\includegraphics{figure_4} A uniform lamina of weight 15 N is in the form of a trapezium \(ABCD\) with dimensions as shown in the diagram. The lamina is freely hinged at \(A\) to a fixed point. One end of a light inextensible string is attached to the lamina at \(B\). The lamina is in equilibrium with \(AB\) horizontal; the string is taut and in the same vertical plane as the lamina, and makes an angle of \(30°\) upwards from the horizontal (see diagram). Find the tension in the string. [5]

Weight split is 9N:6N

For lamina $9 \times 0.75 + 6 \times 0.5$ $T = 1 \times 1.5\sin30°$ Tension is 13N | B1 M1 A1ft A1 A1 [5] | For taking moments about A

**Alternatively**

$[(1.5^2 + 1.5 \times 2)\bar{x} = 1.5^2 \times 0.75 + \frac{1}{2} 1.5 \times 2 \times 0.5]$ $\bar{x} = 0.65$

$15 \times 0.65 = T \times 1.5\sin30°$ Tension is 13N | M1 A1 M1 A1ft A1 [5] | For using $A\bar{x} = A_1x_1 + A_2x_2$. For taking moments about A

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\includegraphics{figure_4}

A uniform lamina of weight 15 N is in the form of a trapezium $ABCD$ with dimensions as shown in the diagram. The lamina is freely hinged at $A$ to a fixed point. One end of a light inextensible string is attached to the lamina at $B$. The lamina is in equilibrium with $AB$ horizontal; the string is taut and in the same vertical plane as the lamina, and makes an angle of $30°$ upwards from the horizontal (see diagram). Find the tension in the string.
[5]

\hfill \mbox{\textit{CAIE M2 2010 Q4 [5]}}