WAEC 2010 · Paper 2 · Q17

Not to scale.

(a)
A particle $Q$ is acted upon by three coplanar forces $40$ N, $15\sqrt3$ N and $20$ N, as shown in the diagram. Find the force required to prevent $Q$ from moving.
(b)
A body of mass 5 kg at rest is acted upon by forces $F_1 = (10 \text{ N}, 090^\circ)$ , $F_2 = (20 \text{ N}, 210^\circ)$ and $F_3 = (4 \text{ N}, 330^\circ)$ . Find, correct to one decimal place, the magnitude and direction of the resultant force.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Resolve horizontally (to the right): $40\cos30^\circ + 15\sqrt3 = 20\sqrt3 + 15\sqrt3$
$= 35\sqrt3$ N.
Resolve vertically (upwards): $40\sin30^\circ - 20 = 20 - 20 = 0$ .
So the three forces add up to $35\sqrt3$ N horizontally to the right.
The force that stops $Q$ moving is equal and opposite: $35\sqrt3 \approx 60.6$ N horizontally to the left (bearing $270^\circ$ ).

Resolve east using $F\sin(\text{bearing})$ : $10\sin90^\circ + 20\sin210^\circ + 4\sin330^\circ = 10 - 10 - 2$
$= -2$ .
Resolve north using $F\cos(\text{bearing})$ : $0 + 20\cos210^\circ + 4\cos330^\circ = -17.321 + 3.464$
$= -13.856$ .
Magnitude: $\sqrt{(-2)^2 + (-13.856)^2} = \sqrt{196}$
$= 14.0$ N.
Both parts are negative, so the resultant points south-west of south: angle west of south $= \tan^{-1}\dfrac{2}{13.856} = 8.2^\circ$ .
Bearing $= 180^\circ + 8.2^\circ = 188.2^\circ$ .