Question

55.
Two horizontal forces of magnitudes 10 N and P N act on
a particle. The force of magnitude 10 N 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 this resultant.
PN
309
10 N
(1) 10/2 N
(3) 12.15 N
(2) 1513 N
(4) 10/3 N
​

Answer 1

Answer:

option 4 is our choice my friend check your answer...

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Resultant\:force=10\sqrt{3}\:N}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Angle \: between \: resultant \: vector \: and \: P \: N = 30 \degree \\ \\ \tt: \implies Force \: acting \: in \: west = 10 \: N\\ \\ \red{\underline \bold{To \: find :}} \\ \tt: \implies Magnitude \: of \: resultant =?$

• According to given question :

$\bold{Breaking \: force \: P\: in \: component : } \\ \tt: \implies Force \:acting \: in \: north \: direction =P\: cos \theta \\ \\ \tt: \implies Force \: acting \: in \: east \: direction = P \: sin \: \theta \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies P\:sin \: \theta = 10 \\ \\ \tt: \implies P \: sin \: 30 \degree = 10 \\ \\ \tt: \implies P \times \frac{1}{2} = 10 \\ \\ \green{\tt: \implies P = 20 \: N} \\ \\ \bold{For \: Resultant \: force: } \\ \tt: \implies Resultant \: force= P \: cos \: \theta \\ \\ \tt: \implies Resultant \: force = 20 \times cos \: 30 \degree \\ \\ \tt: \implies Resultant \: force= 20 \times \frac{ \sqrt{3} }{2} \\ \\ \green{\tt: \implies Resultant \: force =10 \sqrt{3} \: N}$