Problems

The lines \(l_1\), \(l_2\) and \(l_3\) lie in an inclined plane \(P\) and pass through a common point \(A\). The line \(l_2\) is a line of greatest slope in \(P\). The line \(l_1\) is perpendicular to \(l_3\) and makes an acute angle \(\alpha\) with \(l_2\). The angles between the horizontal and \(l_1\), \(l_2\) and \(l_3\) are \(\pi/6\), \(\beta\) and \(\pi/4\), respectively. Show that \(\cos\alpha\sin\beta = \frac12\,\) and find the value of \(\sin\alpha \sin\beta\,\). Deduce that \(\beta = \pi/3\,\). The lines \(l_1\) and \(l_3\) are rotated in \(P\) about \(A\) so that \(l_1\) and \(l_3\) remain perpendicular to each other. The new acute angle between \(l_1\) and \(l_2\) is \(\theta\). The new angles which \(l_1\) and \(l_3\) make with the horizontal are \(\phi\) and \(2\phi\), respectively. Show that \[ \tan^2\theta = \frac{3+\sqrt{13}}2\;. \]

A tall container made of light material of negligible thickness has the form of a prism, with a square base of area \(a^2\). It contains a volume \(ka^3\) of fluid of uniform density. The container is held so that it stands on a rough plane, which is inclined at angle \(\theta\) to the horizontal, with two of the edges of the base of the container horizontal. In the case \(k > \frac12 \tan\theta\), show that the centre of mass of the fluid is at a distance \(x\) from the lower side of the container and at a distance \(y\) from the base of the container, where \[ \frac x a = \frac12 - \frac {\tan\theta}{12k}\;, \ \ \ \ \ \ \frac y a = \frac k 2 + \frac{\tan^2\theta}{24k}\;. \] Determine the corresponding coordinates in the case \(k < \frac12 \tan\theta\). The container is now released. Given that \(k < \frac12\), show that the container will topple if \(\theta >45^\circ\).

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Solution:

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The fluid can be divided into a cuboid parallel to the slope and a right-angled triangle. If the height of the water on the longer side is \(\ell a\), then we have \(ka^3 = (\ell a - a\tan \theta)a^2 + \frac12 a^3\tan \theta \Rightarrow \ell = k + \frac12 \tan \theta\) This is acceptable when \(k > \frac12 \tan \theta\). The centre of mass of the cuboid will be \((\frac{a}{2}, \frac12 (k - \frac12 \tan \theta))\) and of the triangle will be \((\frac13 a, \frac13 \tan \theta + (k - \frac12 \tan \theta) )\) Therefore we have: \begin{align*} && \text{COM} && \text{mass} \\ \text{cuboid} && (\frac{a}{2}, \frac{a}2 (k - \frac12 \tan \theta)) && a^3(k - \frac12 \tan \theta) \\ \text{triangle} && (\frac13 a, \frac{a}3 \tan \theta + a(k - \frac12 \tan \theta) ) && a^3\frac12 \tan \theta \\ \text{whole system} && (x, y) && a^3k \end{align*} Therefore \begin{align*} && a^3k x &= \frac{a}{2} \cdot a^3(k - \frac12 \tan \theta) + \frac13 a \cdot a^3\frac12 \tan \theta \\ &&&= a^4 \frac{k}{2} - \frac{1}{12}a^4 \tan \theta \\ \Rightarrow && \frac{x}{a} &= \frac12 - \frac{\tan \theta}{12 k} \\ \\ && a^3k y &= \frac{a}2 (k - \frac12 \tan \theta) \cdot a^3(k - \frac12 \tan \theta) + \\ &&& \qquad\qquad \cdots + \l \frac{a}3 \tan \theta + a(k - \frac12 \tan \theta) \r \cdot a^3\frac12 \tan \theta \\ &&&= \frac{a^4k^2}{2} -\frac{a^4k \tan \theta}{2} + \frac{a^4 \tan^2 \theta}{8} - \frac{a^4 \tan^2 \theta}{12} + \frac{a^4k \tan \theta}{2} \\ \Rightarrow && \frac{y}{a} &= \frac{k}2 + \frac{\tan^2 \theta}{24k} \end{align*}

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If the water only fills up a prism, it's sides must be \(b\) and \(b\tan \theta \), therefore the volume is \(\frac12 ab^2 \tan \theta = ka^3 \Rightarrow b = a\sqrt{\frac{2k}{\tan \theta}}\) The centre of mass will be at \(\l \frac13 a\sqrt{\frac{2k}{\tan \theta}}, a\sqrt{2k \tan \theta}\r\) The container will topple if the centre of mass is outside the base, ie if the centre of mass \((x,y)\) lies above the line \(y = \tan (90^\circ- x) = \frac{1}{\tan \theta} x\). If \(\theta > 45^\circ\) then \(\tan \theta > 1\) and so we are in the \(\frac12 \tan \theta > \frac12 > k\) and so we are in the second case. \begin{align*} \frac{y}{x} &= \frac{\frac 13 a\sqrt{2k \tan \theta}}{\frac13 a\sqrt{\frac{2k}{\tan \theta}}} \\ &= \tan \theta \end{align*} \(\tan \theta > \frac{1}{\tan \theta} \Leftrightarrow \tan \theta > 1 \Leftrightarrow \theta > 45^\circ\).

Arthur and Bertha stand at a point \(O\) on an inclined plane. The steepest line in the plane through \(O\) makes an angle \(\theta\) with the horizontal. Arthur walks uphill at a steady pace in a straight line which makes an angle \(\alpha\) with the steepest line. Bertha walks uphill at the same speed in a straight line which makes an angle \(\beta\) with the steepest line (and is on the same side of the steepest line as Arthur). Show that, when Arthur has walked a distance \(d\), the distance between Arthur and Bertha is \(2d \vert\sin\frac12(\alpha-\beta)\vert\). Show also that, if \(\alpha\ne\beta\), the line joining Arthur and Bertha makes an angle \(\phi\) with the vertical, where \[ \cos\phi = \sin\theta \sin \frac12(\alpha+\beta). \]

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*}

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A cannon is situated at the bottom of a plane inclined at angle \(\beta\) to the horizontal. A (small) cannon ball is fired from the cannon at an initial speed \(u.\) Ignoring air resistance, find the angle of firing which will maximise the distance up the plane travelled by the cannon ball and show that in this case the ball will land at a distance \[ \frac{u^{2}}{g(1+\sin\beta)} \] from the cannon.

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1. Find the condition which \(m,M\) and \(\theta\) must satisfy to ensure that the ball will fall to the ground. Assuming that this condition is satisfied, show that the velocity \(v\) of the ball when it hits the ground satisfies \[ v^{2}=\frac{2g(M-m\sin\theta)d\sin\theta}{M+m}. \]
2. Find the condition which \(m,M\) and \(\theta\) must satisfy if the wagon is not to collide with the pulley at the top of the incline.

The points \(A,B\) and \(C\) lie on the surface of the ground, which is an inclined plane. The point \(B\) is 100m due north of \(A,\) and \(C\) is 60m due east of \(B\). The vertical displacements from \(A\) to \(B,\) and from \(B\) to \(C\), are each 5m downwards. A plane coal seam lies below the surface and is to be located by making vertical bore-holes at \(A,B\) and \(C\). The bore-holes strike the coal seam at 95m, 45m and 76m below \(A,B\) and \(C\) respectively. Show that the coal seam is inclined at \(\cos^{-1}(\frac{4}{5})\) to the horizontal. The coal seam comes to the surface along a line. Find the bearing of this line.

A rough ring of radius \(a\) is fixed so that it lies in a plane inclined at an angle \(\alpha\) to the horizontal. A uniform heavy rod of length \(b(>a)\) has one end smoothly pivoted at the centre of the ring, so that the rod is free to move in any direction. It rests on the circumference of the ring, making an angle \(\theta\) with the radius to the highest point on the circumference. Find the relation between \(\alpha,\theta\) and the coefficient of friction, \(\mu,\) which must hold when the rod is in limiting equilibrium.

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Solution:

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It is important to define clear coordinate axes, so let the \(x\)-axis point up the line of greatest slope of the ring. The \(z\)-axis perpendicular to the ring, and the \(y\)-axis perpendicular to both of these. Our method is going to be to take moments about \(O\) to avoid worrying about the force at the pivot. There are \(3\) forces we need to worry about:

• The mass of the rod
• The reaction where it meets the ring
• The friction at the ring

In our coordinate frame, the reaction will act in the \(z\)-direction, \(\displaystyle \begin{pmatrix} 0 \\ 0 \\ R \end{pmatrix}\), the friction force will act in the \(x-y\) plane: \(\displaystyle \begin{pmatrix} \mu R \sin \theta \\ -\mu R \cos \theta \\ 0 \end{pmatrix}\). We don't know the mass, but we know it will be acting "vertically", so \(\cos \alpha\) of it will act in the \(z\)-axis and \(\sin \alpha\) will act in the \(y\)-axis, ie it will act parallel to \(\displaystyle \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix}\). When taking moments, we need to consider \(\mathbf{r}\) the direction of the rod. This will be \(\displaystyle \begin{pmatrix} \cos \theta \\ \sin \theta \\ 0 \end{pmatrix}\). The moment of the weight will all be parallel to \(\mathbf{r} \times \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix}\). Similarly the moments of the contact forces will be \(\mathbf{r} \times \begin{pmatrix} \mu R \sin \theta \\ -\mu R \cos \theta \\ R \end{pmatrix}\). Since these moments sum to \(\mathbf{0}\) as we are in equilibrium, these vectors must be parallel. Therefore it is sufficient to check the vector triple product, \begin{align*} && 0 &= \begin{pmatrix} \cos \theta \\ \sin \theta \\ 0 \end{pmatrix} \cdot \left ( \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix} \times \begin{pmatrix} \mu \sin \theta \\ -\mu \cos \theta \\ 1 \end{pmatrix} \right ) \\ &&&= \cos \theta (\mu \cos \theta \cos \alpha)-\sin \theta (\sin \alpha - \mu \sin \theta \cos \alpha) \\ &&&= \mu((\sin^2 \theta+\cos^2 \theta) \cos \alpha) -\sin \theta \sin \alpha \\ \Rightarrow && \mu &= \tan \alpha \sin \theta \end{align*}