WAEC 2018 · Paper 2 · Q15

Statics: forces, equilibrium & moments

A uniform beam $XY$ , $4\text{ m}$ long and weighing $350\text{ N}$ , rests on two pivots $P$ and $Q$ . It is kept in equilibrium by weights of $80\text{ N}$ attached at $X$ and $1000\text{ N}$ attached at a point between $P$ and $Q$ such that it is $0.6\text{ m}$ from $Q$ . $|XP| = 0.8\text{ m}$ and $|PQ| = 2.2\text{ m}$ .

(a)
Calculate the reactions at $P$ and $Q$ .
Separate values with commas, e.g. 3, −2
(b)
If the $1000\text{ N}$ weight is replaced with a $1200\text{ N}$ weight, at what point from $Q$ should it be placed in order to maintain this equilibrium?
Show the answer
With the weight $x\text{ m}$ from $Q$ : $R_Q = \dfrac{2996 - 1200x}{2.2}$ and $R_P = \dfrac{590 + 1200x}{2.2}$

Worked solution (try it first)

(a)

From $X$ : $P$ at $0.8$ m, the centre at $2$ m, the $1000\text{ N}$ weight at $2.4$ m and $Q$ at $3$ m.
Moments about $P$ : $2.2R_Q + 80(0.8) = 350(1.2) + 1000(1.6)$ .
$2.2R_Q = 420 + 1600 - 64 = 1956$ , so $R_Q \approx 889.09\text{ N}$ .
Up = down: $R_P = 80 + 350 + 1000 - 889.09$
$\approx 540.91\text{ N}$ .

(b)

Put the $1200\text{ N}$ weight $x$ m from $Q$ , so $2.2 - x$ m from $P$ .
Moments about $P$ : $2.2R_Q + 64 = 420 + 1200(2.2 - x)$ , so $R_Q = \dfrac{2996 - 1200x}{2.2}$ .
Up = down: $R_P = 1630 - R_Q$
$= \dfrac{590 + 1200x}{2.2}$ .
Both reactions stay positive for any point between $P$ and $Q$ , so the beam balances wherever the weight goes there.

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