A rhombus of uniform rods \(ABCD\) freely jointed together rests symmetrically with \(AC\) horizontal, the rods \(AB, BC\) resting on two smooth pegs at \(E\) and \(F\). By means of the principle of virtual work or otherwise shew that in equilibrium the vertical through \(A\) and the horizontal through \(B\) meet on the normal at \(E\) to \(AB\). Find also the reactions at \(A\), \(B\) and \(E\).
A uniform circular cylinder rests on a rough horizontal plane (coefficient of friction \(\mu_1\)). A second cylinder rests partly on the former, touching it along one generator (the coefficient of friction between them being \(\mu_2\)), and partly on an inclined plane of inclination \(2\beta\), touching this along another generator (the coefficient of friction here being \(\mu_3\)). The plane through the axes of the cylinders makes with the vertical an angle \(2\alpha\). Shew that for equilibrium to be possible \[ \mu_3 > \tan\beta, \quad \mu_2 > \tan\alpha, \] and \[ \mu_1 \ge \frac{\tan\beta\tan\alpha}{\tan\beta+\frac{W}{W'}(\tan\alpha+\tan\beta)}, \] where \(W\) and \(W'\) are the weights of the upper and lower cylinders, respectively.
The corners \(A, B, C, D\) of a rigid rectangular platform are attached to and rest in a horizontal plane on four vertical springs of slight compressibility, the compressions of which for unit load are \(\lambda_1, \lambda_2, \lambda_3, \lambda_4\) respectively. Shew that a weight \(P\) placed on the platform at \(A\) will produce loads on the four springs given by \[ \frac{P_1}{\lambda_2+\lambda_3+\lambda_4} = \frac{P_2}{-\lambda_3} = \frac{P_3}{\lambda_4} = \frac{P_4}{-\lambda_2} = \frac{P}{\lambda_1+\lambda_2+\lambda_3+\lambda_4}. \] Note: The image for the above relations is blurry, this is a best-effort transcription. Hence shew that a weight \(W\) placed on the platform at a point \(O\), the distances of which from the parallel sides \(AB, DC\) are \(a_1, a_2\) and from the sides \(AD, BC\) are \(b_1, b_2\), will produce loads on the springs equal to \[ \frac{Wa_2b_2}{ab}-w, \quad \frac{Wa_2b_1}{ab}+w, \quad \frac{Wa_1b_1}{ab}-w, \quad \frac{Wa_1b_2}{ab}+w, \] where \[ w = \frac{W}{W_r} = \frac{\lambda_1\lambda_3 a_2 b_2 - \lambda_2\lambda_4 a_1 b_1 + \lambda_3\lambda_4 a_1 b_2 - \lambda_1\lambda_2 a_2 b_1}{(\lambda_1+\lambda_2+\lambda_3+\lambda_4)ab}, \] where \(a=a_1+a_2\) and \(b=b_1+b_2\).
You are given a number of unequal particles and a number of unequal pieces of elastic string. Explain how, from a knowledge of the accelerations produced in the particles by means of the strings and without any prior assumptions as to mass or force, to establish Newton's Law of Motion, \(P=Mf\), and Hooke's Law for an elastic string.
Shew that the time of swing of a simple pendulum is independent of the amplitude if the cube of the ratio of the amplitude to the length is neglected. A pendulum of length 32 feet is drawn aside a distance of 1 foot and the bob is then projected towards the position of equilibrium with a velocity of 1 foot per second. Find the point at which the bob will first come to rest and the time from the moment of projection to that point.
A pile weighing 3 tons is driven into the ground by the falling of a weight of 1 ton from a height of 6 feet. At each blow it is driven in a distance of 2 inches. The impact is inelastic. Shew that the resistance to the movement of the pile is equal to a weight of 13 tons. If there were a coefficient of elasticity \(\frac{1}{4}\), the resistance being unaltered, shew that after the first impact the pile would come to rest before a second impact, and find by how much the pile would be driven in at the first impact.
The total mass of a train is 384 tons and the maximum tractive force exerted by the engine at its wheels is 12 tons wt. This force is exerted until the speed of the train from rest reaches 20 feet per second, after which the engine exerts a constant horse power at its wheels. The resistance of the train at various speeds is given by the following table: \begin{center{tabular}{lcccccc} Speed in ft. per sec. & 0 & 5 & 20 & 40 & 60 & 80 \\ Resistance in tons wt. & 2.5 & 0.8 & 1 & 1.4 & 2 & 2.6 \end{{tabular} Shew by means of a curve the acceleration of the train at various speeds. Estimate the time taken to reach a speed of 80 feet per second.
A particle hangs by an inelastic string of length \(a\) from a fixed point, and a second particle of the same mass hangs from the first by an equal string. The whole moves with uniform angular velocity \(\omega\) about the vertical through the point of suspension, the strings making constant angles \(\alpha\) and \(\beta\) with the vertical. Shew that \[ \tan\alpha = \frac{a\omega^2}{g}(\sin\alpha + \frac{1}{2}\sin\beta), \] and \[ \tan\beta = \frac{a\omega^2}{g}(\sin\alpha + \sin\beta). \] Hence shew that if \(\alpha\) and \(\beta\) are small such a steady motion is only possible if \(a\omega^2\) has one of the values \[ (2\pm\sqrt{2})g \] and that \(\beta/\alpha = \pm\sqrt{2}\).
Two particles of masses \(m\) and \(m'\) travelling in the same straight line collide. Shew that the impulse \(I\) between them is given by \[ I\left(\frac{1}{m}+\frac{1}{m'}\right) = U+U', \] where \(U\) is the relative velocity of approach before the impulse and \(U'\) the relative velocity with which they separate. Shew also that the loss of kinetic energy is \(\frac{1}{2}I(U-U')\), and express this in terms of the initial motion and the coefficient of elasticity. Two pendulums of equal length have small spherical bobs of masses \(m\) and \(m'\) which hang in contact with one another. The bob \(m\) is drawn aside through an angle \(\theta\) and allowed to fall so as to strike the second, which comes to rest after turning through an angle \(\theta'\). Shew that the coefficient of elasticity is \(\frac{(m+m')\sin\frac{1}{2}\theta'}{m\sin\frac{1}{2}\theta}-1\).
A flywheel of moment of inertia \(I\) is set in motion from rest by a constant couple \(G\), there being a frictional resistance equal to \(\mu\omega^2\), where \(\omega\) is the angular velocity. Shew that the time taken to attain the angular velocity \(\omega\) is \[ \frac{I\omega_0}{G}\tanh^{-1}\frac{\omega}{\omega_0}, \] where \(\omega_0\) is the limiting velocity \((G/\mu)^{\frac{1}{2}}\). Shew also that the angle turned through from rest is \[ -\frac{I}{2\mu}\log\left(1-\frac{\omega^2}{\omega_0^2}\right). \]