Problems

A rod \(AB\) of length 0.81 m and mass 5 kg is in equilibrium with the end \(A\) on a rough floor and the end \(B\) against a very rough vertical wall. The rod is in a vertical plane perpendicular to the wall and is inclined at \(45^{\circ}\) to the horizontal. The centre of gravity of the rod is at \(G\), where \(AG = 0.21\) m. The coefficient of friction between the rod and the floor is 0.2, and the coefficient of friction between the rod and the wall is 1.0. Show that the friction cannot be limiting at both \(A\) and \(B\). A mass of 5 kg is attached to the rod at the point \(P\) such that now the friction is limiting at both \(A\) and \(B\). Determine the length of \(AP\).

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 \(\mu_2\).

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Solution:

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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*}

A thin beam is fixed at a height \(2a\) above a horizontal plane. A uniform straight rod \(ACB\) of length \(9a\) and mass \(m\) is supported by the beam at \(C\). Initially, the rod is held so that it is horizontal and perpendicular to the beam. The distance \(AC\) is \(3a\), and the coefficient of friction between the beam and the rod is \(\mu\). The rod is now released. Find the minimum value of \(\mu\) for which \(B\) strikes the horizontal plane before slipping takes place at \(C\).

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 \(1/\sqrt{3}\).

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Solution:

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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 \(\mu \geq \frac{1}{\sqrt{3}}\)

By pressing a finger down on it, a uniform spherical marble of radius \(a\) is made to slide along a horizontal table top with an initial linear velocity \(v_0\) and an initial {\em backward} angular velocity \(\omega_0\) about the horizontal axis perpendicular to \(v_0\). The frictional force between the marble and the table is constant (independent of speed). For what value of \(v_0/(a\omega_0)\) does the marble

1. slide to a complete stop,
2. come to a stop and then roll back towards its initial position with linear speed \(v_0/7\).

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Solution:

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If the frictional force is \(F\), then: \begin{align*} L = I\ddot{\theta} && -Fa &= \frac{2}{5}ma^2 \dot{\omega} \\ \text{N2}(\rightarrow) && -F &= m\dot{v} \\ \\ \Rightarrow && \l \frac{2}{5}ma \dot{\omega} - \dot{v} \r &= 0 \\ \Rightarrow && \frac{2}{5}a \omega - v &= c \\ \end{align*}

1. If the ball completely stops, then \(\omega = v = 0 \Rightarrow \frac{2}{5}a \omega_0 - v_0 = 0 \Rightarrow \frac{v_0}{a \omega_0} = \frac25\).
2. If the ball rolls backwards with linear speed \(v_0/7\), \(v = - \frac{v_0}{7}\) and \(a \omega = \frac{v_0}{7}\), \begin{align*} && \frac{2}{5}a \omega_0 - v_0 &= \frac{2}{5} \frac{v_0}{7} + \frac{v_0}{7} \\ && &= \frac{1}{5} v_0 \\ \Rightarrow && \frac{v_0}{a \omega_0} &= \frac{1}{3} \end{align*}

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. \(\mu_{1}\geqslant\left(\dfrac{2W_{1}\cot\alpha}{W_{1}+W_{2}}-1\right)^{-1}\,.\)

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Solution:

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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*}

Two identical uniform cylinders, each of mass \(m,\) lie in contact with one another on a horizontal plane and a third identical cylinder rests symmetrically on them in such a way that the axes of the three cylinders are parallel. Assuming that all the surfaces in contact are equally rough, show that the minimum possible coefficient of friction is \(2-\sqrt{3}.\)

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Solution:

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First observe that many forces are equal by symmetry. Also notice that \(A\) and \(B\) are trying to roll in opposite directions, therefore there is no friction between \(A\) and \(B\). Considering the system as a whole \(R_1 = \frac32 mg\). \begin{align*} \text{N2}(\uparrow,C): && 0 &= -mg +2R_3\cos 30^{\circ} + 2F_{CA} \cos 60^{\circ} \\ \Rightarrow && mg &= \sqrt{3}R_3 + F_{CA} \\ \\ \text{N2}(\uparrow, A): && 0 &= -mg + \frac32mg-R_3\cos 30^{\circ} -F_{AC} \cos 60^\circ \\ \Rightarrow && mg &= \sqrt{3}R_3+F_{AC} \\ \text{N2}(\rightarrow, A): && 0 &= F_{AC}\cos 30^{\circ}+F_1-R_3 \cos 60^\circ -R_2 \\ \Rightarrow && 0 &=F_{AC}\sqrt{3}+2F_1-R_3-2R_2 \\ \overset{\curvearrowleft}{A}: && 0 &= F_1 - F_{AC} \end{align*} Since \(F_1 = F_{AC} = F_{CA}\) we can rewrite everything in terms of \(F= F_1\) so. \begin{align*} && mg &= \sqrt{3}R_3 + F \\ && 0 &= (2+\sqrt{3})F -R_3-2R_2 \\ \Rightarrow && (2+\sqrt3)F &\geq R_3 \\ \Rightarrow && F &\geq (2-\sqrt{3})R_3 \\ \Rightarrow && \mu & \geq 2 - \sqrt{3} \end{align*}

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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

Two particles \(P_{1}\) and \(P_{2}\), each of mass \(m\), are joined by a light smooth inextensible string of length \(\ell.\) \(P_{1}\) lies on a table top a distance \(d\) from the edge, and \(P_{2}\) hangs over the edge of the table and is suspended a distance \(b\) above the ground. The coefficient of friction between \(P_{1}\) and the table top is \(\mu,\) and \(\mu<1\). The system is released from rest. Show that \(P_{1}\) will fall off the edge of the table if and only if \[ \mu<\frac{b}{2d-b}. \] Suppose that \(\mu>b/(2d-b)\) , so that \(P_{1}\) comes to rest on the table, and that the coefficient of restitution between \(P_{2}\) and the floor is \(e\). Show that, if \(e>1/(2\mu),\) then \(P_{1}\) comes to rest before \(P_{2}\) bounces a second time.

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A non-uniform rod \(AB\) of mass \(m\) is pivoted at one end \(A\) so that it can swing freely in a vertical plane. Its centre of mass is a distance \(d\) from \(A\) and its moment of inertia about any axis perpendicular to the rod through \(A\) is \(mk^{2}.\) A small ring of mass \(\alpha m\) is free to slide along the rod and the coefficient of friction between the ring and rod is \(\mu.\) The rod is initially held in a horizontal position with the ring a distance \(x\) from \(A\). If \(k^{2} > xd\), show that when the rod is released, the ring will start to slide when the rod makes an angle \(\theta\) with the downward vertical, where \[ \mu\tan\theta=\frac{3\alpha x^{2}+k^{2}+2xd}{k^{2}-xd}. \] Explain what will happen if (i) \(k^{2}=xd\) and (ii) \(k^{2} < xd\).