2 \theta &= \frac{8(1-e)}{(1+e)^2} \\ \Rightarrow && \cosec^2 \theta &= \frac{8(1-e)}{(1+e)^2} + 1 \\ &&&= \frac{8-8e+1+2e+e^2}{(1+e)^2} \\ &&&= \frac{9-6e+e^2}{(1+e)^2} \\ &&&= \frac{(3-e)^2}{(1+e)^2} \\ \Rightarrow && \theta &= \sin^{-1} \left ( \frac{1+e}{3-e}\right) \end{align*}

A rubber band band of length \(2\pi\) and modulus of elasticity \(\lambda\) encircles a smooth cylinder of unit radius, whose axis is horizontal. A particle of mass \(m\) is attached to the lowest point of the band, and hangs in equilibrium at a distance \(x\) below the axis of the cylinder. Obtain an expression in terms of \(x\) for the stretched length of the band in equilibrium. What is the value of \(\lambda\) if \(x=2\)?

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Solution:

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If \(\alpha\) is as labelled then \(\cos \alpha = \frac{1}{x}, \sin \alpha = \frac{\sqrt{x^2-1}}{x}, \tan \alpha = \sqrt{x^2-1}\). We also have the full length of the rubber band is \(2\pi - 2\alpha +2\tan \alpha\) so the extension is \(2 \l \sqrt{x^2-1} - \cos^{-1} \l \frac{1}{x}\r \r\) Therefore \(T = \frac{\l \sqrt{x^2-1} - \cos^{-1} \l \frac{1}{x}\r \r\lambda}{\pi}\). If \(x = 2\), \(T = \frac{\sqrt{3} - \frac{\pi}{3}}{\pi} \lambda, \sin \alpha = \frac{\sqrt{3}}{2}\) \begin{align*} \text{N2}(\uparrow): && 2T\sin \alpha - mg &= 0 \\ \Rightarrow && \frac{\sqrt{3} - \frac{\pi}{3}}{\pi} \lambda \sqrt{3} &= mg \\ \Rightarrow && \lambda &= \frac{\sqrt{3}\pi}{(3\sqrt{3}-\pi)}mg \end{align*}

A train of length \(l_{1}\) and a lorry of length \(l_{2}\) are heading for a level crossing at speeds \(u_{1}\) and \(u_{2}\) respectively. Initially the front of the train and the front of the lorry are at distances \(d_{1}\) and \(d_{2}\) from the crossing. Find conditions on \(u_{1}\) and \(u_{2}\) under which a collision will occur. On a diagram with \(u_{1}\) and \(u_{2}\) measured along the \(x\) and \(y\) axes respectively, shade in the region which represents collision. Hence show that if \(u_{1}\) and \(u_{2}\) are two independent random variables, both uniformly distributed on \((0,V)\), then the probability of a collision in the case when initially the back of the train is nearer to the crossing than the front of the lorry is \[ \frac{l_{1}l_{2}+l_{2}d_{1}+l_{1}d_{2}}{2d_{2}\left(l_{2}+d_{2}\right)}. \] Find the probability of a collision in each of the other two possible cases.

Show that, if the lengths of the diagonals of a parallelogram are specified, then the parallogram has maximum area when the diagonals are perpendicular. Show also that the area of a parallelogram is less than or equal to half the square of the length of its longer diagonal. The set \(A\) of points \((x,y)\) is given by \begin{alignat*}{1} \left|a_{1}x+b_{1}y-c_{1}\right| & \leqslant\delta,\\ \left|a_{2}x+b_{2}y-c_{2}\right| & \leqslant\delta, \end{alignat*} with \(a_{1}b_{2}\neq a_{2}b_{1}.\) Sketch this set and show that it is possible to find \((x_{1},y_{1}),(x_{2},y_{2})\in A\) with \[ (x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}\geqslant\frac{8\delta^{2}}{\left|a_{1}b_{2}-a_{2}b_{1}\right|}. \]

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Solution: In a parallelogram the diagonals meet at their mid points. Fixing one diagonal, we can look at the two triangles formed by the other diagonal. Suppose the angle between them is \(\theta\). Then the area of the triangles will be \(\frac12 \frac{l_1}{2} \frac{l_2}2 \sin \theta+\frac12 \frac{l_1}{2} \frac{l_2}2 \sin (\pi -\theta) = \frac{l_1l_2}{4} \sin \theta\). This will be true on both sides. Therefore we can maximise this area by setting \(\theta = \frac{\pi}{2}\).

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Consider the (darker) shaded area. This is our set \(A\). The area of the set is indifferent to a parallel shift in the lines, so without loss of generality, we can consider \(c_1 = 0, c_2 = 0\), so our lines meet at the origin. Now also consider the linear transformation \(\begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix}^{-1}\) which takes the coordinate axes to these lines. This will take the square \([-\delta, \delta] \times [-\delta, \delta]\) which has area \(4\delta^2\) to our square, which will have area \(\frac{4 \delta^2}{ |a_1b_2 - a_2b_1|}\). If we consider the length of the two diagonals of this area, \(l_1, l_2\) we know that \(\frac{l_1l_2}2 \sin \theta = \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}\), if we consider the larger of \(l_1\) and \(l_2\) (wlog \(l_1\)) we must have that \(\frac{l_1^2}{2} \geq \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}\) and so points on opposite ends of the diagonal will satisfy the inequality in the question.

A woman stands in a field at a distance of \(a\,\mathrm{m}\) from the straight bank of a river which flows with negligible speed. She sees her frightened child clinging to a tree stump standing in the river \(b\,\mathrm{m}\) downstream from where she stands and \(c\,\mathrm{m}\) from the bank. She runs at a speed of \(u\,\mathrm{ms}^{-1}\) and swims at \(v\,\mathrm{ms}^{-1}\) in straight lines. Find an equation to be satisfied by \(x,\) where \(x\,\mathrm{m}\) is the distance upstream from the stump at which she should enter the river if she is to reach the child in the shortest possible time. Suppose now that the river flows with speed \(v\) ms\(^{-1}\) and the stump remains fixed. Show that, in this case, \(x\) must satisfy the equation \[ 2vx^{2}(b-x)=u(x^{2}-c^{2})[a^{2}+(b-x)^{2}]^{\frac{1}{2}}. \] For this second case, draw sketches of the woman's path for the three possibilities \(b>c,\) \(b=c\) and \(b< c\).

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Solution:

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The distance to where she enters the water is \(\sqrt{a^2+(b-x)^2}\) and the distance through the water is \(\sqrt{x^2+c^2}\). The total time will be \(\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{\sqrt{x^2+c^2}}{v}\). To minimise this, we can differentiate. \begin{align*} \frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} + \frac{x}{v \sqrt{x^2+c^2}} &= 0 \\ \Rightarrow && v(b-x)(x^2+c^2)^{\frac12} &= xu(a^2+(b-x)^2)^{\frac12} \end{align*} When she is in the water, she can will move with velocity \(\begin{pmatrix} v \cos \theta \\ v \sin \theta -v \end{pmatrix}\). She needs to travel a distance \(\begin{pmatrix} c \\ -x \end{pmatrix}\), so we must have that \begin{align*} && \frac{x}{c} &= \frac{1-\sin \theta}{\cos \theta} \\ \Rightarrow && \sec \theta - \tan \theta &= \frac{x}{c} \\ \Rightarrow && \sec \theta &= \tan \theta + \frac{x}{c} \\ \Rightarrow && \sec^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\ \Rightarrow && 1 + \tan^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\ \Rightarrow && \tan \theta &=\frac{c^2-x^2}{2xc} \\ \Rightarrow && \sin \theta &= \frac{c^2-x^2}{c^2+x^2} \\ && \cos \theta &= \frac{2xc}{c^2+x^2} \\ \end{align*} (where we have taken the positive value for \(\cos \theta\) since we must be heading towards the child). Since \(v \cos \theta t = c\) the time taken to reach the child in the water is \(\frac{c}{v} \frac{c^2+x^2}{2xc} = \frac{c^2+x^2}{2xv}\). So the total time is: \(\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{c^2+x^2}{2xv}\). To minimise this, we can differentiate. \begin{align*} \frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} -\frac{c^2}{2vx^2} + \frac{x^2}{2vx^2}&= 0 \\ \Rightarrow && u(x^2-c^2)\sqrt{a^2+(b-x)^2}&= 2vx^2(b-x) \end{align*} as required. When \(b = c\), the shortest path will be running directly to the bank (there's no quicker way to get to the bank) then swimming directly out (and letting the current take you downstream exactly as far as you need)). Therefore the path will be:

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If \(b > c\) then she should run a little downstream first.

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and if \(c > b\) she should actually run a little upstream to take advantage of the current:

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A patient arrives with blue thumbs at the doctor's surgery. With probability \(p\) the patient is suffering from Fenland fever and requires treatment costing \(\pounds 100.\) With probability \(1-p\) he is suffering from Steppe syndrome and will get better anyway. A test exists which infallibly gives positive results if the patient is suffering from Fenland fever but also has probability \(q\) of giving positive results if the patient is not. The test cost \(\pounds 10.\) The doctor decides to proceed as follows. She will give the test repeatedly until either the last test is negative, in which case she dismisses the patient with kind words, or she has given the test \(n\) times with positive results each time, in which case she gives the treatment. In the case \(n=0,\) she treats the patient at once. She wishes to minimise the expected cost \(\pounds E_{n}\) to the National Health Service.

1. Show that \[ E_{n+1}-E_{n}=10p-10(1-p)q^{n}(9-10q), \] and deduce that if \(p=10^{-4},q=10^{-2},\) she should choose \(n=3.\)
2. Show that if \(q\) is larger than some fixed value \(q_{0},\) to be determined explicitly, then whatever the value of \(p,\) she should choose \(n=0.\)

1. \(X_{1},X_{2},\ldots,X_{n}\) are independent identically distributed random variables drawn from a uniform distribution on \([0,1].\) The random variables \(A\) and \(B\) are defined by \[ A=\min(X_{1},\ldots,X_{n}),\qquad B=\max(X_{1},\ldots,X_{n}). \] For any fixed \(k\), such that \(0< k< \frac{1}{2},\) let \[ p_{n}=\mathrm{P}(A\leqslant k\mbox{ and }B\geqslant1-k). \] What happens to \(p_{n}\) as \(n\rightarrow\infty\)? Comment briefly on this result.
2. Lord Copper, the celebrated and imperious newspaper proprietor, has decided to run a lottery in which each of the \(4,000,000\) readers of his newspaper will have an equal probability \(p\) of winning \(\pounds 1,000,000\) and their changes of winning will be independent. He has fixed all the details leaving to you, his subordinate, only the task of choosing \(p\). If nobody wins \(\pounds 1,000,000\), you will be sacked, and if more than two readers win \(\pounds 1,000,000,\) you will also be sacked. Explaining your reasoning, show that however you choose \(p,\) you will have less than a 60\% change of keeping your job.