Moments

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2025 Paper 3 Q10
D: 1500.0 B: 1500.0

A plank \(AB\) of length \(L\) initially lies horizontally at rest along the \(x\)-axis on a flat surface, with \(A\) at the origin. Point \(C\) on the plank is such that \(AC\) has length \(sL\), where \(0 < s < 1\). End \(A\) is then raised vertically along the \(y\)-axis so that its height above the horizontal surface at time \(t\) is \(h(t)\), while end \(B\) remains in contact with the flat surface and on the \(x\)-axis. The function \(h(t)\) satisfies the differential equation $$\frac{d^2h}{dt^2} = -\omega^2 h, \text{ with } h(0) = 0 \text{ and } \frac{dh}{dt} = \omega L \text{ at } t = 0$$ where \(\omega\) is a positive constant. A particle \(P\) of mass \(m\) remains in contact with the plank at point \(C\).

  1. Show that the \(x\)-coordinate of \(P\) is \(sL\cos\omega t\), and find a similar expression for its \(y\)-coordinate.
  2. Find expressions for the \(x\)- and \(y\)-components of the acceleration of the particle.
  3. \(N\) and \(F\) are the upward normal and frictional components, respectively, of the force of the plank on the particle. Show that $$N = mg(1 - k\sin\omega t)\cos\omega t$$ and that $$F = mgsk\frac{\omega^2}{g}\tan\omega t$$ where \(k = \frac{L\omega^2}{g}\).
  4. The coefficient of friction between the particle and the plank is \(\tan\alpha\), where \(\alpha\) is an acute angle. Show that the particle will not slip initially, provided \(sk < \tan\alpha\). Show further that, in this case, the particle will slip
    • while \(N\) is still positive,
    • when the plank makes an angle less than \(\alpha\) to the horizontal.

Show Solution
  1. Since we have \(h'' + \omega^2 h = 0\) we must have that \(h(t) = A \cos \omega t + B \sin \omega t\). The initial conditions tell us that \(A = 0\) and \(B = L\), so \(h(t) = L \sin \omega t\).
    TikZ diagram
    Therefore we can see the angle at \(B\) is \(\omega t\) and so \(P\) has \(y\)-coordinate \((1-s)L \sin \omega t\) and \(x\)-coordinate \(sL \cos \omega t\)
  2. If the position is \(\binom{sL \cos \omega t}{(1-s) L \sin \omega t}\) then the acceleration is \(-\omega^2 \binom{sL \cos \omega t}{(1-s) L \sin \omega t}\)
  3. TikZ diagram
    \begin{align*} \text{N2}(\rightarrow): && - F\cos \omega t + N \sin \omega t &= -m\omega^2 sL \cos \omega t\\ \text{N2}(\uparrow): && -mg + F\sin \omega t + N \cos \omega t &= -m\omega^2 (1-s) L \sin \omega t \\ \Rightarrow && \begin{pmatrix} \cos \omega t & -\sin \omega t \\ \sin \omega t & \cos \omega t \end{pmatrix} \begin{pmatrix} F \\ N \end{pmatrix} &= \begin{pmatrix} m\omega^2 s L \cos \omega t \\ mg - m\omega^2(1-s)L \sin \omega t \end{pmatrix} \\ \Rightarrow && \begin{pmatrix}F \\ N \end{pmatrix} &= \begin{pmatrix} \cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t \end{pmatrix} \begin{pmatrix} m\omega^2 s L \cos \omega t \\ mg - m\omega^2(1-s)L \sin \omega t \end{pmatrix} \\ \Rightarrow && N &= m \omega^2 s L (-\sin \omega t \cos \omega t) + mg \cos \omega t - m \omega^2 (1-s)L \sin \omega t \cos \omega t \\ &&&=mg \cos \omega t - m \omega^2 L \sin \omega t \cos \omega t \\ &&&= mg \cos \omega t \left (1 - \frac{L \omega^2}{g} \sin \omega t \right) \\ &&&= mg (1 - k \sin \omega t) \cos \omega t \\ \Rightarrow && F &= m \omega^2 s L \cos^2 \omega t + mg \sin \omega t - m \omega^2 (1-s) L \sin ^2 \omega t \\ &&&= m \omega^2 s L + mg \sin \omega t - m \omega^2 L \sin^2 \omega t \\ &&&= mg \frac{\omega^2 L}{g} s + mg(1-\frac{\omega^2 L}{g} \sin \omega t)\sin \omega t \\ &&&= mg sk + mg(1-k \sin \omega t) \cos \omega t \tan \omega t \\ &&&= mgsk + N \tan \omega t \end{align*}
  4. The particle will not slip if \(F < \tan \alpha N\). When \(t = 0\), \(N = mg, F = mgsk\), but clearly \(sk < \tan \alpha \Rightarrow mgsk = F < \tan \alpha mg = \tan \alpha N\). The particle will slip when: \(F > \tan \alpha N\), but we have \(F = mgsk + N \tan \omega t\). Clearly when \(\omega t = \alpha\) we have reached a point where \(F > \tan \alpha N\). Therefore we must slip before we reach this point, ie at a point where the plank makes an angle of less than \(\alpha\) to the horizontal. Notice also that \(N\) changes sign when \(1-k \sin \omega t = 0\), however, to do this \(N\) must become very small, smaller than \(mgsk\), therefore we must slip before this point too. Since we slip before either condition occurs, we must be in a position when \(N\) is positive AND the plank still makes a shallow angle.
2024 Paper 3 Q10
D: 1500.0 B: 1500.0

A cube of uniform density \(\rho\) is placed on a horizontal plane and a second cube, also of uniform density \(\rho\), is placed on top of it. The lower cube has side length \(1\) and the upper cube has side length \(a\), with \(a \leqslant 1\). The centre of mass of the upper cube is vertically above the centre of mass of the lower cube and all the edges of the upper cube are parallel to the corresponding edges of the lower cube. The contacts between the two cubes, and between the lower cube and the plane, are rough, with the same coefficient of friction \(\mu < 1\) in each case. The midpoint of the base of the upper cube is \(X\) and the midpoint of the base of the lower cube is \(Y\). A horizontal force \(P\) is exerted, perpendicular to one of the vertical faces of the upper cube, at a point halfway between the two vertical edges of this face, and a distance \(h\), with \(h < a\), above the lower edge of this face.

  1. Show that, if the two cubes remain in equilibrium, the normal reaction of the plane on the lower cube acts at a point which is a distance \[\frac{P(1+h)}{(1+a^3)\rho g}\] from \(Y\), and find a similar expression for the distance from \(X\) of the point at which the normal reaction of the lower cube on the upper cube acts.
The force \(P\) is now gradually increased from zero.
  1. Show that, if neither cube topples, equilibrium will be broken by the slipping of the upper cube on the lower cube, and not by the slipping of the lower cube on the ground.
  2. Show that, if \(a = 1\), then equilibrium will be broken by the slipping of the upper cube on the lower cube if \(\mu(1+h) < 1\) and by the toppling of the lower and upper cube together if \(\mu(1+h) > 1\).
  3. Show that, in a situation where \(a < 1\) and \(h\bigl(1 + a^3(1-a)\bigr) > a^4\), and no slipping occurs, equilibrium will be broken by the toppling of the upper cube.
  4. Show, by considering \(a = \frac{1}{2}\) and choosing suitable values of \(h\) and \(\mu\), that the situation described in (iv) can in fact occur.

2023 Paper 3 Q10
D: 1500.0 B: 1500.0

A thin uniform beam \(AB\) has mass \(3m\) and length \(2h\). End \(A\) rests on rough horizontal ground and the beam makes an angle of \(2\beta\) to the vertical, supported by a light inextensible string attached to end \(B\). The coefficient of friction between the beam and the ground at \(A\) is \(\mu\). The string passes over a small frictionless pulley fixed to a point \(C\) which is a distance \(2h\) vertically above \(A\). A particle of mass \(km\), where \(k < 3\), is attached to the other end of the string and hangs freely.

  1. Given that the system is in equilibrium, find an expression for \(k\) in terms of \(\beta\) and show that \(k^2 \leqslant \dfrac{9\mu^2}{\mu^2 + 1}\).
  2. A particle of mass \(m\) is now fixed to the beam at a distance \(xh\) from \(A\), where \(0 \leqslant x \leqslant 2\). Given that \(k = 2\), and that the system is in equilibrium, show that \[\frac{F^2}{N^2} = \frac{x^2 + 6x + 5}{4(x+2)^2}\,,\] where \(F\) is the frictional force and \(N\) is the normal reaction on the beam at \(A\). By considering \(\dfrac{1}{3} - \dfrac{F^2}{N^2}\), or otherwise, find the minimum value of \(\mu\) for which the beam can be in equilibrium whatever the value of \(x\).

2022 Paper 2 Q9
D: 1500.0 B: 1500.0

A rectangular prism is fixed on a horizontal surface. A vertical wall, parallel to a vertical face of the prism, stands at a distance \(d\) from it. A light plank, making an acute angle \(\theta\) with the horizontal, rests on an upper edge of the prism and is in contact with the wall below the level of that edge of the prism and above the level of the horizontal plane. You may assume that the plank is long enough and the prism high enough to make this possible. The contact between the plank and the prism is smooth, and the coefficient of friction at the contact between the plank and the wall is \(\mu\). When a heavy point mass is fixed to the plank at a distance \(x\), along the plank, from its point of contact with the wall, the system is in equilibrium.

  1. Show that, if \(x = d\sec^3\theta\), then there is no frictional force acting between the plank and the wall.
  2. Show that, if \(x > d\sec^3\theta\), it is necessary that \[\mu \geqslant \frac{x - d\sec^3\theta}{x\tan\theta}\] and give the corresponding inequality if \(x < d\sec^3\theta\).
  3. Show that \[\frac{x}{d} \geqslant \frac{\sec^3\theta}{1 + \mu\tan\theta}\,.\] Show also that, if \(\mu < \cot\theta\), then \[\frac{x}{d} \leqslant \frac{\sec^3\theta}{1 - \mu\tan\theta}\,.\]
  4. Show that if \(x\) is such that the point mass is fixed to the plank somewhere between the edge of the prism and the wall, then \(\tan\theta < \mu\).

2020 Paper 3 Q9
D: 1500.0 B: 1500.0

Two inclined planes \(\Pi_1\) and \(\Pi_2\) meet in a horizontal line at the lowest points of both planes and lie on either side of this line. \(\Pi_1\) and \(\Pi_2\) make angles of \(\alpha\) and \(\beta\), respectively, to the horizontal, where \(0 < \beta < \alpha < \frac{1}{2}\pi\). A uniform rigid rod \(PQ\) of mass \(m\) rests with \(P\) lying on \(\Pi_1\) and \(Q\) lying on \(\Pi_2\) so that the rod lies in a vertical plane perpendicular to \(\Pi_1\) and \(\Pi_2\) with \(P\) higher than \(Q\).

  1. It is given that both planes are smooth and that the rod makes an angle \(\theta\) with the horizontal. Show that \(2\tan\theta = \cot\beta - \cot\alpha\).
  2. It is given instead that \(\Pi_1\) is smooth, that \(\Pi_2\) is rough with coefficient of friction \(\mu\) and that the rod makes an angle \(\phi\) with the horizontal. Given that the rod is in limiting equilibrium, with \(P\) about to slip down the plane \(\Pi_1\), show that \[ \tan\theta - \tan\phi = \frac{\mu}{(\mu + \tan\beta)\sin 2\beta} \] where \(\theta\) is the angle satisfying \(2\tan\theta = \cot\beta - \cot\alpha\).

2019 Paper 1 Q9
D: 1500.0 B: 1500.0

A box has the shape of a uniform solid cuboid of height \(h\) and with a square base of side \(b\), where \(h > b\). It rests on rough horizontal ground. A light ladder has its foot on the ground and rests against one of the upper horizontal edges of the box, making an acute angle of \(\alpha\) with the ground, where \(h = b \tan \alpha\). The weight of the box is \(W\). There is no friction at the contact between ladder and box. A painter of weight \(kW\) climbs the ladder slowly. Neither the base of the ladder nor the box slips, but the box starts to topple when the painter reaches height \(\lambda h\) above the ground, where \(\lambda < 1\). Show that:

  1. \(R = k\lambda W \cos \alpha\), where \(R\) is the magnitude of the force exerted by the box on the ladder;
  2. \(2k\lambda \cos 2\alpha + 1 = 0\);
  3. \(\mu \geq \frac{\sin 2\alpha}{1 - 3 \cos 2\alpha}\), where \(\mu\) is the coefficient of friction between the box and the ground.

Show Solution
TikZ diagram
At the point we are about to topple, reaction and friction forces will be acting at \(C\)
  1. \(\,\) \begin{align*} \overset{\curvearrowright}{X}:&& kW \cdot \lambda h\cos \alpha - R h &= 0 \\ \Rightarrow && R &= k\lambda W \cos \alpha \\ \end{align*}
  2. \(\,\) \begin{align*} \overset{\curvearrowright}{C}:&& R \sin \alpha \cdot h-R\cos \alpha \cdot b-W\frac{b}{2} &= 0 \\ && k\lambda W \cos \alpha \sin \alpha \cdot b \tan \alpha- k\lambda W \cos \alpha\cos \alpha \cdot h-W\frac{b}{2} &= 0 \\ && k \lambda (\cos^2 \alpha - \sin^2 \alpha) +\frac12 &= 0 \\ \Rightarrow && 2k \lambda \cos 2\alpha + 1 &= 0 \end{align*}
  3. \(\,\) \begin{align*} \text{N2}(\uparrow): && R_b -W-R\cos \alpha &= 0 \\ \Rightarrow && R_b &= W + k\lambda W \cos^2 \alpha\\ \text{N2}(\rightarrow): && R\sin \alpha - F_b &= 0 \\ \Rightarrow && F_b &= R \sin \alpha \\ \\ && F_b &\leq \mu R \\ \Rightarrow && k\lambda W \cos \alpha \sin \alpha &= \mu (W + k\lambda W \cos^2 \alpha) \\ \Rightarrow && \mu &\geq \frac{k\lambda \cos \alpha \sin \alpha}{1 + k\lambda \cos^2 \alpha} \\ &&&= \frac{k\lambda \sin 2\alpha}{2 + 2k\lambda cos^2 \alpha} \\ &&&= \frac{k\lambda \sin 2\alpha}{2 + k\lambda (\cos 2 \alpha+1)} \\ &&&= \frac{k\lambda \sin 2\alpha}{-4k\lambda \cos 2 \alpha + k\lambda (\cos 2 \alpha+1)} \\ &&&= \frac{\sin 2 \alpha}{1 -3 \cos 2\alpha} \end{align*}
2017 Paper 1 Q11
D: 1516.0 B: 1500.0

A plane makes an acute angle \(\alpha\) with the horizontal. A box in the shape of a cube is fixed onto the plane in such a way that four of its edges are horizontal and two of its sides are vertical. A uniform rod of length \(2L\) and weight \(W\) rests with its lower end at \(A\) on the bottom of the box and its upper end at \(B\) on a side of the box, as shown in the diagram below. The vertical plane containing the rod is parallel to the vertical sides of the box and cuts the lowest edge of the box at \(O\). The rod makes an acute angle~\(\beta\) with the side of the box at \(B\). The coefficients of friction between the rod and the box at the two points of contact are both \(\tan \gamma\), where \(0 < \gamma < \frac12\pi\). %The frictional force on the rod at \(A\) acts toward \(O\), %and the frictional force on the rod at~\(B\) %acts away from \(O\). The rod is in limiting equilibrium, with the end at \(A\) on the point of slipping in the direction away from \(O\) and the end at \(B\) on the point of slipping towards \(O\). Given that \(\alpha < \beta\), show that \(\beta = \alpha + 2\gamma\). [\(Hint\): You may find it helpful to take moments about the midpoint of the rod.]

TikZ diagram

Show Solution
TikZ diagram
Since we're at limiting equilibrium and about to slip \(Fr_B = \mu R_B\) and \(Fr_A = \mu R_A\) \begin{align*} \text{N2}(\parallel OB): && \mu R_B + R_A - W \cos \alpha &= 0 \\ \text{N2}(\parallel OA): && R_B - \mu R_A - W \sin \alpha &= 0 \\ \\ \Rightarrow && \sin\alpha \l \mu R_B + R_A \r - \cos \alpha \l R_B - \mu R_A \r &= 0 \\ \Leftrightarrow && R_A(\sin \alpha + \mu \cos \alpha) - R_B(\cos \alpha - \mu \sin \alpha) &= 0 \\ \Rightarrow && \frac{\tan \alpha + \mu}{1 - \mu \tan \alpha} R_A &= R_B\\ && \tan (\alpha + \gamma) R_A &= R_B \\ \\ \\ \overset{\curvearrowleft}{\text{midpoint}}: && R_A \sin \beta - \mu R_A \cos \beta - R_B \cos \beta - \mu R_B \sin \beta &= 0\\ \Rightarrow && \tan \beta - \mu - \tan (\alpha + \gamma) - \mu \tan (\alpha + \gamma) \tan \beta &= 0\\ \Rightarrow && \tan \beta \l 1 - \mu \tan (\alpha + \gamma) \r - \mu - \tan (\alpha + \gamma) &= 0\\ \Rightarrow && \frac{\mu + \tan (\alpha + \gamma)}{1 - \mu \tan (\alpha + \gamma)} &= \tan \beta \\ \Rightarrow && \tan (\alpha + 2\gamma) &= \tan \beta \end{align*} Since \(\alpha < \beta\) and \(\gamma < \frac{\pi}{4}\) we must have \(\alpha + 2\gamma = \beta\)
2004 Paper 1 Q11
D: 1500.0 B: 1500.0

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 \(\mu\)?

Show Solution
TikZ diagram
\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 \[ \mu = \frac{F_C}{R_C} = \frac{\frac{29}{12\sqrt{3}}W}{\frac{53}{12}W} = \boxed{\frac{29}{53\sqrt{3}}}\]
2000 Paper 3 Q10
D: 1700.0 B: 1500.0

A sphere of radius \(a\) and weight \(W\) rests on horizontal ground. A thin uniform beam of weight \(3\sqrt3\,W\) and length \(2a\) is freely hinged to the ground at \(X\), which is a distance \({\sqrt 3} \, a\) from the point of contact of the sphere with the ground. The beam rests on the sphere, lying in the same vertical plane as the centre of the sphere. The coefficients of friction between the beam and the sphere and between the sphere and the ground are \(\mu_1\) and \(\mu_2\) respectively. Given that the sphere is on the point of slipping at its contacts with both the ground and the beam, find the values of \(\mu_1\) and \(\mu_2\).

Show Solution
TikZ diagram
The first important thing to observe is the angle at \(X\) is \(60^{\circ}\). Now we can start resolving: \begin{align*} \overset{\curvearrowleft}{X}: && 3\sqrt{3} W \cos 60^{\circ} a - R_1\sqrt{3}a &= 0 \tag{\(1\)}\\ \overset{\curvearrowleft}{O}: && \mu_2 R_2 a - \mu_1R_1a &= 0 \tag{\(2\)} \\ \text{N2}(\rightarrow): && \mu_2 R_2 + \mu_1R_1 \cos 60^{\circ} - R_1 \cos 30^{\circ} &= 0 \tag{\(3\)} \\ \text{N2}(\uparrow): && R_2 - W - \mu_1 R_1 \cos 30^{\circ} - R_1 \cos 60^{\circ} &= 0 \tag{\(4\)} \\ \Rightarrow && \frac{3}{2}W &= R_1 \tag{\((5)\) from \((1)\)} \\ && \mu_1 R_1 &= \mu_2 R_2 \tag{\(2\)}\\ && \mu_1 R_1 \l 1 + \frac{1}{2} \r - R_1 \frac{\sqrt{3}}2 &= 0 \tag{\((3)\) and \((2)\)} \\ && \mu_1 &= \frac{1}{\sqrt3} \\ \\ && R_2 - W - \frac{1}{\sqrt3} \frac{3}{2}W \frac{\sqrt3}{2} - \frac{3}2W \frac12 &= 0 \\ \Rightarrow && R_2 &= W \l 1 + \frac{3}{2}\r \tag{\(6\)} \\ \Rightarrow && \mu_2 &= \frac{\mu_1 R_1}{R_2} = \frac{1}{\sqrt{3}} \frac{3}{5} = \frac{\sqrt3}{5} \tag{\((5)\) and \((6)\)} \end{align*}
1997 Paper 2 Q9
D: 1600.0 B: 1475.3

A uniform solid sphere of diameter \(d\) and mass \(m\) is drawn very slowly and without slipping from horizontal ground onto a step of height \(d/4\) by a horizontal force which is always applied to the highest point of the sphere and is always perpendicular to the vertical plane which forms the face of the step. Find the maximum horizontal force throughout the movement, and prove that the coefficient of friction between the sphere and the edge of the step must exceed \(1/\sqrt{3}\).

Show Solution
TikZ diagram
The ball is on the ground when \(\cos \theta = \frac12 \Rightarrow \theta = 60^\circ\) and ball will make it over the step when \(\theta = 0^\circ\). It is also worth emphasising we are moving \emph{very slowly}, so we can treat the system as static at any given point. \begin{align*} \overset{\curvearrowleft}{X}: && mg \frac{d}{2}\sin \theta - F \frac{d}{2} \l 1 + \cos \theta \r &= 0\\ \Rightarrow && \frac{mg \sin \theta}{1 + \cos \theta} &= F& \\ \Rightarrow && mg \tan \frac{\theta}{2} &= F& \\ \end{align*} Therefore \(F\) is maximised when \(\theta = 60^\circ\), ie \(F_{max} = \frac{mg}{\sqrt{3}}\) \begin{align*} \text{N2}(\parallel OX): && mg \cos \theta - R + F \sin \theta &= 0 \\ \Rightarrow && mg \cos \theta - R + \frac{mg\sin \theta}{1 + \cos \theta} \sin \theta &= 0 \\ \Rightarrow && mg &= R \\ \\ \text{N2}(\perp OX): && F_X - mg \sin \theta + F \cos \theta &= 0 \\ \Rightarrow && mg \sin \theta - \frac{mg\sin \theta}{1 + \cos \theta} \cos \theta &= F_X \\ \Rightarrow && \frac{mg\sin \theta}{1 + \cos \theta} &= F_X \tag{We could also see this taking moments about \(O\)}\\ % \text{N2}(\rightarrow): && F + \mu R \cos \theta - R \sin \theta &\geq 0 \\ % \text{N2}(\uparrow): && -mg +\mu R \sin \theta + R \cos \theta &\geq 0 \\ % \Rightarrow && R \l \sin \theta - \mu \cos \theta\r &\leq F \\ % \Rightarrow && R \l \mu \sin \theta + \cos \theta\r &\geq mg \\ % \Rightarrow && \l \frac{\sin \theta - \mu \cos \theta}{\mu \sin \theta + \cos \theta} \r mg & \leq F \\ % \Rightarrow && \l \frac{\tan \theta - \mu }{1+\mu \tan \theta} \r mg & \leq F \\ % \Rightarrow && \tan \l \theta - \alpha \r mg & \leq F \tag{where \(\tan \alpha = \mu\)} \end{align*} Therefore since \(F_X \leq \mu R\), \(\displaystyle \frac{mg\sin \theta}{1 + \cos \theta} \leq \mu mg \Rightarrow \mu \geq \tan \frac{\theta}{2}\) which is maximised at \(\theta = 60^\circ\) and implies \(\mu \geq \frac{1}{\sqrt{3}}\)
1996 Paper 3 Q10
D: 1700.0 B: 1500.0

Two rough solid circular cylinders, of equal radius and length and of uniform density, lie side by side on a rough plane inclined at an angle \(\alpha\) to the horizontal, where \(0<\alpha<\pi/2\). Their axes are horizontal and they touch along their entire length. The weight of the upper cylinder is \(W_1\) and the coefficient of friction between it and the plane is \(\mu_1\). The corresponding quantities for the lower cylinder are \(W_2\) and \(\mu_2\) respectively and the coefficient of friction between the two cylinders is \(\mu\). Show that for equilibrium to be possible:

  1. \(W_1\ge W_2\);
  2. \(\mu\geqslant\dfrac{W_1+W_2}{W_1-W_2}\);
  3. \(\mu_{1}\geqslant\left(\dfrac{2W_{1}\cot\alpha}{W_{1}+W_{2}}-1\right)^{-1}\,.\)
Find the similar inequality to (iii) for \(\mu_2\).

Show Solution
TikZ diagram
  1. \begin{align*} \overset{\curvearrowright}{O_2}: && 0 &= F_2 - F \\ \Rightarrow && F_2 &= F \\ \overset{\curvearrowright}{O_1}: && 0 &= F_1- F \\ \Rightarrow && F_1 &= F \\ \text{N2}(\swarrow, 2): && 0 &= R+W_2\sin\alpha -F \tag{1}\\ \text{N2}(\swarrow, 1): && 0 &= W_1\sin\alpha -F-R\tag{2}\\ \Rightarrow && W_1 \sin \alpha-R &= W_2 \sin \alpha+R \\ \Rightarrow && W_1 &\geq W_2 \end{align*}
  2. \begin{align*} (1)+(2)\Rightarrow && F &= \frac12 \sin \alpha (W_1 + W_2) \\ (1)-(2) \Rightarrow && R &= \frac12 \sin \alpha (W_1-W_2) \\ \Rightarrow && \frac{F}{R} &= \frac{W_1+W_2}{W_1-W_2} \\ \underbrace{\Rightarrow}_{F \leq \mu R} && \mu &\geq \frac{W_1+W_2}{W_1-W_2}\\ \end{align*}
  3. \begin{align*} \text{N2}(\nwarrow, 1): && 0 &= F+R_1-W_1\cos \alpha \\ \Rightarrow && R_1 &= W_1\cos \alpha - F \\ &&&= W_1 \cos \alpha - \frac12 \sin \alpha (W_1 + W_2) \\ \Rightarrow && \frac{R_1}{F_1} &= \frac{R_1}{F} \\ &&&= \frac{W_1 \cos \alpha - \frac12 \sin \alpha (W_1 + W_2)}{\frac12 \sin \alpha (W_1 + W_2) } \\ &&&= \frac{2W_1 \cot \alpha}{W_1+W_2} - 1 \\ \Rightarrow && \mu_1 & \geq \left ( \frac{2W_1 \cot \alpha}{W_1+W_2} - 1 \right)^{-1} \end{align*}
\begin{align*} \text{N2}(\nwarrow, 2): && 0 &= -F+R_2-W_2\cos \alpha \\ \Rightarrow && R_2 &= W_2\cos \alpha + F \\ &&&= W_2 \cos \alpha + \frac12 \sin \alpha (W_1 + W_2) \\ \Rightarrow && \frac{R_2}{F_2} &= \frac{R_2}{F} \\ &&&= \frac{ W_2 \cos \alpha + \frac12 \sin \alpha (W_1 + W_2)}{\frac12 \sin \alpha (W_1 + W_2) } \\ &&&= \frac{2W_2 \cot \alpha}{W_1+W_2} + 1 \\ \Rightarrow && \mu_2 & \geq \left ( \frac{2W_1 \cot \alpha}{W_1+W_2} + 1 \right)^{-1} \end{align*}
1993 Paper 2 Q12
D: 1600.0 B: 1484.0

TikZ diagram
A uniform sphere of mass \(M\) and radius \(r\) rests between a vertical wall \(W_{1}\) and an inclined plane \(W_{2}\) that meets \(W_{1}\) at an angle \(\alpha.\) \(Q_{1}\) and \(Q_{2}\) are the points of contact of the sphere with \(W_{1}\) and \(W_{2}\) resectively, as shown in the diagram. A particle of mass \(m\) is attached to the sphere at \(P\), where \(PQ_{1}\) is a diameter, and the system is released. The sphere is on the point of slipping at \(Q_{1}\) and at \(Q_{2}.\) Show that if the coefficients of friction between the sphere and \(W_{1}\) and \(W_{2}\) are \(\mu_{1}\) and \(\mu_{2}\) respectively, then \[ m=\frac{\mu_{2}+\mu_{1}\cos\alpha-\mu_{1}\mu_{2}\sin\alpha}{(2\mu_{1}\mu_{2}+1)\sin\alpha+(\mu_{2}-2\mu_{1})\cos\alpha-\mu_{2}}M. \] If the sphere is on the point of rolling about \(Q_{2}\) instead of slipping, show that \[ m=\frac{M}{\sec\alpha-1}. \]

Show Solution
TikZ diagram
Since the sphere is on the point of slipping at both \(Q_1\) and \(Q_2\), \(F_{r1} = \mu_1 R_1\) and \(F_{r2} = \mu_2 R_2\) \begin{align*} \text{N2}(\uparrow): && -mg-Mg-\mu_1 R_1 + R_2 \sin \alpha + \mu_2 R_2 \cos \alpha &= 0 \\ \text{N2}(\rightarrow): && -R_1 + R_2 \cos \alpha - \mu_2 R_2 \sin \alpha &= 0 \\ \\ \Rightarrow && R_2 \cos \alpha - \mu_2 R_2 \sin \alpha &= R_1 \\ % && -mg-Mg+\mu_1 (R_2 \cos \alpha - \mu_2 R_2 \sin \alpha) + R_2 \sin \alpha + \mu_2 R_2 \cos \alpha &= 0 \\ % \\ \overset{\curvearrowleft}{O}: && mg - \mu_1 R_1 - \mu_2R_2 &= 0 \\ \Rightarrow && \mu_1 R_2 \l \cos \alpha - \mu_2 \sin \alpha \r - \mu_2 R_2 &= -mg \\ && \mu_1 (R_2 \cos \alpha - \mu_2 R_2 \sin \alpha) + R_2 \sin \alpha + \\ && \quad \quad \mu_2 R_2 \cos \alpha - \mu_1 R_2 \l \cos \alpha - \mu_2 \sin \alpha \r - \mu_2 R_2 &= Mg \\ \Rightarrow && \frac{\mu_2+\mu_1 \l \cos \alpha - \mu_2 \sin \alpha \r }{\mu_1 ( \cos \alpha - \mu_2 \sin \alpha) + \sin \alpha + \mu_2 \cos \alpha - \mu_1 \l \cos \alpha - \mu_2 \sin \alpha \r - \mu_2 } &= \frac{m}{M} \\ && \frac{\mu_2+\mu_1 \cos \alpha - \mu_1\mu_2 \sin \alpha }{\cos \alpha (-2\mu_1+\mu_2) + \sin \alpha (1 +2\mu_1\mu_2) -\mu_2} &= \frac{m}{M} \end{align*} If instead the sphere is about to roll about \(Q_2\), then the forces at \(Q_1\) will be \(0\), we can then take moments about \(Q_2\).
TikZ diagram
Looking at perpendicular distances from \(Q_2\) to \(O\) and \(P\) we have \(r \cos \alpha\) and \(r(1-\cos \alpha)\) \begin{align*} \overset{\curvearrowleft}{Q_2}: && mg (1 - \cos \alpha) - Mg \cos \alpha &= 0 \\ \Rightarrow && \frac{1}{\sec \alpha-1} &= \frac{m}{M} \end{align*}
1992 Paper 1 Q12
D: 1484.0 B: 1471.5

The diagram shows a crude step-ladder constructed by smoothly hinging-together two light ladders \(AB\) and \(AC,\) each of length \(l,\) at \(A\). A uniform rod of wood, of mass \(m\), is pin-jointed to \(X\) on \(AB\) and to \(Y\) on \(AC\), where \(AX=\frac{3}{4}l=AY.\) The angle \(\angle XAY\) is \(2\theta.\) \noindent

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The rod \(XY\) will break if the tension in it exceeds \(T\). The step-ladder stands on rough horizontal ground (coefficient of friction \(\mu\)). Given that \(\tan\theta>\mu,\) find how large a mass \(M\) can safely be placed at \(A\) and show that if \[ \tan\theta>\frac{6T}{mg}+4\mu \] the step-ladder will fail under its own weight. {[}You may assume that friction is limiting at the moment of collapse.{]}

1991 Paper 1 Q12
D: 1484.0 B: 1500.0

\(\ \)\vspace{-1.5cm} \noindent

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The above diagram illustrates a makeshift stepladder, made from two equal light planks \(AB\) and \(CD\), each of length \(2l\). The plank \(AB\) is smoothly hinged to the ground at \(A\) and makes an angle of \(\alpha\) with the horizontal. The other plank \(CD\) has its bottom end \(C\) resting on the same horizontal ground and makes an angle \(\beta\) with the horizontal. It is pivoted smoothly to \(B\) at a point distance \(2x\) from \(C\). The coefficient of friction between \(CD\) and the ground is \(\mu.\) A painter of mass \(M\) stands on \(CD\), half between \(C\) and \(B\). Show that, for equilibrium to be possible, \[ \mu\geqslant\frac{\cot\alpha\cot\beta}{2\cot\alpha+\cot\beta}. \] Suppose now that \(B\) coincides with \(D\). Show that, as \(\alpha\) varies, the maximum distance from \(A\) at which the painter will be standing is \[ l\sqrt{\frac{1+81\mu^{2}}{1+9\mu^{2}}}. \]

1991 Paper 3 Q11
D: 1700.0 B: 1484.0

TikZ diagram
A uniform circular cylinder of radius \(2a\) with a groove of radius \(a\) cut in its central cross-section has mass \(M\). It rests, as shown in the diagram, on a rough plane inclined at an acute angle \(\alpha\) to the horizontal. It is supported by a light inextensible string would round the groove and attached to the cylinder at one end. The other end of the string is attached to the plane at \(Q\), the free part of the string, \(PQ,\) making an angle \(2\alpha\) with the inclined plane. The coefficient of friction at the contact between the cylinder and the plane is \(\mu.\) Show that \(\mu\geqslant\frac{1}{3}\tan\alpha.\) The string \(PQ\) is now detached from the plane and the end \(Q\) is fastened to a particle of mass \(3M\) which is placed on the plane, the position of the string remain unchanged. Given that \(\tan\alpha=\frac{1}{2}\) and that the system remains in equilibrium, find the least value of the coefficient of friction between the particle and the plane.

1990 Paper 1 Q13
D: 1500.0 B: 1484.0

A rough circular cylinder of mass \(M\) and radius \(a\) rests on a rough horizontal plane. The curved surface of the cylinder is in contact with a smooth rail, parallel to the axis of the cylinder, which touches the cylinder at a height \(a/2\) above the plane. Initially the cylinder is held at rest. A particle of mass \(m\) rests in equilibrium on the cylinder, and the normal reaction of the cylinder on the particle makes an angle of \(\theta\) with the upward vertical. The particle is on the same side of the centre of the cylinder as the rail. The coefficient of friction between the cylinder and the particle and between the cylinder and the plane are both \(\mu\). Obtain the condition on \(\theta\) for the particle to rest in equilibrium. Show that, if the cylinder is released, equilibrium of both particle and cylinder is possible provided another inequality involving \(\mu\) and \(\theta\) (which should be found explicitly) is satisfied. Determine the largest possible value of \(\theta\) for equilibrium, if \(m=7M\) and \(\mu=0.75\).

Show Solution
TikZ diagram
\begin{align*} \text{N2}(\nwarrow): && R -mg \cos \theta &= 0 \\ \text{N2}(\rightarrow): && -R \sin \theta + F \cos \theta &= 0 \\ \\ \Rightarrow && F &= \tan \theta R \\ \\ && F & \leq \mu R \\ \Rightarrow && \tan \theta R &\leq \mu R \\ \Rightarrow && \tan \theta &\leq \mu \end{align*} (Notice also \(F = mg \sin \theta\)) Once everything is released, we have the following situation. (Red forces act on the cylinder, blue forces on the particle).
TikZ diagram
\begin{align*} \text{N2}(\uparrow): && 0 &= R_g - Mg - \underbrace{mg}_{R_p \text{ and } F_p} + \frac{1}{\sqrt{2}}R_r \\ \text{N2}(\rightarrow): && 0 &= \frac{1}{\sqrt{2}}R_r - F_g \\ \overset{\curvearrowleft}{O}: && 0 &= aF_p - aF_g \\ \Rightarrow && F_g &= mg \sin \theta \\ \Rightarrow && R_r &= \sqrt{2} mg \sin \theta \\ \Rightarrow && R_g &=(M+m)g + mg \sin \theta \\ \\ && F_g &\leq \mu R_g \\ \Rightarrow && mg \sin \theta &\leq \mu (M+m(1+\sin \theta))g \\ \Rightarrow && \mu &\geq \frac{m \sin \theta}{M+m(1+\sin \theta)} \end{align*} If \(m = 7M\) and \(\mu = \frac34\) we have: \begin{align*} && \tan \theta &\leq \frac34 \\ && 3(M+7M(1 + \sin \theta)) &\geq 4 \cdot 7 M \sin \theta \\ \Rightarrow && 10 + 7 \sin \theta & \geq 28 \sin \theta \\ \Rightarrow && 10 &\geq 21 \sin \theta \\ \Rightarrow && \sin \theta &\leq \frac{10}{21} \end{align*} If \(\tan \theta = \frac{3}{4}, \sin \theta = \frac35 > \frac{10}{21}\), so the critical bound is \(\sin \theta \leq \frac{10}{21}\), ie \( \theta \leq \sin^{-1} \frac{10}{21} \approx 30^{\circ}\)
1990 Paper 3 Q14
D: 1700.0 B: 1484.0

The edges \(OA,OB,OC\) of a rigid cube are taken as coordinate axes and \(O',A',B',C'\) are the vertices diagonally opposite \(O,A,B,C,\) respectively. The four forces acting on the cube are \[ \begin{pmatrix}\alpha\\ \beta\\ \gamma \end{pmatrix}\mbox{ at }O\ (0,0,0),\ \begin{pmatrix}\lambda\\ 0\\ 1 \end{pmatrix}\mbox{ at }O'\ (a,a,a),\ \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix}\mbox{ at }B\ (0,a,0),\ \mbox{ and }\begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix}\mbox{ at }B'\ (a,0,a). \] The moment of the system about \(O\) is zero: find \(\lambda,\mu\) and \(\nu\).

  1. Given that \(\alpha=\beta=\gamma=0,\) find the system consisting of a single force at \(B\) together with a couple which is equivalent to the given system.
  2. Given that \(\alpha=2,\beta=3\) and \(\gamma=2,\) find the equation of the locus about each point of which the moment of the system is zero. Find the number of units of work done on the cube when it moves (without rotation) a distance in the direction of this line under the action of the given forces only.

Show Solution
\begin{align*} &&\mathbf{M} &= \begin{pmatrix}\lambda \\ 0\\ 1 \end{pmatrix} \times \begin{pmatrix}a\\ a \\ a \end{pmatrix} + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} \times \begin{pmatrix} 0 \\ a \\ 0 \end{pmatrix} + \begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix} \times \begin{pmatrix} a \\ 0 \\ a \end{pmatrix} \\ &&&= \begin{pmatrix} -a \\ -a(\lambda -1) \\ \lambda a \end{pmatrix} + \begin{pmatrix} -2a \\ 0 \\ -a \end{pmatrix} + \begin{pmatrix} \mu a \\ -a(1-\nu) \\ -a \mu \end{pmatrix} \\ &&&=a \begin{pmatrix} \mu - 3 \\ \nu - \lambda \\ \lambda-1-\mu \end{pmatrix} \\ \Rightarrow && \mu &= 3, \lambda = 4, \nu = 4 \end{align*}
  1. To find the force we add all vectors: \begin{align*} \mathbf{F} &= \begin{pmatrix}\lambda \\ 0\\ 1 \end{pmatrix} + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} + \begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix} \\ &= \begin{pmatrix}4\\ 0\\ 1 \end{pmatrix} + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} + \begin{pmatrix}1\\ 3 \\ 4 \end{pmatrix} \\ &= \begin{pmatrix} 4 \\ 3 \\ 7 \end{pmatrix} \end{align*} Since the moment about \(O\) is \(0\), we have the moment about \(B\) is: \begin{align*} \mathbf{M} &= \begin{pmatrix} 0 \\ a \\ 0 \end{pmatrix} \times \begin{pmatrix} 4 \\ 3 \\ 7\end{pmatrix} \\ &= \begin{pmatrix} 7a \\ 0 \\ -4a\end{pmatrix} \end{align*}
  2. \begin{align*} \mathbf{0} &= \mathbf{r} \times \begin{pmatrix} 4 + 2 \\ 3+3 \\ 7+2 \end{pmatrix} \\ &= \mathbf{r} \times \begin{pmatrix} 6 \\ 6 \\ 9 \end{pmatrix} \\ \end{align*} Therefore \(\mathbf{r} = t\begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix}\) (ie a line) \begin{align*} \text{Work done} &= \text{Force} \cdot \text{distance} \end{align*} Since they are parallel, it's just the magnitude of the force, which is \(3\sqrt{2^2+2^2+3^2} = 3\sqrt{17}\)
1989 Paper 1 Q13
D: 1516.0 B: 1540.5

A uniform ladder of mass \(M\) rests with its upper end against a smooth vertical wall, and with its lower end on a rough slope which rises upwards towards the wall and makes an angle of \(\phi\) with the horizontal. The acute angle between the ladder and the wall is \(\theta\). If the ladder is in equilibrium, show that \(N\) and \(F\), the normal reaction and frictional force at the foot of the ladder are given by \[ N=Mg\left(\cos\phi-\frac{\tan\theta\sin\phi}{2}\right), \] \[ F=Mg\left(\sin\phi+\frac{\tan\theta\cos\phi}{2}\right). \] If the coefficient of friction between the ladder and the slope is \(2\), and \(\phi=45^{\circ}\), what is the largest value of \(\theta\) for which the ladder can rest in equilibrium?

Show Solution
\begin{align*} \overset{\curvearrowleft}{X}: && 0&= \frac{l}{2} Mg\sin \theta - l R_1 \cos \theta \\ \Rightarrow && R_1 &= \frac12 \tan \theta Mg \\ \text{N2}(\uparrow): && 0 &= R\cos \phi +F \sin \phi - Mg \\ \text{N2}(\rightarrow):&& 0&=R_1-F \cos \phi + R \sin \phi \\ \Rightarrow && \frac12 \tan \theta Mg &= F \cos \phi- R \sin \phi \\ && Mg &= F \sin \phi +R \cos \phi \\ \Rightarrow && F &= Mg \left ( \sin \phi + \frac12 \tan \theta \cos \phi \right) \\ && N &= Mg \left (\cos \phi - \frac12 \tan \theta \sin \phi \right ) \end{align*} If \(\mu = 2\) and \(\phi = 45^{\circ}\), we must have \(F \leq 2 N\), so: \begin{align*} && Mg \left ( \sin \phi + \frac12 \tan \theta \cos \phi \right) &\leq 2 Mg \left (\cos \phi - \frac12 \tan \theta \sin \phi \right ) \\ \Rightarrow && 1 + \frac12 \tan \theta \leq 2-\tan \theta \\ \Rightarrow && \frac 32 \tan \theta \leq 1 \\ \Rightarrow && \tan \theta \leq \frac23 \\ \Rightarrow && \theta \leq \tan^{-1} \frac23 \end{align*}
1988 Paper 3 Q11
D: 1700.0 B: 1484.0

A uniform ladder of length \(l\) and mass \(m\) rests with one end in contact with a smooth ramp inclined at an angle of \(\pi/6\) to the vertical. The foot of the ladder rests, on horizontal ground, at a distance \(l/\sqrt{3}\) from the foot of the ramp, and the coefficient of friction between the ladder and the ground is \(\mu.\) The ladder is inclined at an angle \(\pi/6\) to the horizontal, in the vertical plane containing a line of greatest slope of the ramp. A labourer of mass \(m\) intends to climb slowly to the top of the ladder.

TikZ diagram
  1. Find the value of \(\mu\) if the ladder slips as soon as the labourer reaches the midpoint.
  2. Find the minimum value of \(\mu\) which will ensure that the labourer can reach the top of the ladder.

Show Solution
TikZ diagram
  1. \begin{align*} \text{N2}(\uparrow): && R_1 + R_2\sin(\frac{\pi}{6})-2mg &= 0 \\ \text{N2}(\rightarrow): && R_2 \cos (\frac{\pi}{6})-F_r &= 0 \\ \overset{\curvearrowleft}{X}: && lmg \cos \tfrac{\pi}{6} - l R_2 \cos \tfrac{\pi}{6} &= 0 \\ \\ \Rightarrow && R_2 &= mg \\ \Rightarrow && R_1 &= 2mg - \frac12mg \\ &&&=\frac32mg \\ \Rightarrow && \frac{\sqrt{3}}2mg - \mu\frac32mg &= 0 \\ \Rightarrow && \mu &= \frac{1}{\sqrt{3}} \end{align*}
  2. \begin{align*} \text{N2}(\uparrow): && R_1 + R_2\sin(\frac{\pi}{6})-2mg &= 0 \\ \overset{\curvearrowleft}{X}: && \frac12 lmg \cos \tfrac{\pi}{6}+xmg \cos \tfrac{\pi}{6} - l R_2 \cos \tfrac{\pi}{6} &= 0 \\ \\ \Rightarrow && R_2 &= mg(\frac{1}2+\frac{x}{l}) \\ \Rightarrow && R_1 &= 2mg - \frac12mg(\frac{1}2+\frac{x}{l}) \\ &&&=(\frac74 - \frac{x}{2l})mg \\ &&&\geq \frac{5}{4}mg\\ \text{N2}(\rightarrow): && R_2 \cos (\frac{\pi}{6})-\mu R_1& \leq 0 \\ \Rightarrow && \frac{\sqrt{3}}2mg - \mu\frac54mg &\leq 0 \\ \Rightarrow && \mu &\geq \frac{2\sqrt{3}}{5} \end{align*}
1987 Paper 2 Q11
D: 1500.0 B: 1500.0

A rough ring of radius \(a\) is fixed so that it lies in a plane inclined at an angle \(\alpha\) to the horizontal. A uniform heavy rod of length \(b(>a)\) has one end smoothly pivoted at the centre of the ring, so that the rod is free to move in any direction. It rests on the circumference of the ring, making an angle \(\theta\) with the radius to the highest point on the circumference. Find the relation between \(\alpha,\theta\) and the coefficient of friction, \(\mu,\) which must hold when the rod is in limiting equilibrium.

Show Solution
TikZ diagram
It is important to define clear coordinate axes, so let the \(x\)-axis point up the line of greatest slope of the ring. The \(z\)-axis perpendicular to the ring, and the \(y\)-axis perpendicular to both of these. Our method is going to be to take moments about \(O\) to avoid worrying about the force at the pivot. There are \(3\) forces we need to worry about:
  • The mass of the rod
  • The reaction where it meets the ring
  • The friction at the ring
In our coordinate frame, the reaction will act in the \(z\)-direction, \(\displaystyle \begin{pmatrix} 0 \\ 0 \\ R \end{pmatrix}\), the friction force will act in the \(x-y\) plane: \(\displaystyle \begin{pmatrix} \mu R \sin \theta \\ -\mu R \cos \theta \\ 0 \end{pmatrix}\). We don't know the mass, but we know it will be acting "vertically", so \(\cos \alpha\) of it will act in the \(z\)-axis and \(\sin \alpha\) will act in the \(y\)-axis, ie it will act parallel to \(\displaystyle \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix}\). When taking moments, we need to consider \(\mathbf{r}\) the direction of the rod. This will be \(\displaystyle \begin{pmatrix} \cos \theta \\ \sin \theta \\ 0 \end{pmatrix}\). The moment of the weight will all be parallel to \(\mathbf{r} \times \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix}\). Similarly the moments of the contact forces will be \(\mathbf{r} \times \begin{pmatrix} \mu R \sin \theta \\ -\mu R \cos \theta \\ R \end{pmatrix}\). Since these moments sum to \(\mathbf{0}\) as we are in equilibrium, these vectors must be parallel. Therefore it is sufficient to check the vector triple product, \begin{align*} && 0 &= \begin{pmatrix} \cos \theta \\ \sin \theta \\ 0 \end{pmatrix} \cdot \left ( \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix} \times \begin{pmatrix} \mu \sin \theta \\ -\mu \cos \theta \\ 1 \end{pmatrix} \right ) \\ &&&= \cos \theta (\mu \cos \theta \cos \alpha)-\sin \theta (\sin \alpha - \mu \sin \theta \cos \alpha) \\ &&&= \mu((\sin^2 \theta+\cos^2 \theta) \cos \alpha) -\sin \theta \sin \alpha \\ \Rightarrow && \mu &= \tan \alpha \sin \theta \end{align*}
1987 Paper 2 Q12
D: 1500.0 B: 1500.0

A long, inextensible string passes through a small fixed ring. One end of the string is attached to a particle of mass \(m,\) which hangs freely. The other end is attached to a bead also of mass \(m\) which is threaded on a smooth rigid wire fixed in the same vertical plane as the ring. The curve of the wire is such that the system can be in static equilibrium for all positions of the bead. The shortest distance between the wire and the ring is \(d(>0).\) Using plane polar coordinates centred on the ring, find the equation of the curve. The bead is set in motion. Assuming that the string remains taut, show that the speed of the bead when it is a distance \(r\) from the ring is \[ \left(\frac{r}{2r-d}\right)^{\frac{1}{2}}v, \] where \(v\) is the speed of the bead when \(r=d.\)

Show Solution
TikZ diagram
Assume the total length of the string is \(l\). Then the total energy of the system (when nothing is moving) for a given \(\theta\) is: \(mg(r-l) + mgr \sin \theta\) Since for a point in static equilibrium, the derivative of this must be \(0\), this must be constant. So: \(r\l \sin \theta + 1\r = C \Rightarrow r = \frac{C}{1+\sin \theta}\) \(r\) will be smallest when \(\sin \theta = 1\), ie in polar coordinates, the equation should be \(r = \frac{2d}{1+\sin \theta}\) Alternatively, by considering forces, the shape must be a parabola with the ring at the focus. Considering the bead, it will have speed of \(r \dot{\theta}\) tangentially, and \(-\dot{r}\). The other particle will have speed \(\dot{r}\). Differentiating wrt to \(t\) \begin{align*} && 0 &= \dot{r}(\sin \theta + 1) + r \dot{\theta} \cos \theta \\ \Rightarrow && \dot{\theta} &= \frac{-\dot{r}(1+\sin \theta)}{r \cos \theta} \\ &&&= \frac{-\dot{r} 2d}{r^2 \sqrt{1-\l \frac{2d}{r}-1\r^2}} \\ &&&= \frac{-2d\dot{r}}{r^2\sqrt{\frac{r^2-(2d-r)^2}{r^2}}} \\ &&&= \frac{-d\dot{r}}{r\sqrt{dr-d^2}} \end{align*} By conservation of energy (since GPE is constant throughout the system, KE must be constant): \begin{align*} && \frac12 m (r^2 \dot{\theta}^2+\dot{r}^2) +\frac12 m \dot{r}^2 &= \frac12mv^2 \\ \Rightarrow && v^2 &= r^2 \dot{\theta}^2 + 2\dot{r}^2 \\ &&&= r^2 \frac{d^2\dot{r}^2}{r^2(dr-d^2)} + 2\dot{r}^2 \\ &&&= \dot{r}^2 \l \frac{d }{r-d} + 2 \r \\ &&&= \dot{r}^2 \l \frac{2r-d}{r-d} \r \\ \Rightarrow && v &= \dot{r} \l \frac{2r-d}{r-d} \r^{\frac12} \\ \Rightarrow && \dot{r} &= \l \frac{r-d}{2r-d} \r^{\frac12} v \\ \Rightarrow && u^2 &= r^2 \dot{\theta}^2+\dot{r}^2\\ &&&= \dot{r}^2 \l \frac{d }{r-d} + 1 \r \\ &&&= \l \frac{r-d}{2r-d} \r \l \frac{r}{r-d} \r v^2 \\ &&&= \l \frac{d}{2r-d} \r v^2 \\ \Rightarrow && u &= \l \frac{d}{2r-d} \r^{\frac12} v \end{align*}