Problems

A box has the shape of a uniform solid cuboid of height \(h\) and with a square base of side \(b\), where \(h > b\). It rests on rough horizontal ground. A light ladder has its foot on the ground and rests against one of the upper horizontal edges of the box, making an acute angle of \(\alpha\) with the ground, where \(h = b \tan \alpha\). The weight of the box is \(W\). There is no friction at the contact between ladder and box. A painter of weight \(kW\) climbs the ladder slowly. Neither the base of the ladder nor the box slips, but the box starts to topple when the painter reaches height \(\lambda h\) above the ground, where \(\lambda < 1\). Show that:

1. \(R = k\lambda W \cos \alpha\), where \(R\) is the magnitude of the force exerted by the box on the ladder;
2. \(2k\lambda \cos 2\alpha + 1 = 0\);
3. \(\mu \geq \frac{\sin 2\alpha}{1 - 3 \cos 2\alpha}\), where \(\mu\) is the coefficient of friction between the box and the ground.

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Solution:

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At the point we are about to topple, reaction and friction forces will be acting at \(C\)

1. \(\,\) \begin{align*} \overset{\curvearrowright}{X}:&& kW \cdot \lambda h\cos \alpha - R h &= 0 \\ \Rightarrow && R &= k\lambda W \cos \alpha \\ \end{align*}
2. \(\,\) \begin{align*} \overset{\curvearrowright}{C}:&& R \sin \alpha \cdot h-R\cos \alpha \cdot b-W\frac{b}{2} &= 0 \\ && k\lambda W \cos \alpha \sin \alpha \cdot b \tan \alpha- k\lambda W \cos \alpha\cos \alpha \cdot h-W\frac{b}{2} &= 0 \\ && k \lambda (\cos^2 \alpha - \sin^2 \alpha) +\frac12 &= 0 \\ \Rightarrow && 2k \lambda \cos 2\alpha + 1 &= 0 \end{align*}
3. \(\,\) \begin{align*} \text{N2}(\uparrow): && R_b -W-R\cos \alpha &= 0 \\ \Rightarrow && R_b &= W + k\lambda W \cos^2 \alpha\\ \text{N2}(\rightarrow): && R\sin \alpha - F_b &= 0 \\ \Rightarrow && F_b &= R \sin \alpha \\ \\ && F_b &\leq \mu R \\ \Rightarrow && k\lambda W \cos \alpha \sin \alpha &= \mu (W + k\lambda W \cos^2 \alpha) \\ \Rightarrow && \mu &\geq \frac{k\lambda \cos \alpha \sin \alpha}{1 + k\lambda \cos^2 \alpha} \\ &&&= \frac{k\lambda \sin 2\alpha}{2 + 2k\lambda cos^2 \alpha} \\ &&&= \frac{k\lambda \sin 2\alpha}{2 + k\lambda (\cos 2 \alpha+1)} \\ &&&= \frac{k\lambda \sin 2\alpha}{-4k\lambda \cos 2 \alpha + k\lambda (\cos 2 \alpha+1)} \\ &&&= \frac{\sin 2 \alpha}{1 -3 \cos 2\alpha} \end{align*}

A bus has the shape of a cuboid of length \(a\) and height \(h\). It is travelling northwards on a journey of fixed distance at constant speed \(u\) (chosen by the driver). The maximum speed of the bus is \(w\). Rain is falling from the southerly direction at speed \(v\) in straight lines inclined to the horizontal at angle \(\theta\), where \(0<\theta<\frac12\pi\). By considering first the case \(u=0\), show that for \(u>0\) the total amount of rain that hits the roof and the back or front of the bus in unit time is proportional to \[ h\big \vert v\cos\theta - u \big\vert + av\sin\theta \,. \] Show that, in order to encounter as little rain as possible on the journey, the driver should choose \( u=w\) if either \(w< v\cos\theta\) or \( a\sin\theta > h\cos\theta\). How should the speed be chosen if \(w>v\cos\theta\) and \( a\sin\theta < h\cos\theta\)? Comment on the case \( a\sin\theta = h\cos\theta\). How should the driver choose \(u\) on the return journey?

The cuboid \(ABCDEFGH\) is such \(AE\), \(BF\), \(CG\), \(DH\) are perpendicular to the opposite faces \(ABCD\) and \(EFGH\), and \(AB =2, BC=1, AE={\lambda}\). Show that if \(\alpha\) is the acute angle between the diagonals \(AG\) and \(BH\) then $$\cos {\alpha} = |\frac {3-{\lambda}^2} {5+{\lambda}^2} |$$ Let \(R\) be the ratio of the volume of the cuboid to its surface area. Show that \(R<\frac{1}{3}\) for all possible values of \(\lambda\). Prove that, if \(R\ge \frac{1}{4}\), then \(\alpha \le \arccos \frac{1}{9}\).

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Solution:

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Set \(A\) to be the origin, then \(B = \langle 2, 0, 0 \rangle, G = \langle 2, 1, \lambda \rangle, H = \langle 0, 1, \lambda \rangle\), in particular \begin{align*} && AG&= \langle 2, 1, \lambda \rangle \\ && BH &= \langle -2, 1, \lambda \rangle \\ \Rightarrow && \cos \alpha &= |\frac{-4+1+\lambda^2}{\sqrt{2^2+1^2+\lambda^2}\sqrt{(-2)^2+1^2+\lambda^2}}| \\ &&&= |\frac{-3+\lambda^2}{5+\lambda^2}| \end{align*} \begin{align*} && \text{Volume} &= 2\lambda \\ && \text{Surface area} &= 2\cdot2\lambda + 2\cdot\lambda + 2\cdot2 \\ \Rightarrow && R&= \frac{\lambda}{3\lambda + 2} < \frac{1}{3} \\ && \frac14 &\leq R \\ \Rightarrow && 3\lambda +2 &\leq 4\lambda \\ \Rightarrow &&2 & \leq \lambda \end{align*} Then \(\frac{\lambda^2-3}{5+\lambda^2}\) is increasing as \(\lambda\) increases, in particularly the smallest value is \(\frac{1}{9}\).