A framework of six equal light rods forms a regular hexagon \(ABCDEF\), which is stiffened by light rods \(BF, CE, CF\). Weights \(W, 2W\) are suspended from \(A, B\) and the framework is supported, with \(AB\) horizontal, by vertical forces at \(D\) and \(E\). Draw the stress diagram, and find the stress in \(CF\).
A fixed spherical shell has a small hole in it at an angular distance \(\alpha\) from the highest point. A uniform rod whose length is four times the radius of the shell passes through the hole and rests with one end in contact with the inner surface of the shell, all the surfaces being smooth. Write down the potential energy of the rod when it is in the same vertical plane as the highest point of the shell, and makes an angle \(\theta\) with the horizontal. If \(\alpha = 30^\circ\), show that the acute angle which the rod makes with the horizontal in equilibrium may be \(20^\circ, 40^\circ, \text{ or } 60^\circ\).
An engine working at 500 H.P. pulls a train of 200 tons along a level track, the resistances being 16 lb. per ton. When the velocity of the train is 30 miles per hour, find its acceleration. At what steady speed will the engine pull the train up an incline of 1 in 100 with the same expenditure of power against the same resistances?
A pile driver weighing 2 cwt. falls through 5 feet and drives a pile weighing 6 cwt. through a distance of 4 inches. Find the average resistance to the pile in cwt., assuming the two to remain in contact. Find in foot-pounds the energy dissipated in one stroke.
A particle is projected under gravity from \(A\) so as to pass through \(B\). Show that for a given velocity of projection there are two paths. Show that if \(B\) has horizontal and vertical coordinates \(x, y\) referred to \(A\), and the velocity of projection is \(\sqrt{2gh}\), the angle between the two paths at \(B\) is a right angle if \(B\) lies on the ellipse \(x^2 + 2y^2 = 2hy\).
A mass is suspended by a light elastic string from a point \(A\) and produces an extension \(c\), the natural length being \(a\). Show that if the mass is raised through a vertical distance less than \(c\) and is then let go from rest, it makes oscillations of the same period as a simple pendulum of length \(c\). If the mass is raised up to \(A\) and let fall, show that the maximum extension of the string is equal to \(c(1+\sec\alpha)\), where \(\alpha\) is the acute angle given by \(\tan\alpha=2\sqrt{a/c}\); and that this maximum extension is attained at a time \((\pi-\alpha)\sqrt{(c/g)}\) after the string first becomes taut.
Find the acceleration of a point describing a circle with variable velocity. Two beads connected by a string are held at rest on a vertical circular wire with the string horizontal, and above the centre. Their masses are \(m, m'\), and the string subtends an angle \(2\alpha\) at the centre. If the beads are released, show that the tension of the string when it makes an angle \(\theta\) with the horizontal is \[ \frac{2mm'g \tan\alpha \cos\theta}{m+m'}. \]
\(ABD, CAE, CBF\) are three circles touching each other at \(A, B, C\). The common tangent at \(C\) passes through \(D\), and \(DAE, DBF\) are straight lines. Prove that \(EF\) touches the circles at \(E\) and \(F\).
Solve the equations \[ \frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 1\frac{2}{3}, \quad \frac{y}{x} + \frac{z}{y} + \frac{x}{z} = -3\frac{2}{3}, \quad xyz = 72. \]
52 different cards (of which 4 are aces) are distributed equally among 4 players. Shew that in nearly three-fifths of the possible distributions one player will have two aces and two other players one ace each.
Solution: This is \begin{align*} && P &= \frac{\binom{4}{2}\binom{48}{11}}{\binom{52}{13}} \cdot \frac{\binom{2}{1}\binom{37}{12}}{\binom{39}{13}}\cdot\frac{\binom{1}{1}\binom{25}{12}}{\binom{26}{13}} \cdot 4 \cdot 3 \\ &&&= \frac{6 \cdot 48! \cdot 13! \cdot 39!}{52! \cdot 11! \cdot 37!} \cdot \frac{2 \cdot 37! \cdot 13! \cdot 26!}{39! \cdot 12! \cdot 25!} \cdot \frac{25! \cdot 13! \cdot 13!}{26! \cdot 12! \cdot 13!} \cdot 4 \cdot 3 \\ &&&= \frac{6 \cdot 13 \cdot 12 }{52 \cdot 51 \cdot 50 \cdot 49 } \cdot 26 \cdot 13 \cdot 4 \cdot 3 \\ &&&= \frac{6 \cdot 13 \cdot 12 }{17\cdot 25\cdot 49 } \cdot 13\\ &&&= \frac{3}{5} \frac{2 \cdot 12 \cdot 13\cdot 13}{5 \cdot 17 \cdot 49}\\ &&&= \frac35 \cdot \frac{4056}{4165} \\ &&&\approx \frac35 \end{align*}