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2014 Paper 2 Q9
D: 1600.0 B: 1484.0
centre of mass lamina rectangular friction limiting equilibrium moments rough wall string tension rigid body trigonometric identity

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 \(b(2\tan\theta+ \tan \phi) < a\).

Solution:

TikZ diagram
  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!

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2010 Paper 1 Q9
D: 1500.0 B: 1500.0
centre of mass lamina rectangular friction limiting equilibrium moments trigonometric identity rough wall

TikZ diagram
The diagram shows a uniform rectangular lamina with sides of lengths \(2a\) and \(2b\) leaning against a rough vertical wall, with one corner resting on a rough horizontal plane. The plane of the lamina is vertical and perpendicular to the wall, and one edge makes an angle of \(\alpha\) with the horizontal plane. Show that the centre of mass of the lamina is a distance \(a\cos\alpha + b\sin\alpha\) from the wall. The coefficients of friction at the two points of contact are each \(\mu\) and the friction is limiting at both contacts. Show that \[ a\cos(2\lambda +\alpha) = b\sin\alpha \,, \] where \(\tan\lambda = \mu\). Show also that if the lamina is square, then \(\lambda = \frac{1}{4}\pi -\alpha\).

Solution:

TikZ diagram
The horizontal distance to \(X\) is \(a\cos \alpha\). The horizontal distance to \(G\) from \(X\) is \(b \sin \alpha\), therefore the centre of mass is a distance \(a \cos \alpha + b \sin \alpha\) from the wall.
TikZ diagram
\begin{align*} \text{lim eq}: && F_W &= \mu R_W \\ && F_G &= \mu R_G\\ \text{N2}(\rightarrow): && \mu R _G &= R_W \\ \text{N2}(\uparrow): && \mu R_W + R_G &= W \\ \Rightarrow && (1+\mu^2)R_G &= W \\ \overset{\curvearrowleft}{Y}: && R_G 2a \cos \alpha - F_G 2a \sin \alpha - W (a \cos \alpha + b \sin \alpha) &= 0 \\ \Leftrightarrow && 2a R_G \cos \alpha -2a \mu R_G \sin \alpha - (1+\mu^2)R_G(a \cos \alpha + b \sin \alpha) &= 0 \\ \Leftrightarrow && a(1-\mu^2)\cos \alpha - (b(1+\mu^2)+2a\mu) \sin \alpha &= 0 \\ \Leftrightarrow && a(1-\tan^2 \lambda )\cos \alpha - (b(1-\tan^2 \lambda)+2a\tan \lambda) \sin \alpha &= 0 \\ \Leftrightarrow&& a(2-\sec^2 \lambda) \cos \alpha - (b\sec^2 \lambda+2a\mu) \sin \alpha &= 0 \\ \Leftrightarrow && a (2\cos \lambda - 1)\cos \alpha - 2a \sin \lambda \cos \lambda \sin \alpha &= b \sin \alpha \\ \Leftrightarrow && a\cos 2 \lambda \cos \alpha - a\sin 2 \lambda \sin \alpha &= b \sin \alpha \\ \Leftrightarrow && a\cos (2 \lambda +\alpha) &= b \sin \alpha \end{align*} as required. If the lamina is a square, \(a = b\), so \begin{align*} && \cos(2\lambda + \alpha) &= \sin \alpha \\ \Rightarrow && 0 &= \cos(2\lambda + \alpha) -\sin \alpha \\ &&&= \sin \left (\frac{\pi}{2} - 2 \lambda - \alpha \right )-\sin \alpha \\ &&&= 2 \cos\left ( \frac{\frac{\pi}{2} - 2 \lambda - \alpha +\alpha}{2} \right) \sin\left ( \frac{\frac{\pi}{2} - 2 \lambda - \alpha -\alpha}{2} \right) \\ &&&= 2 \cos\left ( \frac{\pi}4 -\lambda\right) \sin\left ( \frac{\pi}4 -\lambda-\alpha \right) \\ \Rightarrow && \lambda -\frac{\pi}{4} = -\frac{\pi}{2} & \text{ or } \frac{\pi}{4} - \lambda - \alpha = 0 \\ \Rightarrow && \alpha &= \frac{\pi}{4}-\lambda \end{align*}

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2009 Paper 2 Q9
D: 1600.0 B: 1484.0
centre of mass trapezium lamina composite shape 3D geometry thin shell open tank moments

  1. A uniform lamina \(OXYZ\) is in the shape of the trapezium shown in the diagram. It is right-angled at \(O\) and \(Z\), and \(OX\) is parallel to \(YZ\). The lengths of the sides are given by \(OX=9\,\)cm, \(XY=41\,\)cm, \(YZ=18\,\)cm and \(ZO=40\,\)cm. Show that its centre of mass is a distance \(7\,\)cm from the edge \(OZ\).
    TikZ diagram
  2. The diagram shows a tank with no lid made of thin sheet metal. The base \(OXUT\), the back \(OTWZ\) and the front \(XUVY\) are rectangular, and each end is a trapezium as in part (i). The width of the tank is \(d\,\)cm.
    TikZ diagram
    Show that the centre of mass of the tank, when empty, is a distance \[ \frac {3(140+11d)}{5(12+d)}\,\text{cm} \] from the back of the tank. The tank is then filled with a liquid. The mass per unit volume of this liquid is \(k\) times the mass per unit area of the sheet metal. In the case \(d=20\), find an expression for the distance of the centre of mass of the filled tank from the back of the tank.

Solution:

  1. TikZ diagram
    \begin{array}{c|c|c|c} & OXX'Z & XX'Y & OXYZ \\ \hline \text{Area} & 360 & 180 & 540\\ \text{COM} & \binom{4.5}{20} & \binom{12}{\frac{80}{3}} & \binom{\overline{x}}{\overline{y}} \end{array} \begin{align*} && 2 \binom{3}{20} + \binom{12}{\frac{80}{3}} &= 3 \binom{\overline{x}}{\overline{y}} \\ \Rightarrow && \binom{\overline{x}}{\overline{y}} &= \frac13 \binom{21}{\frac{200}{3}} \\ &&&= \binom{7}{\frac{200}{9}} \end{align*} ie, the centre of mass is \(7\text{ cm}\) from \(OZ\)
  2. \begin{align*} && \underbrace{540 \cdot 7}_{OXYZ} + \underbrace{540 \cdot 7}_{TUVW} + \underbrace{40d\cdot 0}_{OTWZ} + \underbrace{9d\cdot 4.5}_{OXUT} + \underbrace{41d \cdot 13.5}_{XUVY} &= (540+540+40d+9d+41d) \overline{x} \\ \Rightarrow && \overline{x} &= \frac{540\cdot 14 + 50d \cdot 4.5 + 41d \cdot 9}{1080 + 90d} \\ &&&= \frac{90 \cdot 84 + 225d + 369d}{1080+90d} \\ &&&= \frac{90 \cdot 84 + 594d}{1080+90d} \\ &&&= \frac{54(140+11d)}{90(12+d)} \\ &&&= \frac{3(140+11d)}{5(12+d)} \end{align*} The volume of the prizm is \(540d\), it's center of mass is \(7\). For the tank, it COM is \(\frac{3(140+11\cdot20)}{5(12+20)} = \frac{27}4\) and area is \(2880\) Therefore for the combined shape we have: \begin{align*} && 540dk \cdot 7 + 2880 \cdot \frac{27}{4} &= (540 \cdot20 k+2880) \overline{x} \\ \Rightarrow && \overline{x} &= \frac{720(150k+27)}{720(15k + 4)} \\ &&&= \frac{3(50k+9)}{15k+4} \end{align*} \begin{align*} && \end{align*}

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2003 Paper 1 Q10
D: 1500.0 B: 1500.0
centre of mass lamina trapezium composite body suspension equilibrium quadratic rectangular lamina

\(ABCD\) 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.

Solution:

TikZ diagram
\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*}
TikZ diagram
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
TikZ diagram
which definitely wouldn't have \(G\) hanging below \(A\).

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1991 Paper 1 Q13
D: 1516.0 B: 1484.0
circular motion conservation of momentum semicircular groove lamina impulse smooth constraint particle in groove average force

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

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A heavy smooth lamina of mass \(M\) is free to slide without rotation along a straight line on a fixed smooth horizontal table. A smooth groove \(ABC\) is inscribed in the lamina, as indicated in the above diagram. The tangents to the groove at \(A\) and at \(B\) are parallel to the line. When the lamina is stationary, a particle of mass \(m\) (where \(m < M\)) enters the groove at \(A\). The particle is travelling, with speed \(V\), parallel to the line and in the plane of the lamina and table. Calculate the speeds of the particle and of the lamina, when the particle leaves the groove at \(C\). Suppose now that the lamina is held fixed by a peg attached to the line. Supposing that the groove \(ABC\) is a semicircle of radius \(r\), obtain the value of the average force per unit time exerted on the peg by the lamina between the instant that the particle enters the groove and the instant that it leaves it.

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1989 Paper 2 Q12
D: 1600.0 B: 1500.0
centre of mass lamina rectangular equilibrium smooth pegs trigonometric identity rigid body geometric proof inequality

A uniform rectangular lamina of sides \(2a\) and \(2b\) rests in a vertical plane. It is supported in equilibrium by two smooth pegs fixed in the same horizontal plane, a distance \(d\) apart, so that one corner of the lamina is below the level of the pegs. Show that if the distance between this (lowest) corner and the peg upon which the side of length \(2a\) rests is less than \(a\), then the distance between this corner and the other peg is less than \(b\). Show also that \[ b\cos\theta-a\sin\theta=d\cos2\theta, \] where \(\theta\) is the acute angle which the sides of length \(2b\) make with the horizontal.

Solution:

TikZ diagram
We must have \(G\) between the two pegs (vertically), otherwise we will induce a moment. Considering moments about the peg, if the second peg is outside the centre then we must induce a moment and therefore we cannot be in equilibrium. \begin{align*} \text{N2}(\nearrow):&& 0 &= R_1-mg\sin\theta \\ \text{N2}(\nwarrow):&&0&= R_2-mg \cos \theta \\ \Rightarrow && R_1 &= mg \sin\theta \\ && R_2 &= mg \cos\theta \\ \\ \overset{\curvearrowleft}{G}: && 0 &= R_1(a-d\sin\theta) -R_2(b-d \cos\theta) \\ \Rightarrow && 0&= a \sin\theta -d \sin^2\theta - b\cos \theta+d \cos^2 \theta \\ \Rightarrow && b \cos \theta - a \sin \theta &= d \cos 2 \theta \end{align*}

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