WAEC 2022 · Paper 2 · Q14

(a)
An object of mass $40\text{ kg}$ sits at the end $T$ of a see-saw that consists of a uniform beam $ST$ of length $7\text{ m}$ . Another object of mass $50\text{ kg}$ sits at a distance $y\text{ m}$ from $S$ . Given that the beam is supported at the point $P$ such that $|SP| : |ST| = 3 : 5$ and the mass of the beam is $17\text{ kg}$ , (i) illustrate the information on a diagram; (ii) calculate, correct to one decimal place, the value of $y$ such that the see-saw is kept in equilibrium. $[\text{Take } g = 10\text{ m s}^{-2}]$
(b)
If the angle between $(\mathbf i + k\mathbf j)$ and $(3\mathbf i - 4\mathbf j)$ is $\cos^{-1}\left(\frac{11\sqrt5}{25}\right)$ , find the value of $k$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Draw the beam $ST$ with the support $P$ , 40 kg at $T$ , 50 kg at $y$ m from $S$ , and the beam's weight at its midpoint.

$|SP| = \frac35 \times 7 = 4.2\text{ m}$ and $|PT| = 2.8\text{ m}$ .
The midpoint is 3.5 m from $S$ , so the beam's weight ( $170\text{ N}$ ) acts $0.7\text{ m}$ from $P$ on the $S$ side.
Moments about $P$ : $500(4.2 - y) + 170(0.7) = 400(2.8)$ .
$2100 - 500y + 119 = 1120$ , so $500y = 1099$ and $y \approx 2.2\text{ m}$ .

$(\mathbf i + k\mathbf j) \cdot (3\mathbf i - 4\mathbf j) = 3 - 4k$ , and the lengths are $\sqrt{1 + k^2}$ and 5.
So $\dfrac{3 - 4k}{5\sqrt{1 + k^2}} = \dfrac{11\sqrt5}{25}$ .
Square both sides: $\dfrac{(3 - 4k)^2}{25(1 + k^2)} = \dfrac{121}{125}$ , so $5(3 - 4k)^2 = 121(1 + k^2)$ .
Expand: $45 - 120k + 80k^2 = 121 + 121k^2$ , so $41k^2 + 120k + 76 = 0$ .
Factorise: $(41k + 38)(k + 2) = 0$ , so $k = -2$ or $k = -\frac{38}{41}$ .
Both make $3 - 4k$ positive, as the positive cosine needs, so both are valid.