WAEC 2009 · Paper 2 · Q18✱

(a)
The position vectors of points $A$ , $B$ and $C$ are $\mathbf i + 5\mathbf j$ , $3\mathbf i + 9\mathbf j$ and $-\mathbf i + \mathbf j$ respectively. (i) Show that the points $A$ , $B$ and $C$ are collinear. (ii) Determine the ratio $|AB| : |BC|$ .
Show the answer
(ii) $|AB| : |BC| = 1 : 2$
(b)
A uniform beam $XY$ of mass 10 kg and length 24 m is hung horizontally from a cross bar by two vertical inextensible strings, one attached to $X$ and the other at a point $M$ , 4 m away from $Y$ . A mass of 50 kg is suspended at a point $N$ which is 8 m from $X$ . If the system remains in equilibrium, calculate the tensions in the strings. [Take $g = 10\text{ m s}^{-2}$ ]
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$\overrightarrow{AB} = (3 - 1)\mathbf i + (9 - 5)\mathbf j$
$= 2\mathbf i + 4\mathbf j$ .
$\overrightarrow{AC} = (-1 - 1)\mathbf i + (1 - 5)\mathbf j$
$= -2\mathbf i - 4\mathbf j$
$= -\overrightarrow{AB}$ .
$\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel and share the point $A$ , so $A$ , $B$ and $C$ are collinear.

$|AB| = \sqrt{2^2 + 4^2}$
$= \sqrt{20}$
$= 2\sqrt5$ .
$\overrightarrow{BC} = -4\mathbf i - 8\mathbf j$ , so $|BC| = \sqrt{16 + 64}$
$= \sqrt{80}$
$= 4\sqrt5$ .
So $|AB| : |BC| = 2\sqrt5 : 4\sqrt5 = 1 : 2$ .

Weights: beam $10 \times 10 = 100$ N at its midpoint, 12 m from $X$ .
Load $50 \times 10 = 500$ N at $N$ , 8 m from $X$ .
The string at $M$ is $24 - 4 = 20$ m from $X$ .
Take moments about $X$ (this removes $T_1$ ): $T_2 \times 20 = 500 \times 8 + 100 \times 12$
$= 5200$ .
So $T_2 = 260$ N.
Resolve vertically: $T_1 + T_2 = 500 + 100$ , so $T_1 = 600 - 260 = 340$ N.
The tension at $X$ is 340 N and the tension at $M$ is 260 N.