The resultant of two forces P N and 3 N is a force of 7 N. If the direction of the 3 N force were reversed, the resultant would be \[\sqrt{19}N\]. The value of P is
AIEEE Solved Paper-2007
A) 5 N
B) 6 N
C) 3 N
D) 4 N
Correct Answer:
A
Solution :
\[{{7}^{2}}={{P}^{2}}+{{3}^{2}}+2\times 3\times p\cos \theta \] ... (i) and\[{{(\sqrt{19})}^{2}}={{p}^{2}}+{{(-3)}^{2}}+2\times (-3)\times p\cos \theta \]...(ii) On adding Eqs. (i) and (ii), we get \[68\,=2{{P}^{2}}\,+18\Rightarrow P=5\]