Question

Two horizontal forces of magnitudes $10N$ and $PN$ act on a particle. The force of magnitude $10N$ acts due west and the force of magnitude $PN$acts on a bearing of ${30}^{\circ }$ east of north as shown in figure. The resultant of these two force acts due north. Find the magnitude of this resultant.

• A. $10\sqrt{3}N$
  
• B. $12\sqrt{5}N$
  
• C. $15\sqrt{3}N$
  
• D. $10\sqrt{2}N$
  

Answer 1

The correct option is A $10\sqrt{3}N$


As given the resultant force acts due north.
So there will be no horizontal force.The horizontal components of force $PN$cancel $10N$
$10=P\sin 30$
$\frac{P}{2}=10$
$P=20N$
Resultant
$R=P\cos 30=20\times \frac{\sqrt{3}}{2}=10\sqrt{3}N$
Final Answer: (d)