WAEC 2019 · Paper 2 · Q15

(a)
Forces $(5\text{ N}, 030^\circ)$ , $(P\text{ N}, 060^\circ)$ , $(Q\text{ N}, 150^\circ)$ , $(3\text{ N}, 180^\circ)$ and $(5\text{ N}, 270^\circ)$ act on a body. If the system is in equilibrium, find, correct to one decimal place, the values of $P$ and $Q$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

East: $5\sin30^\circ + P\sin60^\circ + Q\sin150^\circ + 3\sin180^\circ + 5\sin270^\circ = 0$ .
So $2.5 + 0.866P + 0.5Q - 5 = 0$ , which gives $0.866P + 0.5Q = 2.5$ .
North: $5\cos30^\circ + P\cos60^\circ + Q\cos150^\circ + 3\cos180^\circ + 5\cos270^\circ = 0$ .
So $4.330 + 0.5P - 0.866Q - 3 = 0$ , which gives $0.5P - 0.866Q = -1.330$ .
Multiply the first by 0.866 and the second by 0.5, then add: $P = 2.165 - 0.665 = 1.5\text{ N}$ .
Then $0.5Q = 2.5 - 1.299$ , so $Q \approx 2.4\text{ N}$ .