Problems

A uniform triangular table with a leg at each corner \(A, B, C\) is placed on a rough horizontal plane. Show that the pressure at each point of support is equal to one-third the weight of the table. A gradually increasing couple in a horizontal plane is applied to the table until it begins to turn. Show that the point \(I\) about which it begins to turn is such that \[ AI+BI+CI \] is a minimum. Under what circumstances will the table begin to turn about one of the points of support?

Show that, if forces acting along the sides of a tetrahedron are in equilibrium, then they are all zero. Show that a given force may be resolved into six components acting along the sides of a given tetrahedron, and that this resolution is unique.

An aeroplane is flying horizontally at height \(k\) with velocity \(U\). An anti-aircraft gun is situated on the ground at a distance \(h\) from the vertical plane in which the aeroplane is flying. The gun can fire shells with velocity \(V\). Prove that the aeroplane is within range of the gun for a time \[ \frac{2}{gU} (V^4 - 2V^2gk - g^2h^2)^{\frac{1}{2}}, \] provided that \[ g^2h^2+2V^2gk < V^4. \]

A wedge of mass \(M\) and angle \(\alpha\) is sliding along a smooth horizontal plane with velocity \(V\). A smooth uniform sphere of mass \(m\) is dropped vertically and strikes the wedge. Show that if the coefficient of restitution between the wedge and the table is zero and between the sphere and the wedge is \(e\), then the sphere must strike the wedge with velocity \[ \frac{2V(M-me \sin^2\alpha)}{m(1+e)\sin 2\alpha} \] in order to stop the wedge. What happens if the masses of the wedge and of the sphere satisfy the equation \[ M=me \sin^2\alpha? \]

One end \(A\) of a uniform rod \(AB\), of mass \(ml\) and length \(l\), is freely hinged to a horizontal rod of length \(a\). The horizontal rod is forced to rotate with uniform angular velocity \(\omega\). Show that the angle \(\beta\) which the rod \(AB\) makes with the vertical is given by the equation \[ \omega^2(3a+2l \sin\beta) = 3g \tan\beta. \]

The weight of a man, as measured by a spring balance, at the equator is 196 lb. Prove that his weight, as measured by a spring balance, is increased or diminished by about 0.4 oz. if he travels on a train going at 20 m.p.h. along the equator, according as the train travels W. or E. respectively. (Take \(g=32\) ft. per sec. per sec. at the equator.)

A waggon of mass \(M\) carries a simple pendulum of mass \(m\) and length \(l\) which can swing in the direction of motion of the waggon. If \(V\) be the velocity of the waggon and \(\theta\) the inclination of the pendulum to the vertical, measured in a suitable sense, prove that the kinetic energy \(T\) of the system is given by the equation \[ 2T=(M+m)V^2+2mlV\dot{\theta}\cos\theta+ml^2\dot{\theta}^2. \] Show that, if the waggon is jolted into motion with initial velocity \(V\), then the initial value of \(\dot{\theta}\) is equal to the value of \(\omega\) which makes the quadratic form \[ (M+m)V^2+2mlV\omega+ml^2\omega^2 \] a minimum.

A uniform cylinder is pulled over a rough horizontal plane by a force \(P\) making an angle \(\alpha\) with the horizontal and whose direction passes through the axis of the cylinder. Prove that, in order that the cylinder may roll, the force \(P\) must satisfy the inequality \[ P \left( \sin\alpha \sin\phi + \frac{k^2}{a^2+k^2} \cos\alpha \cos\phi \right) < W \sin\phi, \] where \(\phi\) is the angle of friction between the cylinder and the plane, \(a\) is the radius of the cylinder, \(W\) its weight and \(k\) its radius of gyration about its axis.

Coplanar forces of magnitudes \(kA_1A_2, kA_2A_3, \dots, kA_nA_1\) act at the middle points of, and perpendicular to, the sides of a polygon \(A_1A_2\dots A_n\); the polygon is convex and all forces act outwards. If the coordinates of each vertex \(A_r\) are \((x_r, y_r)\) referred to orthogonal Cartesian axes \(Ox, Oy\), find the moment about \(O\) of the force \(kA_rA_{r+1}\), and prove that the system of forces is in equilibrium. If the lines of action of all the forces are rotated in the same direction through an angle \(\alpha\), the points of application being unchanged, find the resultant of the new system.

A uniform solid cube is at rest on a rough plane (coefficient of friction \(\mu\)) inclined at an angle \(\alpha\) to the horizontal, two of the edges of the cube being along lines of greatest slope of the plane. A slowly increasing horizontal force is then applied at the middle point of, and perpendicular to, the highest horizontal edge of the cube so that the cube tends to move up the plane. Find how equilibrium will be broken (i) if \(\mu=0.4\), and (ii) if \(\mu=0.25\); explain why the result in case (ii) does not depend on \(\alpha\).

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Solution:

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On the point of sliding, we have \(F = \mu R\), \begin{align*} \text{N2}(\nearrow): && D\cos \alpha - F-W\sin\alpha &= 0 \\ && D\cos \alpha- W \sin \alpha &= \mu R\\ \text{N2}(\nwarrow): && R - W \cos\alpha - D \sin \alpha &= 0 \\ && R &= W \cos \alpha + D \sin \alpha \\ \Rightarrow && D\cos \alpha- W \sin \alpha &= \mu(W \cos \alpha + D \sin \alpha) \\ && D - W \tan \alpha &= \mu(W + D \tan \alpha) \\ && D(1-\mu\tan \alpha) &= W(\mu + \tan \alpha) \\ \Rightarrow && D &= \left ( \frac{\mu + \tan \alpha}{1 - \mu\tan \alpha} \right) W \\ \end{align*} On the point of toppling \(R\) will be acting at \(X\) \begin{align*} \overset{\curvearrowright}{X}:&& 0 &= D \cos \alpha l - W \cos(45^{\circ}-\alpha) \frac{l}{\sqrt{2}} \\ \Rightarrow && \cos \alpha D &= \frac12 \left ( \cos \alpha + \sin \alpha \right)W \\ \Rightarrow && D &= \frac12 \left ( 1+ \tan \alpha \right)W \end{align*} Therefore we topple before sliding if \(D_{\text{top}} < D_{\text{slide}}\), ie \begin{align*} &&\frac12 \left ( 1+ \tan \alpha \right)W &< \left ( \frac{\mu + \tan \alpha}{1 - \mu\tan \alpha} \right) W \\ \Rightarrow && \frac12 (1+ \tan \alpha) &< \frac{\mu + \tan \alpha}{1 - \mu \tan \alpha}\\ \mu = \tfrac25: && \frac12 (1+ \tan \alpha) &< \frac{2+ 5\tan \alpha}{5 - 2 \tan \alpha}\\ \Rightarrow && (5-2\tan \alpha) (1 + \tan \alpha) & < 4 + 10 \tan \alpha \\ \Rightarrow && 5 +3 \tan \alpha - 2 \tan^2 \alpha & < 4 + 10 \tan \alpha \\ \Rightarrow && 1 -7 \tan \alpha - 2 \tan^2 \alpha & <0 \\ \Rightarrow && \tan \alpha &< \frac{\sqrt{49+8}-7}{4} = \frac{\sqrt{57}-7}{4}\\ \\ \mu = \tfrac14: && \frac12 (1+ \tan \alpha) &< \frac{1 + 4\tan \alpha}{4 -\tan \alpha} \\ && 4+3\tan \alpha-\tan^2 \alpha &<2 + 8 \tan \alpha \\ && 0 &< -2+5\tan \alpha + \tan^2 \alpha \\ && \tan \alpha & > \frac{-5 \pm \sqrt{25+8}}{2} = \frac{\sqrt{33}-5}{2} > \frac14 \end{align*} For the cube to be at rest before the cube starts sliding, we must have \(\tan \alpha < \mu\), therefore for all values of \(\alpha\) the cube always slides