Two horizontal forces of magnitudes of `10N` and `P N` act on a particle. The force of magnitude `10N` acts due west and the force of magnitude `P N` acts on a bearing of `30^(@)` east of north as shown in (figure) The resultant of these two force acts due north. Find the magnitude of the resultant.

Answer

`10sqrt(3)N`

Explanation

Resultant acts along y-axis
`rArr P_(x)=Pcos 60^(@)=10N` or `P/2=10` or `P=20`
Resultant, `P_(y)= P sin 60^(@)=20xxsqrt(3)/2=10sqrt(3)N`