WAEC 2016 · Paper 2 · Q14

(a)
A block of mass $0.5\text{ kg}$ rests on the floor of a lift which is moving upwards with an acceleration of $2\text{ m s}^{-2}$ . Calculate the reaction between the block and the lift. $[\text{Take } g = 10\text{ m s}^{-2}]$
(b)
A uniform beam $XY$ of length $4\text{ m}$ and mass $M\text{ kg}$ rests on two supports $A$ and $B$ , where $|AX| = 0.4\text{ m}$ and $|BY| = 1.2\text{ m}$ . Masses of $8\text{ kg}$ and $10\text{ kg}$ are suspended at points $P$ and $Q$ respectively, where $|XP| = 1.1\text{ m}$ and $|QY| = 0.3\text{ m}$ . If the reaction at $A$ is $90\text{ N}$ and the system remains in equilibrium, find, correct to one decimal place, the: (i) value of $M$ ; (ii) reaction at $B$ . $[\text{Take } g = 10\text{ m s}^{-2}]$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

From $X$ : $A$ at $0.4$ , $P$ at $1.1$ , the centre at $2.0$ , $B$ at $2.8$ and $Q$ at $3.7$ m.
Moments about $B$ : $90(2.4) + 100(0.9) = 80(1.7) + 10M(0.8)$ .
$216 + 90 = 136 + 8M$ , so $M = 21.25 \approx 21.3\text{ kg}$ .
Up = down: $90 + R_B = 80 + 212.5 + 100$ , so $R_B = 302.5\text{ N}$ .