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2009 Paper 2 Q9
D: 1600.0
B: 1484.0
- 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 \(OZ\).
- 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.
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:
- \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 \(OZ\)
- \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*}