Problems

\(AB\) is a uniform rod of weight \(W\,\). The point \(C\) on \(AB\) is such that \(AC>CB\,\). The rod is in contact with a rough horizontal floor at \(A\,\) and with a cylinder at \(C\,\). The cylinder is fixed to the floor with its axis horizontal. The rod makes an angle \({\alpha}\) with the horizontal and lies in a vertical plane perpendicular to the axis of the cylinder. The coefficient of friction between the rod and the floor is \(\tan \lambda_1\) and the coefficient of friction between the rod and the cylinder is \(\tan \lambda_2\,\). Show that if friction is limiting both at \(A\) and at \(C\), and \({\alpha} \ne {\lambda}_2 - {\lambda}_1\,\), then the frictional force acting on the rod at \(A\) has magnitude $$ \frac{ W\sin {\lambda}_1 \, \sin({\alpha}-{\lambda}_2)} {\sin ({\alpha}+{\lambda}_1-{\lambda}_2)} \;.$$ %and that %$$ %p=\frac{\cos{\alpha} \, \sin({\alpha}+{\lambda}_1-{\lambda}_2)} %{2\cos{\lambda}_1 \, \sin {\lambda}_2}\;. %$$

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1. If the normal reactions on the lower and higher wheels of the lorry are equal, show that the sum of the frictional forces between the wheels and the ground is zero.
2. If \(P\) is such that the lorry does not tip over (but the normal reactions on the lower and higher wheels of the lorry need not be equal), show that \[ W\tan(\alpha - \beta) \le P \le W\tan(\alpha+\beta)\;, \] where \(\tan\beta = d/h\,\).

A rigid straight beam \(AB\) has length \(l\) and weight \(W\). Its weight per unit length at a distance \(x\) from \(B\) is \(\alpha Wl^{-1} (x/l)^{\alpha-1}\,\), where \(\alpha\) is a positive constant. Show that the centre of mass of the beam is at a distance \(\alpha l/(\alpha+1)\) from \(B\). The beam is placed with the end \(A\) on a rough horizontal floor and the end \(B\) resting against a rough vertical wall. The beam is in a vertical plane at right angles to the plane of the wall and makes an angle of \(\theta\) with the floor. The coefficient of friction between the floor and the beam is \(\mu\) and the coefficient of friction between the wall and the beam is also \(\mu\,\). Show that, if the equilibrium is limiting at both \(A\) and \(B\), then \[ \tan\theta = \frac{1-\alpha \mu^2}{(1+\alpha)\mu}\;. \] Given that \(\alpha =3/2\,\) and given also that the beam slides for any \(\theta<\pi/4\,\) find the greatest possible value of \(\mu\,\).

\(B_1\) and \(B_2\) are parallel, thin, horizontal fixed beams. \(B_1\) is a vertical distance \(d \sin \alpha \) above \(B_2\), and a horizontal distance \(d\cos\alpha \) from \(B_2\,\), where \(0<\alpha<\pi/2\,\). A long heavy plank is held so that it rests on the two beams, perpendicular to each, with its centre of gravity at \(B_1\,\). The coefficients of friction between the plank and \(B_1\) and \(B_2\) are \(\mu_1\) and \(\mu_2\,\), respectively, where \(\mu_1<\mu_2\) and \(\mu_1+\mu_2=2\tan\alpha\,\). The plank is released and slips over the beams experiencing a force of resistance from each beam equal to the limiting frictional force (i.e. the product of the appropriate coefficient of friction and the normal reaction). Show that it will come to rest with its centre of gravity over \(B_2\) in a time \[ \pi \left(\frac{d}{g(\mu_2-\mu_1)\cos\alpha }\right)^{\!\frac12}\;. \]

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Solution:

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\begin{align*} \overset{\curvearrowright}{B_2} : && mgx\cos \alpha - R_1d &= 0 \\ && \frac{mgx \cos \alpha}{d} &= R_1 \\ \overset{\curvearrowright}{B_1} : && -mg(d-x)\cos \alpha + R_2d &= 0 \\ && \frac{mg(d-x) \cos \alpha}{d} &= R_2 \\ % \text{N2}(\perp B_1B_2): && R_1 + R_2 - mg\cos \alpha &=0 \\ \text{N2}(\parallel B_1B_2): && mg\sin \alpha - \mu_1R_1 - \mu_2R_2 &= m\ddot{x} \\ && mg \sin \alpha - \mu_1 \frac{mgx \cos \alpha}{d} - \mu_2\frac{mg(d-x) \cos \alpha}{d} &= m \ddot{x} \\ && gd \sin \alpha - \mu_2 gd \cos \alpha - (\mu_1 - \mu_2) x g \cos \alpha &= d \ddot{x} \\ && gd \frac12 \l \mu_1 + \mu_2 \r \cos \alpha - \mu_2 gd \cos \alpha - (\mu_1 - \mu_2) x g \cos \alpha &= d \ddot{x} \\ && gd \frac12 \l \mu_1 - \mu_2 \r \cos \alpha - (\mu_1 - \mu_2) x g \cos \alpha &= d \ddot{x} \\ && \frac12 d C &= d \ddot{x} + Cx \\ && \Big ( C &= g(\mu_1 - \mu_2) \cos \alpha \Big ) \\ \end{align*} We can recognise this differential equation from SHM as having the solution: \[x = A\sin \l \l \frac{d}{C} \r^{\frac12} t \r + B\cos \l \l \frac{d}{C} \r^{\frac12} t \r + \frac12 d\] Since when \(t = 0, x = d, \dot{x} = 0, A = 0, B = \frac{1}{2}d\). We will reach \(B_2, (x = 0)\) when \(\cos \l \l \frac{d}{C} \r^{\frac12} T \r = -1\) (at which point the speed will be zero) and \begin{align*} && \l \frac{d}{C} \r^{\frac12} T &= \pi \\ \Rightarrow && T&= \pi \l \frac{d}{g(\mu_1 - \mu_2) \cos \alpha} \r^{\frac12} \end{align*}

A rod \(AB\) of length 0.81 m and mass 5 kg is in equilibrium with the end \(A\) on a rough floor and the end \(B\) against a very rough vertical wall. The rod is in a vertical plane perpendicular to the wall and is inclined at \(45^{\circ}\) to the horizontal. The centre of gravity of the rod is at \(G\), where \(AG = 0.21\) m. The coefficient of friction between the rod and the floor is 0.2, and the coefficient of friction between the rod and the wall is 1.0. Show that the friction cannot be limiting at both \(A\) and \(B\). A mass of 5 kg is attached to the rod at the point \(P\) such that now the friction is limiting at both \(A\) and \(B\). Determine the length of \(AP\).

The random variable \(X\) is uniformly distributed on the interval \([-1,1]\). Find \(\E(X^2)\) and \(\var (X^2)\). A second random variable \(Y\), independent of \(X\), is also uniformly distributed on \([-1,1]\), and \(Z=Y-X\). Find \(\E(Z^2)\) and show that \(\var (Z^2) = 7 \var (X^2)\).

A sphere of radius \(a\) and weight \(W\) rests on horizontal ground. A thin uniform beam of weight \(3\sqrt3\,W\) and length \(2a\) is freely hinged to the ground at \(X\), which is a distance \({\sqrt 3} \, a\) from the point of contact of the sphere with the ground. The beam rests on the sphere, lying in the same vertical plane as the centre of the sphere. The coefficients of friction between the beam and the sphere and between the sphere and the ground are \(\mu_1\) and \(\mu_2\) respectively. Given that the sphere is on the point of slipping at its contacts with both the ground and the beam, find the values of \(\mu_1\) and \(\mu_2\).

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Solution:

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The first important thing to observe is the angle at \(X\) is \(60^{\circ}\). Now we can start resolving: \begin{align*} \overset{\curvearrowleft}{X}: && 3\sqrt{3} W \cos 60^{\circ} a - R_1\sqrt{3}a &= 0 \tag{\(1\)}\\ \overset{\curvearrowleft}{O}: && \mu_2 R_2 a - \mu_1R_1a &= 0 \tag{\(2\)} \\ \text{N2}(\rightarrow): && \mu_2 R_2 + \mu_1R_1 \cos 60^{\circ} - R_1 \cos 30^{\circ} &= 0 \tag{\(3\)} \\ \text{N2}(\uparrow): && R_2 - W - \mu_1 R_1 \cos 30^{\circ} - R_1 \cos 60^{\circ} &= 0 \tag{\(4\)} \\ \Rightarrow && \frac{3}{2}W &= R_1 \tag{\((5)\) from \((1)\)} \\ && \mu_1 R_1 &= \mu_2 R_2 \tag{\(2\)}\\ && \mu_1 R_1 \l 1 + \frac{1}{2} \r - R_1 \frac{\sqrt{3}}2 &= 0 \tag{\((3)\) and \((2)\)} \\ && \mu_1 &= \frac{1}{\sqrt3} \\ \\ && R_2 - W - \frac{1}{\sqrt3} \frac{3}{2}W \frac{\sqrt3}{2} - \frac{3}2W \frac12 &= 0 \\ \Rightarrow && R_2 &= W \l 1 + \frac{3}{2}\r \tag{\(6\)} \\ \Rightarrow && \mu_2 &= \frac{\mu_1 R_1}{R_2} = \frac{1}{\sqrt{3}} \frac{3}{5} = \frac{\sqrt3}{5} \tag{\((5)\) and \((6)\)} \end{align*}

A uniform solid sphere of diameter \(d\) and mass \(m\) is drawn very slowly and without slipping from horizontal ground onto a step of height \(d/4\) by a horizontal force which is always applied to the highest point of the sphere and is always perpendicular to the vertical plane which forms the face of the step. Find the maximum horizontal force throughout the movement, and prove that the coefficient of friction between the sphere and the edge of the step must exceed \(1/\sqrt{3}\).

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Solution:

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The ball is on the ground when \(\cos \theta = \frac12 \Rightarrow \theta = 60^\circ\) and ball will make it over the step when \(\theta = 0^\circ\). It is also worth emphasising we are moving \emph{very slowly}, so we can treat the system as static at any given point. \begin{align*} \overset{\curvearrowleft}{X}: && mg \frac{d}{2}\sin \theta - F \frac{d}{2} \l 1 + \cos \theta \r &= 0\\ \Rightarrow && \frac{mg \sin \theta}{1 + \cos \theta} &= F& \\ \Rightarrow && mg \tan \frac{\theta}{2} &= F& \\ \end{align*} Therefore \(F\) is maximised when \(\theta = 60^\circ\), ie \(F_{max} = \frac{mg}{\sqrt{3}}\) \begin{align*} \text{N2}(\parallel OX): && mg \cos \theta - R + F \sin \theta &= 0 \\ \Rightarrow && mg \cos \theta - R + \frac{mg\sin \theta}{1 + \cos \theta} \sin \theta &= 0 \\ \Rightarrow && mg &= R \\ \\ \text{N2}(\perp OX): && F_X - mg \sin \theta + F \cos \theta &= 0 \\ \Rightarrow && mg \sin \theta - \frac{mg\sin \theta}{1 + \cos \theta} \cos \theta &= F_X \\ \Rightarrow && \frac{mg\sin \theta}{1 + \cos \theta} &= F_X \tag{We could also see this taking moments about \(O\)}\\ % \text{N2}(\rightarrow): && F + \mu R \cos \theta - R \sin \theta &\geq 0 \\ % \text{N2}(\uparrow): && -mg +\mu R \sin \theta + R \cos \theta &\geq 0 \\ % \Rightarrow && R \l \sin \theta - \mu \cos \theta\r &\leq F \\ % \Rightarrow && R \l \mu \sin \theta + \cos \theta\r &\geq mg \\ % \Rightarrow && \l \frac{\sin \theta - \mu \cos \theta}{\mu \sin \theta + \cos \theta} \r mg & \leq F \\ % \Rightarrow && \l \frac{\tan \theta - \mu }{1+\mu \tan \theta} \r mg & \leq F \\ % \Rightarrow && \tan \l \theta - \alpha \r mg & \leq F \tag{where \(\tan \alpha = \mu\)} \end{align*} Therefore since \(F_X \leq \mu R\), \(\displaystyle \frac{mg\sin \theta}{1 + \cos \theta} \leq \mu mg \Rightarrow \mu \geq \tan \frac{\theta}{2}\) which is maximised at \(\theta = 60^\circ\) and implies \(\mu \geq \frac{1}{\sqrt{3}}\)

The random variable \(X\) is uniformly distributed on \([0,1]\). A new random variable \(Y\) is defined by the rule \[ Y=\begin{cases} 1/4 & \mbox{ if }X\leqslant1/4,\\ X & \mbox{ if }1/4\leqslant X\leqslant3/4\\ 3/4 & \mbox{ if }X\geqslant3/4. \end{cases} \] Find \({\mathrm E}(Y^{n})\) for all integers \(n\geqslant 1\). Show that \({\mathrm E}(Y)={\mathrm E}(X)\) and that \[{\mathrm E}(X^{2})-{\mathrm E}(Y^{2})=\frac{1}{24}.\] By using the fact that \(4^{n}=(3+1)^{n}\), or otherwise, show that \({\mathrm E}(X^{n}) > {\mathrm E}(Y^{n})\) for \(n\geqslant 2\). Suppose that \(Y_{1}\), \(Y_{2}\), \dots are independent random variables each having the same distribution as \(Y\). Find, to a good approximation, \(K\) such that \[{\rm P}(Y_{1}+Y_{2}+\cdots+Y_{240000} < K)=3/4.\]

Two rough solid circular cylinders, of equal radius and length and of uniform density, lie side by side on a rough plane inclined at an angle \(\alpha\) to the horizontal, where \(0<\alpha<\pi/2\). Their axes are horizontal and they touch along their entire length. The weight of the upper cylinder is \(W_1\) and the coefficient of friction between it and the plane is \(\mu_1\). The corresponding quantities for the lower cylinder are \(W_2\) and \(\mu_2\) respectively and the coefficient of friction between the two cylinders is \(\mu\). Show that for equilibrium to be possible:

1. \(W_1\ge W_2\);
2. \(\mu\geqslant\dfrac{W_1+W_2}{W_1-W_2}\);
3. \(\mu_{1}\geqslant\left(\dfrac{2W_{1}\cot\alpha}{W_{1}+W_{2}}-1\right)^{-1}\,.\)

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Solution:

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1. \begin{align*} \overset{\curvearrowright}{O_2}: && 0 &= F_2 - F \\ \Rightarrow && F_2 &= F \\ \overset{\curvearrowright}{O_1}: && 0 &= F_1- F \\ \Rightarrow && F_1 &= F \\ \text{N2}(\swarrow, 2): && 0 &= R+W_2\sin\alpha -F \tag{1}\\ \text{N2}(\swarrow, 1): && 0 &= W_1\sin\alpha -F-R\tag{2}\\ \Rightarrow && W_1 \sin \alpha-R &= W_2 \sin \alpha+R \\ \Rightarrow && W_1 &\geq W_2 \end{align*}
2. \begin{align*} (1)+(2)\Rightarrow && F &= \frac12 \sin \alpha (W_1 + W_2) \\ (1)-(2) \Rightarrow && R &= \frac12 \sin \alpha (W_1-W_2) \\ \Rightarrow && \frac{F}{R} &= \frac{W_1+W_2}{W_1-W_2} \\ \underbrace{\Rightarrow}_{F \leq \mu R} && \mu &\geq \frac{W_1+W_2}{W_1-W_2}\\ \end{align*}
3. \begin{align*} \text{N2}(\nwarrow, 1): && 0 &= F+R_1-W_1\cos \alpha \\ \Rightarrow && R_1 &= W_1\cos \alpha - F \\ &&&= W_1 \cos \alpha - \frac12 \sin \alpha (W_1 + W_2) \\ \Rightarrow && \frac{R_1}{F_1} &= \frac{R_1}{F} \\ &&&= \frac{W_1 \cos \alpha - \frac12 \sin \alpha (W_1 + W_2)}{\frac12 \sin \alpha (W_1 + W_2) } \\ &&&= \frac{2W_1 \cot \alpha}{W_1+W_2} - 1 \\ \Rightarrow && \mu_1 & \geq \left ( \frac{2W_1 \cot \alpha}{W_1+W_2} - 1 \right)^{-1} \end{align*}

\begin{align*} \text{N2}(\nwarrow, 2): && 0 &= -F+R_2-W_2\cos \alpha \\ \Rightarrow && R_2 &= W_2\cos \alpha + F \\ &&&= W_2 \cos \alpha + \frac12 \sin \alpha (W_1 + W_2) \\ \Rightarrow && \frac{R_2}{F_2} &= \frac{R_2}{F} \\ &&&= \frac{ W_2 \cos \alpha + \frac12 \sin \alpha (W_1 + W_2)}{\frac12 \sin \alpha (W_1 + W_2) } \\ &&&= \frac{2W_2 \cot \alpha}{W_1+W_2} + 1 \\ \Rightarrow && \mu_2 & \geq \left ( \frac{2W_1 \cot \alpha}{W_1+W_2} + 1 \right)^{-1} \end{align*}

Two identical uniform cylinders, each of mass \(m,\) lie in contact with one another on a horizontal plane and a third identical cylinder rests symmetrically on them in such a way that the axes of the three cylinders are parallel. Assuming that all the surfaces in contact are equally rough, show that the minimum possible coefficient of friction is \(2-\sqrt{3}.\)

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Solution:

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First observe that many forces are equal by symmetry. Also notice that \(A\) and \(B\) are trying to roll in opposite directions, therefore there is no friction between \(A\) and \(B\). Considering the system as a whole \(R_1 = \frac32 mg\). \begin{align*} \text{N2}(\uparrow,C): && 0 &= -mg +2R_3\cos 30^{\circ} + 2F_{CA} \cos 60^{\circ} \\ \Rightarrow && mg &= \sqrt{3}R_3 + F_{CA} \\ \\ \text{N2}(\uparrow, A): && 0 &= -mg + \frac32mg-R_3\cos 30^{\circ} -F_{AC} \cos 60^\circ \\ \Rightarrow && mg &= \sqrt{3}R_3+F_{AC} \\ \text{N2}(\rightarrow, A): && 0 &= F_{AC}\cos 30^{\circ}+F_1-R_3 \cos 60^\circ -R_2 \\ \Rightarrow && 0 &=F_{AC}\sqrt{3}+2F_1-R_3-2R_2 \\ \overset{\curvearrowleft}{A}: && 0 &= F_1 - F_{AC} \end{align*} Since \(F_1 = F_{AC} = F_{CA}\) we can rewrite everything in terms of \(F= F_1\) so. \begin{align*} && mg &= \sqrt{3}R_3 + F \\ && 0 &= (2+\sqrt{3})F -R_3-2R_2 \\ \Rightarrow && (2+\sqrt3)F &\geq R_3 \\ \Rightarrow && F &\geq (2-\sqrt{3})R_3 \\ \Rightarrow && \mu & \geq 2 - \sqrt{3} \end{align*}

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Suppose \(X\) is a random variable with probability density \[ \mathrm{f}(x)=Ax^{2}\exp(-x^{2}/2) \] for \(-\infty < x < \infty.\) Find \(A\). You belong to a group of scientists who believe that the outcome of a certain experiment is a random variable with the probability density just given, while other scientists believe that the probability density is the same except with different mean (i.e. the probability density is \(\mathrm{f}(x-\mu)\) with \(\mu\neq0\)). In each of the following two cases decide whether the result given would shake your faith in your hypothesis, and justify your answer.

1. A single trial produces the result 87.3.
2. 1000 independent trials produce results having a mean value \(0.23.\)

A light rod of length \(2a\) is hung from a point \(O\) by two light inextensible strings \(OA\) and \(OB\) each of length \(b\) and each fixed at \(O\). A particle of mass \(m\) is attached to the end \(A\) and a particle of mass \(2m\) is attached to the end \(B.\) Show that, in equilibrium, the angle \(\theta\) that the rod makes the horizontal satisfies the equation \[ \tan\theta=\frac{a}{3\sqrt{b^{2}-a^{2}}}. \] Express the tension in the string \(AO\) in terms of \(m,g,a\) and \(b\).

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Solution:

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The centre of mass of the rod will be at a point \(G\) which divides the rod in a ratio \(1:2\). Let \(M\) be the midpoint of \(AB\), so \(|AM| = a\) To be in equilibrium \(G\) must lie directly below \(O\). Note that \(OM^2 +a^2 = b^2 \Rightarrow OM = \sqrt{b^2-a^2}\) Notice that \(AG = \frac{4}{3}a\) and \(AM = a\), so \(|MG| = \frac13 a\). Therefore \(\displaystyle \tan \theta = \frac{\tfrac{a}{3}}{\sqrt{b^2-a^2}} \Rightarrow \tan \theta = \frac{a}{3\sqrt{b^2-a^2}}\).

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Notice that \(\frac{\sin \beta}{\frac43a} = \frac{\sin \angle OGA}{b} = \frac{\sin \alpha}{\frac23 a} \Rightarrow \sin \beta = 2 \sin \alpha\) \begin{align*} \text{N2}(\rightarrow, A): && C\cos \theta - T_A \sin \beta&= 0 \\ \text{N2}(\rightarrow, B): && T_B \sin \alpha - C\cos \theta &= 0 \\ \Rightarrow && T_B \sin \alpha &= T_A\sin \beta \\ \Rightarrow && T_B &= 2T_A \\ \text{N2}(\uparrow, A): && T_A \cos \beta+C\sin \theta-mg &= 0 \\ \text{N2}(\uparrow, B): && T_B \cos \alpha- C\sin \theta-2mg &= 0 \\ \Rightarrow && T_A \cos \beta+2T_A \cos \alpha&=3mg \\ \end{align*} Using the cosine rule: \((\frac23a)^2 = b^2 + OG^2 - 2b|OG|\cos \alpha\) and \((\frac43a)^2 = b^2 + OG^2-2b|OG|\cos \beta\). \(|OG|^2 = b^2 + (\frac43a)^2-\frac83ab \cos \angle A = b^2 +\frac{16}{9}a^2-\frac83a^2 = b^2-\frac89a^2\). Therefore \(\cos \alpha = \frac{2b^2-\frac43a^2}{2b |OG|}\), \(\cos \beta = \frac{2b^2-\frac83a^2}{2b|OG|}\) Therefore \(\cos \beta + 2 \cos \alpha = \frac{18b^2-16a^2}{6b|OG|} = \frac{9b^2-8a^2}{b\sqrt{9b^2-a^2}} = \frac{\sqrt{9b^2-a^2}}b\) Therefore \(\displaystyle T_A = \frac{3bmg}{\sqrt{9b^2-8a^2}}\)

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*}