Problems

A uniform rectangular lamina of sides \(2a\) and \(2b\) rests in a vertical plane. It is supported in equilibrium by two smooth pegs fixed in the same horizontal plane, a distance \(d\) apart, so that one corner of the lamina is below the level of the pegs. Show that if the distance between this (lowest) corner and the peg upon which the side of length \(2a\) rests is less than \(a\), then the distance between this corner and the other peg is less than \(b\). Show also that \[ b\cos\theta-a\sin\theta=d\cos2\theta, \] where \(\theta\) is the acute angle which the sides of length \(2b\) make with the horizontal.

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Solution:

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We must have \(G\) between the two pegs (vertically), otherwise we will induce a moment. Considering moments about the peg, if the second peg is outside the centre then we must induce a moment and therefore we cannot be in equilibrium. \begin{align*} \text{N2}(\nearrow):&& 0 &= R_1-mg\sin\theta \\ \text{N2}(\nwarrow):&&0&= R_2-mg \cos \theta \\ \Rightarrow && R_1 &= mg \sin\theta \\ && R_2 &= mg \cos\theta \\ \\ \overset{\curvearrowleft}{G}: && 0 &= R_1(a-d\sin\theta) -R_2(b-d \cos\theta) \\ \Rightarrow && 0&= a \sin\theta -d \sin^2\theta - b\cos \theta+d \cos^2 \theta \\ \Rightarrow && b \cos \theta - a \sin \theta &= d \cos 2 \theta \end{align*}

The points \(A,B,C,D\) and \(E\) lie on a thin smooth horizontal table and are equally spaced on a circle with centre \(O\) and radius \(a\). At each of these points there is a small smooth hole in the table. Five elastic strings are threaded through the holes, one end of each beging attached at \(O\) under the table and the other end of each being attached to a particle \(P\) of mass \(m\) on top of the table. Each of the string has natural length \(a\) and modulus of elasticity \(\lambda.\) If \(P\) is displaced from \(O\) to any point \(F\) on the table and released from rest, show that \(P\) moves with simple harmonic motion of period \(T\), where \[ T=2\pi\sqrt{\frac{am}{5\lambda}}. \] The string \(PAO\) is replaced by one of natural length \(a\) and modulus \(k\lambda.\) \(P\) is displaced along \(OA\) from its equilibrium position and released. Show that \(P\) still moves in a straight line with simple harmonic motion, and, given that the period is \(T/2,\) find \(k\).

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Solution:

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The extension of \(OAP\) is \(|AP|\) and so the tension \(T_a = \frac{\lambda}{a} |AP|\). To simplify calculations, let \(A = a, B = a \omega, C = a \omega^2, \cdots\) where \(\omega = e^{2 \pi i/5}\) and let \(P = z\). then we can calculate the force as: \begin{align*} &&\sum_{p}T_p \mathbf{n}_{z \to p} &= \sum_{p} \frac{\lambda}{a} |z-p| \frac{p-z}{|p-z|} \\ &&&= \frac{\lambda}{a} \sum_{p} ( p - z) \\ &&&= -\frac{5\lambda}{a}z \end{align*} Therefore the force has magnitude \(\frac{5 \lambda}{a} |OP|\) directly towards the origin. Therefore if we set up our coordinate axis such that \(OP\) is the \(x\) axis, the particle will remain on the \(x\) axis and will move under the equation: \[ m \ddot{x} + \frac{5 \lambda}{a} x = 0 \] But then we can say that \(P\) moves under SHM with period \(\displaystyle 2 \pi \sqrt{\frac{am}{5 \lambda}}\) as required. Now suppose that \(PAO\) has been replaced with the string of modulus \(k \lambda\) but that \(P\) is along \(OA\). \begin{align*} F &= \frac{\lambda}{a}\left ( (a \omega - z) + (a \omega^2 - z)+ (a \omega^3 -z)+ (a \omega^4 - z) + k(a -z) \right) \\ &= \frac{\lambda}{a}(-a - 4z+ka -kz) \\ &= \frac{\lambda}{a}((k-1)a-(k+4)z) \end{align*} Notice that if \(z\) is real, this expression is also real, so all forces are acting along \(OA\). Therefore the particle will remain on the line \(OA\). We can also notice that the particle will move under the differential equation \[ m \ddot{x} + \frac{(k+4) \lambda}{a}x = \lambda(k-1) \] Therefore it will move with SHM about a point slightly displaced from the origin. The period will be: \(\displaystyle 2 \pi \sqrt{\frac{ma}{(k+4)\lambda}}\) which is equal to \(T/2\) if \((k+4) = 20 \Rightarrow k = 16\)

Derive a formula for the position of the centre of mass of a uniform circular arc of radius \(r\) which subtends an angle \(2\theta\) at the centre.

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Solution:

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Splitting the arc up into strips of width \(\delta \theta\), then we must have \begin{align*} && \sum r\cos \theta (r \delta \theta) &= \bar{x}\sum (r \delta \theta) \\ \lim_{\delta \theta \to 0}: && \int_{-\theta}^{\theta} r^2 \cos \theta \d \theta &= \bar{x}2 \theta r \\ \Rightarrow && 2r^2 \sin \theta &= \bar{x} 2 \theta r \\ \Rightarrow && \bar{x} &= \frac{r\sin \theta}{\theta} \end{align*}

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The centre of mass will lie on the line of symmetry. It also must lie at the center of the base of the semi-circle (see diagram). Using a coordinate frame where that point is the origin we must have \begin{align*} && 0 &= -2r \cdot 2h - 4h \cdot h + \pi r \frac{r}{\frac{\pi}{2}} \\ &&&= -4rh-4h^2+2r^2\\ \Rightarrow && 0 &= r^2-2rh-h^2 \\ \Rightarrow && \frac{r}{h} &= 1 \pm \sqrt{3} \\ \Rightarrow && r &= (1+\sqrt{3})h \\ && h & = \frac12 (\sqrt{3}-1) r \end{align*}

A piece of circus apparatus consists of a rigid uniform plank of mass 1000\(\,\)kg, suspended in a horizontal position by two equal light vertical ropes attached to the ends. The ropes each have natural length 10\(\,\)m and modulus of elasticity 490\(\,\)000 N. Initially the plank is hanging in equilibrium. Nellie, an elephant of mass 4000\(\,\)kg, lands in the middle of the plank while travelling vertically downwards at speed 5\(\,\)ms\(^{-1}.\) While carrying Nellie, the plank comes instantaneously to rest at a negligible height above the floor, and at this instant Nellie steps nimbly and gently off the plank onto the floor. Assuming that the plank remains horizontal, and the rope remain vertical, throughout the motion, find to three significant figures its initial height above the floor. During the motion after Nellie alights, do the ropes ever become slack? {[}Take \(g\) to be \(9.8\mbox{\,\ ms}^{-1}.\){]}

A librarian wishes to pick up a row of identical books from a shelf, by pressing her hands on the outer covers of the two outermost books and lifting the whole row together. The covers of the books are all in parallel vertical planes, and the weight of each book is \(W\). With each arm, the librarian can exert a maximum force of \(P\) in the vertical direction, and, independently, a maximum force of \(Q\) in the horizontal direction. The coefficient of friction between each pair of books and also between each hand and a book is \(\mu.\) Derive an expression for the maximum number of books that can be picked up without slipping, using this method. {[}You may assume that the books are thin enough for the rotational effect of the couple on each book to be ignored.{]}

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Solution:

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The force acting vertically on each of the outer books must be (by symmetry) \(\frac{nW}{2}\). The force acting horizontally on the outer books (and between each book in the horizontal direction) will be the same (we might as well say \(Q\) since increasing this force doesn't make any task less achievable. Looking at an end book, it will have force \(\frac{nW}{2}\) acting on one side, but it this force needs to not slip, ie \(\frac{nW}{2} \leq \mu Q\) \begin{align*} && \frac{nW}{2} &\leq \mu Q \\ \Rightarrow && n &\leq \frac{2\mu Q}{W} \\ && \frac{nW}{2} & \leq P \\ && n & \leq \frac{2P}{W} \\ \Rightarrow && n &\leq \frac2{W}\min \left (P, \mu Q \right) \end{align*}

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 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 long, inextensible string passes through a small fixed ring. One end of the string is attached to a particle of mass \(m,\) which hangs freely. The other end is attached to a bead also of mass \(m\) which is threaded on a smooth rigid wire fixed in the same vertical plane as the ring. The curve of the wire is such that the system can be in static equilibrium for all positions of the bead. The shortest distance between the wire and the ring is \(d(>0).\) Using plane polar coordinates centred on the ring, find the equation of the curve. The bead is set in motion. Assuming that the string remains taut, show that the speed of the bead when it is a distance \(r\) from the ring is \[ \left(\frac{r}{2r-d}\right)^{\frac{1}{2}}v, \] where \(v\) is the speed of the bead when \(r=d.\)

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Solution:

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Assume the total length of the string is \(l\). Then the total energy of the system (when nothing is moving) for a given \(\theta\) is: \(mg(r-l) + mgr \sin \theta\) Since for a point in static equilibrium, the derivative of this must be \(0\), this must be constant. So: \(r\l \sin \theta + 1\r = C \Rightarrow r = \frac{C}{1+\sin \theta}\) \(r\) will be smallest when \(\sin \theta = 1\), ie in polar coordinates, the equation should be \(r = \frac{2d}{1+\sin \theta}\) Alternatively, by considering forces, the shape must be a parabola with the ring at the focus. Considering the bead, it will have speed of \(r \dot{\theta}\) tangentially, and \(-\dot{r}\). The other particle will have speed \(\dot{r}\). Differentiating wrt to \(t\) \begin{align*} && 0 &= \dot{r}(\sin \theta + 1) + r \dot{\theta} \cos \theta \\ \Rightarrow && \dot{\theta} &= \frac{-\dot{r}(1+\sin \theta)}{r \cos \theta} \\ &&&= \frac{-\dot{r} 2d}{r^2 \sqrt{1-\l \frac{2d}{r}-1\r^2}} \\ &&&= \frac{-2d\dot{r}}{r^2\sqrt{\frac{r^2-(2d-r)^2}{r^2}}} \\ &&&= \frac{-d\dot{r}}{r\sqrt{dr-d^2}} \end{align*} By conservation of energy (since GPE is constant throughout the system, KE must be constant): \begin{align*} && \frac12 m (r^2 \dot{\theta}^2+\dot{r}^2) +\frac12 m \dot{r}^2 &= \frac12mv^2 \\ \Rightarrow && v^2 &= r^2 \dot{\theta}^2 + 2\dot{r}^2 \\ &&&= r^2 \frac{d^2\dot{r}^2}{r^2(dr-d^2)} + 2\dot{r}^2 \\ &&&= \dot{r}^2 \l \frac{d }{r-d} + 2 \r \\ &&&= \dot{r}^2 \l \frac{2r-d}{r-d} \r \\ \Rightarrow && v &= \dot{r} \l \frac{2r-d}{r-d} \r^{\frac12} \\ \Rightarrow && \dot{r} &= \l \frac{r-d}{2r-d} \r^{\frac12} v \\ \Rightarrow && u^2 &= r^2 \dot{\theta}^2+\dot{r}^2\\ &&&= \dot{r}^2 \l \frac{d }{r-d} + 1 \r \\ &&&= \l \frac{r-d}{2r-d} \r \l \frac{r}{r-d} \r v^2 \\ &&&= \l \frac{d}{2r-d} \r v^2 \\ \Rightarrow && u &= \l \frac{d}{2r-d} \r^{\frac12} v \end{align*}