Problems

A particle of weight \(W\) is placed on a rough plane inclined at an angle of \(\theta\) to the horizontal. The coefficient of friction between the particle and the plane is \(\mu\). A horizontal force \(X\) acting on the particle is just sufficient to prevent the particle from sliding down the plane; when a horizontal force \(kX\) acts on the particle, the particle is about to slide up the plane. Both horizontal forces act in the vertical plane containing the line of greatest slope. Prove that \[ \left( k-1 \right) \left( 1 + \mu^2 \right) \sin \theta \cos \theta = \mu \left( k + 1 \right) \] and hence that $\displaystyle k \ge \frac{ \left( 1+ \mu \right)^2} { \left( 1 - \mu \right)^2}$ .

A solid figure is composed of a uniform solid cylinder of density \(\rho\) and a uniform solid hemisphere of density \(3\rho\). The cylinder has circular cross-section, with radius \(r\), and height \(3r\), and the hemisphere has radius \(r\). The flat face of the hemisphere is joined to one end of the cylinder, so that their centres coincide. The figure is held in equilibrium by a force \(P\) so that one point of its flat base is in contact with a rough horizontal plane and its base is inclined at an angle \(\alpha\) to the horizontal. The force \(P\) is horizontal and acts through the highest point of the base. The coefficient of friction between the solid and the plane is \(\mu\). Show that \[\mu \ge \left\vert \tfrac98 -\tfrac12 \cot\alpha\right\vert\,. \]

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Solution: The centre of mass of the sphere will be at \((0, \frac{3}{2}r)\) and the centre of mass of the hemisphere will be at \((0, 3r + \frac38r)\), their masses will be \(3\pi r^3 \cdot \rho \) and \(\frac23 \pi r^3 \cdot 3\rho \), meaning the center of mass will be \(\frac{\frac92r + \frac{27}{8} \cdot 2r}{3 + 2} = \frac{45/4}{5}r = \frac{9}{4}r\) above the center of the base.

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\begin{align*} \text{N2}(\uparrow): && R -mg &= 0 \\ \overset{\curvearrowright}{X}: && P\cdot 2r \sin \alpha + mg (r \cos \alpha -\tfrac94 r\sin \alpha) &= 0 \\ \Rightarrow && P &= mg(\tfrac98 - \tfrac12 \cot \alpha) \\ \text{N2}(\rightarrow): && |F| &= |P| \\ (|F| \leq \mu R): && mg|\tfrac98 - \tfrac12 \cot \alpha| & \leq \mu mg \\ \Rightarrow && |\tfrac98 - \tfrac12 \cot \alpha| &\leq \mu \end{align*}

A painter of weight \(kW\) uses a ladder to reach the guttering on the outside wall of a house. The wall is vertical and the ground is horizontal. The ladder is modelled as a uniform rod of weight \(W\) and length \(6a\). The ladder is not long enough, so the painter stands the ladder on a uniform table. The table has weight \(2W\) and a square top of side \(\frac12 a\) with a leg of length \(a\) at each corner. The foot of the ladder is at the centre of the table top and the ladder is inclined at an angle \(\arctan 2\) to the horizontal. The edge of the table nearest the wall is parallel to the wall. The coefficient of friction between the foot of the ladder and the table top is \(\frac12\). The contact between the ladder and the wall is sufficiently smooth for the effects of friction to be ignored.

1. Show that, if the legs of the table are fixed to the ground, the ladder does not slip on the table however high the painter stands on the ladder.
2. It is given that \(k=9\) and that the coefficient of friction between each table leg and the ground is \(\frac13\). If the legs of the table are not fixed to the ground, so that the table can tilt or slip, determine which occurs first when the painter slowly climbs the ladder.

A long, light, inextensible string passes through a small, smooth ring fixed at the point \(O\). One end of the string is attached to a particle \(P\) of mass \(m\) which hangs freely below \(O\). The other end is attached to a bead, \(B\), also of mass \(m\), which is threaded on a smooth rigid wire fixed in the same vertical plane as \(O\). The distance \(OB\) is \(r\), the distance \(OH\) is \(h\) and the height of the bead above the horizontal plane through~\(O\) is \(y\), as shown in the diagram.

A non-uniform rod \(AB\) has weight \(W\) and length \(3l\). When the rod is suspended horizontally in equilibrium by vertical strings attached to the ends \(A\) and \(B\), the tension in the string attached to \(A\) is \(T\). When instead the rod is held in equilibrium in a horizontal position by means of a smooth pivot at a distance \(l\) from \(A\) and a vertical string attached to \(B\), the tension in the string is \(T\). Show that \(5T = 2W\). When instead the end \(B\) of the rod rests on rough horizontal ground and the rod is held in equilibrium at an angle \(\theta\) to the horizontal by means of a string that is perpendicular to the rod and attached to \(A\), the tension in the string is \(\frac12 T\). Calculate \(\theta\) and find the smallest value of the coefficient of friction between the rod and the ground that will prevent slipping.

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Solution:

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Suppose the centre of mass of the rod is \(x\) away from \(A\). \begin{align*} \overset{\curvearrowleft}{B}: && (3l-x)W - 3lT &= 0 \\ \Rightarrow && x &= \frac{3l(W-T)}{W} \tag{1} \end{align*}

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In the second set up we have: \begin{align*} \overset{\curvearrowleft}{\text{pivot}}: && 2lT - (x-l)W &= 0 \\ \Rightarrow && x &= \frac{2lT + lW}{W} \tag{2} \\ \\ (1) \text{ & } (2): && 3l(W-T) &= l(2T+W) \\ \Rightarrow && 2W &= 5T \end{align*}

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\begin{align*} && x&= \frac{3l(W-T)}{W}\\ &&&= \frac{3l(W - \frac25 W)}{W} \\ &&&= \frac{9}{5}l\\ \overset{\curvearrowleft}{B}: && -\frac12 T (3l \sin \theta) + W \frac{6}{5}l \cos \theta &= 0 \\ \Rightarrow && \tan \theta &= \frac{4}{5} \frac{W}{T} \\ &&&= \frac45 \frac52 \\ &&&= 2 \\ \Rightarrow && \theta &= \tan^{-1} 2 \\ \\ \text{N2}(\uparrow): && R &= W \\ \text{N2}(\rightarrow): && F &= \frac12 T \\ \Rightarrow && F & \leq \mu R \\ \Rightarrow && \frac12 T &\leq \mu W \\ \Rightarrow && \mu &\geq \frac12 \frac{T}{W} = \frac12 \frac25 = \frac15 \end{align*}

Two particles, \(A\) and \(B\), of masses \(m\) and \(2m\), respectively, are placed on a line of greatest slope, \(\ell\), of a rough inclined plane which makes an angle of \(30^{\circ}\) with the horizontal. The coefficient of friction between \(A\) and the plane is \(\frac16\sqrt{3}\) and the coefficient of friction between \(B\) and the plane is \(\frac13 \sqrt{3}\). The particles are at rest with \(B\) higher up \(\ell\) than \(A\) and are connected by a light inextensible string which is taut. A force \(P\) is applied to \(B\).

1. Show that the least magnitude of \(P\) for which the two particles move upwards along \(\ell\) is \(\frac{11}8 \sqrt{3}\, mg\) and give, in this case, the direction in which \(P\) acts.
2. Find the least magnitude of \(P\) for which the particles do not slip downwards along~\(\ell\).

A horizontal spindle rotates freely in a fixed bearing. Three light rods are each attached by one end to the spindle so that they rotate in a vertical plane. A particle of mass \(m\) is fixed to the other end of each of the three rods. The rods have lengths \(a\), \(b\) and \(c\), with \(a > b > c\,\) and the angle between any pair of rods is \(\frac23 \pi\). The angle between the rod of length \(a\) and the vertical is \(\theta\), as shown in the diagram. \vspace*{-0.1in}

Two uniform ladders \(AB\) and \(BC\) of equal length are hinged smoothly at \(B\). The weight of \(AB\) is \(W\) and the weight of \(BC\) is \(4W \). The ladders stand on rough horizontal ground with \(\angle ABC=60^\circ\,\). The coefficient of friction between each ladder and the ground is \(\mu\). A decorator of weight \(7W\) begins to climb the ladder \(AB\) slowly. When she has climbed up \(\frac13\) of the ladder, one of the ladders slips. Which ladder slips, and what is the value of \(\mu\)?

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Solution:

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\begin{align*} \text{N2}(\rightarrow): && F_A - F_C &= 0\\ && F_A &= F_C \\ \text{N2}(\uparrow): && R_A + R_C - 7W - W - 4W &= 0\\ && R_A + R_C &= 12W \\ \overset{\curvearrowright}{A}: && \frac{1}{6}7W + \frac{1}{4}W + \frac{3}{4}4W - R_C &= 0 \\ \Rightarrow && R_C &= \frac{53}{12}W\\ \Rightarrow && R_A = 12W - \frac{53}{12}W &= \frac{91}{12}W \\ \overset{\curvearrowleft}{B}(AB): && \frac{1}{2}W + \frac{2}{3}7W - R_A+\sqrt{3}F_A &= 0 \\ \Rightarrow && F_A = \frac{1}{\sqrt{3}} \l \frac{91}{12}-\frac12-\frac{14}3\r W &= \frac{29}{12\sqrt{3}}W \end{align*} We know that the system is about to slip, so equality holds in one of \(F_A \leq \mu R_A\) or \(F_C \leq \mu R_C\). Since \(F_A = F_C\), we know it must occur for whichever of \(\mu R_A\) and \(\mu R_C\) is smaller. Since \(R_C\) is much smaller, this must be the ladder about to slip \(BC\) and \[ \mu = \frac{F_C}{R_C} = \frac{\frac{29}{12\sqrt{3}}W}{\frac{53}{12}W} = \boxed{\frac{29}{53\sqrt{3}}}\]

The base of a non-uniform solid hemisphere, of mass \(M,\) has radius \(r.\) The distance of the centre of gravity, \(G\), of the hemisphere from the base is \(p\) and from the centre of the base is \(\sqrt{p^2 + q^2} \,\). The hemisphere rests in equilibrium with its curved surface on a horizontal plane. A particle of mass \(m,\,\) where \(m\) is small, is attached to \(A\,\), the lowest point of the circumference of the base. In the new position of equilibrium, find the angle, \(\alpha\), that the base makes with the horizontal. The particle is removed and attached to the point \(B\) of the base which is at the other end of the diameter through \(A\,\). In the new position of equilibrium the base makes an angle \({\beta}\) with the horizontal. Show that $$\tan(\alpha-\beta)= \frac{2mMrp} {M^2\left(p^2+q^2\right)-m^2r^2}\;.$$

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Solution:

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In the coordinate system where \((0,0)\) is the centre base of the hemisphere, \(G\) is at \((p, q)\). Once the mass is attached at \(A\), the new centre of mass will satisfy: \(M \begin{pmatrix} p \\ q \end{pmatrix} + m \begin{pmatrix} r \\ 0 \end{pmatrix} = (M+m)\bar{x} \Rightarrow \bar{x} = \frac{1}{M+m} \begin{pmatrix} Mp+mr \\ Mq \end{pmatrix}\) The angle between the horizontal and \(AB\), \(\alpha\) will satisfy: $$\tan \alpha = \frac{Mp + mr}{Mq}$$ Similarly, when the mass is attached at \(B\), the new centre of mass will satisfy: \(M \begin{pmatrix} p \\ q \end{pmatrix} + m \begin{pmatrix} -r \\ 0 \end{pmatrix} = (M+m)\bar{x} \Rightarrow \bar{x} = \frac{1}{M+m} \begin{pmatrix} Mp-mr \\ Mq \end{pmatrix}\) The angle between the horizontal and \(AB\), \(\beta\) will satisfy: $$\tan \beta = \frac{Mp - mr}{Mq}$$ We are trying to find: \begin{align*} \tan \l \alpha - \beta \r &= \frac{\tan \alpha - \tan \beta}{1+ \tan \alpha \tan \beta} \\ &= \frac{\frac{Mp + mr}{Mq} - \frac{Mp - mr}{Mq}}{1 + \frac{Mp + mr}{Mq} \frac{Mp - mr}{Mq}} \\ &= \frac{(Mp + mr)Mq - (Mp - mr)Mq}{M^2q^2 + (Mp + mr)(Mp - mr)} \\ &= \frac{2Mmrp}{M^2(q^2+p^2) -m^2r^2} \\ \end{align*}

A circular hoop of radius \(a\) is free to rotate about a fixed horizontal axis passing through a point \(P\) on its circumference. The plane of the hoop is perpendicular to this axis. The hoop hangs in equilibrium with its centre, \(O\), vertically below \(P\). The point \(A\) on the hoop is vertically below \(O\), so that \(POA\) is a diameter of the hoop. A mouse \(M\) runs at constant speed \(u\) round the rough inner surface of the lower part of the hoop. Show that the mouse can choose its speed so that the hoop remains in equilibrium with diameter \(POA\) vertical. Describe what happens to the hoop when the mouse passes the point at which angle \(AOM = 2 \arctan \mu\,\), where \(\mu\) is the coefficient of friction between mouse and hoop.

\(ABCD\) is a uniform rectangular lamina and \(X\) is a point on \(BC\,\). The lengths of \(AD\), \(AB\) and \(BX\) are \(p\,\), \(q\) and \(r\) respectively. The triangle \(ABX\) is cut off the lamina. Let \((a,b)\) be the position of the centre of gravity of the lamina, where the axes are such that the coordinates of \(A\,\), \(D\) and \(C\) are \((0,0)\,\), \((p,0)\) and \((p,q)\) respectively. Derive equations for \(a\) and \(b\) in terms of \(p\,\), \(q\) and \(r\,\). When the resulting trapezium is freely suspended from the point \(A\,\), the side \(AD\) is inclined at \(45^\circ\) below the horizontal. Show that \(\displaystyle r = q - \sqrt{q^2 - 3pq + 3p^2}\,\). You should justify carefully the choice of sign in front of the square root.

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Solution:

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\begin{array}{c|c|c|c} & ABX & ABCD & AXCD \\ \hline \text{area} & \frac12 q r & pq & q(p - \frac12 r) \\ \text{com} & \binom{\frac{r}{3}}{\frac{2q}{3}} & \binom{p/2}{q/2} & \binom{a}{b} \end{array} \begin{align*} && q(p-\frac12 r) \binom{a}{b} &= pq\binom{p/2}{q/2} - \frac12 q r \binom{\frac{r}{3}}{\frac{2q}{3}} \\ \Rightarrow && \binom{a}{b} &= \frac{2}{2p-r}\binom{p^2/2-\frac16r^2}{pq/2-\frac13qr} \\ &&&= \binom{\frac{p^2-\frac13 r^2}{2p-r}}{\frac{pq-\frac23qr}{2p-r}} \end{align*}

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We must have: \begin{align*} && 1 &= \frac{p^2-\frac13r^2}{pq-\frac23qr} \\ \Rightarrow && pq-\frac23qr &= p^2 - \frac13 r^2 \\ \Rightarrow && 0 &=r^2-2q r + 3p(q-p) \\ \Rightarrow && 0 &= (r-q)^2 -q^2+3pq-3p^2 \\ \Rightarrow && r&= q \pm \sqrt{q^2-3pq+3p^2} \end{align*} Suppose \(r > q\), then \(p > q > r\) and we have a shape which looks like this

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which definitely wouldn't have \(G\) hanging below \(A\).

\(AB\) is a uniform rod of weight \(W\,\). The point \(C\) on \(AB\) is such that \(AC>CB\,\). The rod is in contact with a rough horizontal floor at \(A\,\) and with a cylinder at \(C\,\). The cylinder is fixed to the floor with its axis horizontal. The rod makes an angle \({\alpha}\) with the horizontal and lies in a vertical plane perpendicular to the axis of the cylinder. The coefficient of friction between the rod and the floor is \(\tan \lambda_1\) and the coefficient of friction between the rod and the cylinder is \(\tan \lambda_2\,\). Show that if friction is limiting both at \(A\) and at \(C\), and \({\alpha} \ne {\lambda}_2 - {\lambda}_1\,\), then the frictional force acting on the rod at \(A\) has magnitude $$ \frac{ W\sin {\lambda}_1 \, \sin({\alpha}-{\lambda}_2)} {\sin ({\alpha}+{\lambda}_1-{\lambda}_2)} \;.$$ %and that %$$ %p=\frac{\cos{\alpha} \, \sin({\alpha}+{\lambda}_1-{\lambda}_2)} %{2\cos{\lambda}_1 \, \sin {\lambda}_2}\;. %$$

A particle \(P\) of mass \(m\) is constrained to move on a vertical circle of smooth wire with centre~\(O\) and of radius \(a\). \(L\) is the lowest point of the circle and \(H\) the highest and \(\angle LOP = \theta\,\). The particle is attached to \(H\) by an elastic string of natural length \(a\) and modulus of elasticity~\(\alpha mg\,\), where \(\alpha > 1\,\). Show that, if \(\alpha>2\,\), there is an equilibrium position with \(0<\theta<\pi\,\). Given that \(\alpha =2+\sqrt 2\,\), and that \(\displaystyle \theta = \tfrac{1}{2}\pi + \phi\,\), show that \[ \ddot{\phi} \approx -\frac{g (\sqrt2+1)}{2a }\, \phi \] when \(\phi\) is small. For this value of \(\alpha\), explain briefly what happens to the particle if it is given a small displacement when \( \theta = \frac{1}{2}\pi\).

\(\,\) \vspace*{-0.5in}

1. If the normal reactions on the lower and higher wheels of the lorry are equal, show that the sum of the frictional forces between the wheels and the ground is zero.
2. If \(P\) is such that the lorry does not tip over (but the normal reactions on the lower and higher wheels of the lorry need not be equal), show that \[ W\tan(\alpha - \beta) \le P \le W\tan(\alpha+\beta)\;, \] where \(\tan\beta = d/h\,\).

A rigid straight beam \(AB\) has length \(l\) and weight \(W\). Its weight per unit length at a distance \(x\) from \(B\) is \(\alpha Wl^{-1} (x/l)^{\alpha-1}\,\), where \(\alpha\) is a positive constant. Show that the centre of mass of the beam is at a distance \(\alpha l/(\alpha+1)\) from \(B\). The beam is placed with the end \(A\) on a rough horizontal floor and the end \(B\) resting against a rough vertical wall. The beam is in a vertical plane at right angles to the plane of the wall and makes an angle of \(\theta\) with the floor. The coefficient of friction between the floor and the beam is \(\mu\) and the coefficient of friction between the wall and the beam is also \(\mu\,\). Show that, if the equilibrium is limiting at both \(A\) and \(B\), then \[ \tan\theta = \frac{1-\alpha \mu^2}{(1+\alpha)\mu}\;. \] Given that \(\alpha =3/2\,\) and given also that the beam slides for any \(\theta<\pi/4\,\) find the greatest possible value of \(\mu\,\).