Problems

A uniform rectangular lamina \(ABCD\) rests in equilibrium in a vertical plane with the \(A\) in contact with a rough vertical wall. The plane of the lamina is perpendicular to the wall. It is supported by a light inextensible string attached to the side \(AB\) at a distance \(d\) from \(A\). The other end of the string is attached to a point on the wall above \(A\) where it makes an acute angle \(\theta\) with the downwards vertical. The side \(AB\) makes an acute angle \(\phi\) with the upwards vertical at \(A\). The sides \(BC\) and \(AB\) have lengths \(2a\) and \(2b\) respectively. The coefficient of friction between the lamina and the wall is \(\mu\).

1. Show that, when the lamina is in limiting equilibrium with the frictional force acting upwards, \begin{equation} d\sin(\theta +\phi) = (\cos\theta +\mu \sin\theta)(a\cos\phi +b\sin\phi)\,. \tag{\(*\)} \end{equation}
2. How should \((*)\) be modified if the lamina is in limiting equilibrium with the frictional force acting downwards?
3. Find a condition on \(d\), in terms of \(a\), \(b\), \(\tan\theta\) and \(\tan\phi\), which is necessary and sufficient for the frictional force to act upwards. Show that this condition cannot be satisfied if \(b(2\tan\theta+ \tan \phi) < a\).

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Solution:

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1. \begin{align*} \text{N2}(\uparrow): && T \cos \theta + F -W &= 0 \\ && W &= T\cos \theta + \mu R \tag{1} \\ \text{N2}(\rightarrow): && R-T\sin \theta &= 0 \\ && R &= T \sin \theta \tag{2}\\ \\ (1)+(2): && W&=(\cos \theta + \mu \sin \theta)T \tag{3} \\ \overset{\curvearrowright}{A}: && 0 &= W(b\sin \phi + a \cos \phi) - Td\sin(\phi+\theta) \tag{4} \\ \\ (3)+(4): && 0 &= (\cos \theta + \mu \sin \theta)(b\sin \phi + a \cos \phi)-d\sin(\phi+\theta) \\ \Rightarrow && d\sin(\phi+\theta) &= (\cos \theta + \mu \sin \theta)(b\sin \phi + a \cos \phi) \end{align*} as required.
2. If \(F\) is operating downwards, it's equivalent to \(-\mu\), ie: \[d\sin(\phi+\theta) = (\cos \theta - \mu \sin \theta)(b\sin \phi + a \cos \phi)\]
3. For the frictional force to be acting upwards, we need \begin{align*} && d\sin(\phi+\theta) &\geq \cos \theta(b\sin \phi + a \cos \phi) \\ \Rightarrow && d &\geq \frac{\cos \theta(b\sin \phi + a \cos \phi)}{\sin(\phi + \theta)} \\ &&&= \frac{\cos \theta(b\sin \phi + a \cos \phi)}{\sin\phi \cos\theta+\cos\phi\sin \theta)}\\ &&&= \frac{(b\sin \phi + a \cos \phi)}{\sin\phi+\cos \phi \tan \theta)}\\ &&&= \frac{a+b\tan \phi}{\tan\theta+\tan\phi }\\ \end{align*} We know that \(d < 2b\), so \begin{align*} && 2b &>\frac{a+b\tan \phi}{\tan\theta+\tan\phi }\\ \Rightarrow && 2b \tan \theta + 2b \tan \phi &> a + b \tan \phi \\ \Rightarrow &&b(2 \tan \theta + \tan \phi) &> a\\ \end{align*} Therefore we will have problems if the inequality is reversed!

The diagram shows three identical discs in equilibrium in a vertical plane. Two discs rest, not in contact with each other, on a horizontal surface and the third disc rests on the other two. The angle at the upper vertex of the triangle joining the centres of the discs is \(2\theta\).

1. Show that the normal reaction between the horizontal surface and a disc in contact with the surface is \(\frac32 W\,\).
2. Find the normal reaction between two discs in contact and show that the magnitude of the frictional force between two discs in contact is \(\dfrac{W\sin\theta}{2(1+\cos\theta)}\,\).
3. Show that if \(\mu <2- \surd3\,\) there is no value of \(\theta\) for which equilibrium is possible.

A hollow circular cylinder of internal radius \(r\) is held fixed with its axis horizontal. A uniform rod of length \(2a\) (where \(a < r\)) rests in equilibrium inside the cylinder inclined at an angle of \(\theta\) to the horizontal, where \(\theta\ne0\). The vertical plane containing the rod is perpendicular to the axis of the cylinder. The coefficient of friction between the cylinder and each end of the rod is \(\mu\), where \(\mu > 0\). Show that, if the rod is on the point of slipping, then the normal reactions \(R_1\) and \(R_2\) of the lower and higher ends of the rod, respectively, on the cylinder are related by \[ \mu(R_1+R_2) = (R_1-R_2)\tan\phi \] where \(\phi\) is the angle between the rod and the radius to an end of the rod. Show further that \[ \tan\theta = \frac {\mu r^2}{r^2 -a^2(1+\mu^2)}\,. \] Deduce that \(\lambda <\phi \), where \(\tan\lambda =\mu\).

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Solution:

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Let \(M\) be the midpoint of \(AB\), then \begin{align*} \overset{\curvearrowright}{M}: && R_1 \sin \phi-\mu R_1 \cos \phi &= R_2 \sin \phi+\mu R_2 \cos \phi \\ \Rightarrow && (R_1-R_2) \tan \phi &= \mu(R_1+R_2) \end{align*} As required. \begin{align*} && \cos \phi = \frac{a}{r} &,\,\, \sin \phi = \frac{\sqrt{r^2-a^2}}{r} \\ \text{N2}(\rightarrow): && R_1\cos(\phi + \theta)+\mu R_1 \sin(\phi + \theta) &= R_2 \cos(\theta - \phi) + \mu R_2 \sin(\theta - \phi) \\ \Rightarrow && R_1(\cos \theta \cos \phi - \sin \theta \sin \phi)+ \mu R_1 (\sin \theta \cos \phi + \cos \theta \sin \phi) &= R_2 (\cos\theta \cos \phi + \sin \theta \sin \phi)+ \mu R_2 (\sin \theta \cos \phi - \cos \theta \sin \phi) \\ && R_1 (1 - \tan \theta \tan \phi)+\mu R_1 (\tan \theta + \tan \phi) &= R_2(1 + \tan \theta \tan \phi) +\mu R_2 (\tan \theta - \tan \phi) \\ && 0 &= (R_1-R_2)(1+\mu \tan \theta)+(R_1+R_2)(-\tan \theta \tan\phi+\mu \tan \phi) \\ \Rightarrow && \frac{R_1+R_2}{R_1-R_2} &= \frac{1+\mu \tan \theta}{\tan \phi (\tan \theta - \mu))} \\ \Rightarrow && \frac{\tan \phi}{\mu} &= \frac{1+\mu \tan \theta}{\tan \phi (\tan \theta - \mu))} \\ \Rightarrow && \tan^2 \phi &= \frac{\mu(1+\mu \tan \theta)}{\tan \theta - \mu} \\ \Rightarrow && \frac{r^2-a^2}{a^2} &= \frac{\mu(1+\mu \tan \theta)}{\tan \theta - \mu} \\ \Rightarrow && \tan \theta (r^2-a^2-a^2\mu^2) &= \mu a^2+\mu(r^2-a^2) \\ \Rightarrow && \tan \theta &= \frac{\mu r^2}{r^2-(1+\mu^2)a^2} \end{align*} Since \(\mu r^2 > 0\) we must also have \(r^2 > a^2(1+\mu^2)\) ie \(\\sec^2 \phi > 1 + \mu^2 = \sec^2 \lambda\) and the result follows.

A pulley consists of a disc of radius \(r\) with centre \(O\) and a light thin axle through \(O\) perpendicular to the plane of the disc. The disc is non-uniform, its mass is \(M\) and its centre of mass is at \(O\). The axle is fixed and horizontal. Two particles, of masses \(m_1\) and \(m_2\) where \(m_1>m_2\), are connected by a light inextensible string which passes over the pulley. The contact between the string and the pulley is rough enough to prevent the string sliding. The pulley turns and the vertical force on the axle is found, by measurement, to be~\(P+Mg\).

1. The moment of inertia of the pulley about its axle is calculated assuming that the pulley rotates without friction about its axle. Show that the calculated value is \[ \frac{((m_1 + m_2)P - 4m_1m_2g)r^2} {(m_1 + m_2)g - P}\,. \tag{\(*\)}\]
2. Instead, the moment of inertia of the pulley about its axle is calculated assuming that a couple of magnitude \(C\) due to friction acts on the axle of the pulley. Determine whether this calculated value is greater or smaller than \((*)\). Show that \(C<(m_1-m_2)rg\).

A thin non-uniform bar \(AB\) of length \(7d\) has centre of mass at a point \(G\), where \(AG=3d\). A light inextensible string has one end attached to \(A\) and the other end attached to \(B\). The string is hung over a smooth peg \(P\) and the bar hangs freely in equilibrium with \(B\) lower than~\(A\). Show that \[ 3\sin\alpha = 4\sin\beta\,, \] where \(\alpha\) and \(\beta\) are the angles \(PAB\) and \(PBA\), respectively. Given that \(\cos\beta=\frac45\) and that \(\alpha\) is acute, find in terms of \(d\) the length of the string and show that the angle of inclination of the bar to the horizontal is \(\arctan \frac17\,\).

A thin uniform circular disc of radius \(a\) and mass \(m\) is held in equilibrium in a horizontal plane a distance \(b\) below a horizontal ceiling, where \(b>2a\). It is held in this way by \(n\) light inextensible vertical strings, each of length \(b\); one end of each string is attached to the edge of the disc and the other end is attached to a point on the ceiling. The strings are equally spaced around the edge of the disc. One of the strings is attached to the point \(P\) on the disc which has coordinates \((a,0,-b)\) with respect to cartesian axes with origin on the ceiling directly above the centre of the disc. The disc is then rotated through an angle \(\theta\) (where \(\theta<\pi\)) about its vertical axis of symmetry and held at rest by a couple acting in the plane of the disc. Show that the string attached to~\(P\) now makes an angle \(\phi\) with the vertical, where \[ b\sin\phi = 2a \sin\tfrac12 \theta\,. \] Show further that the magnitude of the couple is \[ \frac {mga^2\sin\theta}{\sqrt{b^2-4a^2\sin^2 \frac12\theta \ } \ }\,. \] The disc is now released from rest. Show that its angular speed, \(\omega\), when the strings are vertical is given by \[ \frac{a^2\omega^2}{4g} = b-\sqrt{b^2 - 4a^2\sin^2 \tfrac12\theta \;}\,. \]

A uniform rod \(AB\) of length \(4L \) and weight \(W\) is inclined at an angle \(\theta\) to the horizontal. Its lower end \(A\) rests on a fixed support and the rod is held in equilibrium by a string attached to the rod at a point \(C\) which is \(3L \) from \(A\). The reaction of the support on the rod acts in a direction \(\alpha\) to \(AC\) and the string is inclined at an angle \(\beta\) to \(CA\). Show that \[ \cot\alpha = 3\tan \theta + 2 \cot \beta\,. \] Given that \(\theta =30^\circ\) and \(\beta = 45^\circ\), show that \(\alpha= 15^\circ\).

A straight uniform rod has mass \(m\). Its ends \(P_1\) and \(P_2\) are attached to small light rings that are constrained to move on a rough rigid circular wire with centre \(O\) fixed in a vertical plane, and the angle \(P_1OP_2\) is a right angle. The rod rests with \(P_1\) lower than \(P_2\), and with both ends lower than \(O\). The coefficient of friction between each of the rings and the wire is \(\mu\). Given that the rod is in limiting equilibrium (i.e. \ on the point of slipping at both ends), show that \[ \tan \alpha = \frac{1-2\mu -\mu^2}{1+2\mu -\mu^2}\,, \] where \(\alpha\) is the angle between \(P_1O\) and the vertical (\(0<\alpha<45^\circ\)). Let \(\theta\) be the acute angle between the rod and the horizontal. Show that \(\theta =2\lambda\), where \(\lambda \) is defined by \(\tan \lambda= \mu\) and \(0<\lambda<22.5^\circ\).

1. A wheel consists of a thin light circular rim attached by light spokes of length \(a\) to a small hub of mass \(m\). The wheel rolls without slipping on a rough horizontal table directly towards a straight edge of the table. The plane of the wheel is vertical throughout the motion. The speed of the wheel is \(u\), where \(u^2
2. Two particles, each of mass \(m/2\), are attached to a light circular hoop of radius \(a\), at the ends of a diameter. The hoop rolls without slipping on a rough horizontal table directly towards a straight edge of the table. The plane of the hoop is vertical throughout the motion. When the centre of the hoop is vertically above the edge of the table it has speed \(u\), where \(u^2

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 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*}

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 \[ \mu = \frac{F_C}{R_C} = \frac{\frac{29}{12\sqrt{3}}W}{\frac{53}{12}W} = \boxed{\frac{29}{53\sqrt{3}}}\]

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.

\(AB\) 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 %$$ %p=\frac{\cos{\alpha} \, \sin({\alpha}+{\lambda}_1-{\lambda}_2)} %{2\cos{\lambda}_1 \, \sin {\lambda}_2}\;. %$$