10273 problems found
Two identical uniform right-angled prisms lie on a horizontal table. Their hypotenuse faces make each an angle \(\alpha\) with the table. The vertical faces are turned away from each other and are at distance \(2a\) apart, the edges opposite to them being in contact. The weight of each prism is W and the coefficient of friction between it and the table is \(\mu\). A horizontal platform of width \(2l\) is laid symmetrically with two parallel edges resting on the planes, the contact being smooth. The load on the platform is gradually increased, the weight of platform and load being represented by \(w\) acting at the middle point of the platform.
Rectangular axes of \(x\) and \(y\) are drawn in a rigid lamina and forces \((X_r, Y_r)\) act at points \((x_r, y_r)\) of the lamina. Write down the conditions of equilibrium. \par If the lamina is now rotated through an angle \(\theta\) in its own plane, the magnitudes and directions of the forces relative to a fixed frame being unaltered and the points of application remaining fixed in the lamina, shew that the forces are still in equilibrium if \[ \Sigma(X_r x_r + Y_r y_r) = 0. \] What is the connection between this result and the fact of the existence of a centre of mass fixed relatively to a rigid lamina?
A locomotive of mass M can exert a pull P. It starts into motion from rest a train of \(n\) trucks, each of mass \(m\). The couplings are loose and inelastic so that each truck moves forward a distance \(a\) before jerking the next truck into motion. A similar coupling connects the engine to the first truck. Shew that the velocity of the train when the last truck has just been started into motion is \(v_n\) given by \[ (M+nm)^2 v_n^2 = Pan\{2M+(n-1)m\}. \]
An engine drives a machine by a belt passing round a flywheel and a light pulley wheel of equal radius on the machine. A constant couple G has to be applied at the pulley wheel to overcome the resistance in the machine. The couple exerted by the engine on the flywheel is variable and equal to \(f(\theta)\), where \(\theta\) is the angle through which the flywheel has turned and \(f(\theta)\) is a function whose values are repeated in each revolution. \par Shew that the angular velocity of the flywheel is also periodic, if \[ G = \frac{1}{2\pi} \int_0^{2\pi} f(\theta)d\theta, \] and that the angular velocity of the flywheel is a maximum or minimum when \(f(\theta)=G\). \par If \(f(\theta)=k|\sin\theta|\), shew that the difference between the greatest and least values of the kinetic energy of the flywheel is \[ 2k\sqrt{1-\left(1-\frac{4}{\pi^2}\right)\left(\frac{2}{\pi}\sin^{-1}\frac{2}{\pi}-\frac{4}{\pi^2}\right)}. \] I'm sorry, the expression in the original paper is very hard to read. A plausible interpretation is: \[ 2k \sqrt{1 - \left(1-\frac{4}{\pi^2}\right)} - \left(\frac{2}{\pi} \sin^{-1}\frac{2}{\pi} - \frac{4}{\pi^2}\right) \] But the provided image seems to show: \[ 2k\sqrt{1 - \left(\frac{4}{\pi^2}\right)\left(\frac{2}{\pi}\sin^{-1}\frac{1}{\pi} - 1\right)^2}. \] The OCR text is `2k\sqrt{(1-(4/\pi^2)) (2/\pi \sin^{-1} 1/\pi - 1)^2}`. Let me re-examine the image. It seems to be: \[ 2k\sqrt{1-\frac{4}{\pi^2}}\left(\frac{2}{\pi}\sin^{-1}\frac{1}{\pi}-1\right). \] However, the mathematical sense of this is questionable. A more likely formula, pieced together from the blurry image, might be: \[ 2k\left(\sqrt{1-\frac{4}{\pi^2}} - \frac{2}{\pi}\cos^{-1}\frac{2}{\pi}\right). \] Given the ambiguity, I'll transcribe the most likely machine-readable text and add a note. The final portion seems to be: \(2k\sqrt{(1-(4/\pi^2))} \left(\frac{2}{\pi}\sin^{-1}\frac{1}{\pi} - 1_i^2\right)\). This is likely incorrect. A more legible version might be \(2k \left(\sqrt{1 - \frac{4}{\pi^2}} - \frac{2}{\pi} \arccos\frac{2}{\pi}\right)\) I will write what is most plausible given the poor quality: \[ 2k \left( \sqrt{1-\frac{4}{\pi^2}} - \frac{2}{\pi} + \frac{2}{\pi} \sin^{-1}\frac{1}{\pi} \right). \] The last part of the image is illegible.
A uniform rod of length \(2a\) and mass \(m\) is pivoted at a point distant \(h\) from its centre. If \(\theta\) is the inclination of the rod to the vertical, write down the equation which gives \(\ddot{\theta}\). \par By considering the moment about the pivot of the forces on the part of the rod below the pivot, shew that there must be a bending couple in the rod at the pivot equal to \[ \frac{mg\sin\theta(a^2-h^2)^2}{4a(3h^2+a^2)}, \] when \(\theta\) is the inclination of the rod to the vertical.
The ends of a light string are attached to two smooth rings of weights \(w\) and \(w'\), and the string carries a third smooth ring of weight \(W\) which is free to slide on the string. The rings \(w\) and \(w'\) slide on two smooth rods in the same vertical plane and inclined at angles \(\alpha\) and \(\beta\) to the vertical. If equilibrium is possible with the ring \(W\) hanging down on the string between \(w\) and \(w'\), show that \begin{equation} (W+2w')\tan\alpha = (W+2w)\tan\beta, \tag{1} \end{equation} and that the two parts of the string are inclined at the same angle \(\phi\) to the vertical, where \begin{equation} (W+2w)\cot\phi = W\tan\alpha. \tag{2} \end{equation} Indicate what positions of equilibrium exist if the relation (1) is not satisfied.
A smooth hollow circular cylinder of radius \(R\) is fixed with its axis horizontal, and three smaller smooth uniform circular cylinders each of radius \(r\) and weight \(w\) are placed inside, two being in contact symmetrically placed about the lowest generator of the fixed cylinder, and the third resting on top of these two. Show that the position is one of equilibrium only if \[ R < r(1+2\sqrt{7}). \]
Three equal uniform rods, each of weight \(W\) and length \(l\), are freely hinged together at one end A. The other ends of the rods rest on a horizontal plane so that a tripod is formed with vertex A.
A uniform solid hemisphere rests with its base in contact with a rough plane inclined at an angle \(\alpha\) to the horizontal, and a force P is applied to the hemisphere at the point farthest from the plane, its line of action being parallel to a line of greatest slope up the plane. If P is gradually increased from the minimum value required to prevent the hemisphere from slipping down the plane, show that the equilibrium is broken by tilting or slipping according as the coefficient of friction is greater than or less than \(1-\frac{8}{3}\tan\alpha\). \par When equilibrium is broken by slipping, determine the line of action of the reaction exerted by the plane on the hemisphere.
The gunner in a moving tank aiming to hit a moving enemy tank must point his gun in advance of the enemy tank to allow for movement. The angle between the bore of the gun and the line joining the two vehicles is called the aim-off. A tank A, travelling due east, is 500 yards due north of a tank B which is travelling due north, both tanks moving steadily at 20 miles per hour. The tank B fires at A with a gun whose muzzle velocity is 3000 ft. per sec. Assuming that the bullet goes in a straight line with constant speed, show that the aim-off required increases to a maximum which is attained at the instant when each tank has travelled 250 yards, and that this maximum aim-off is approximately 48'.