Laws Of Motion Question 64
Question: When forces $ F _{1},F _{2},F _{3} $ are acting on a particle of mass m such that $ F _{2} $ and $ F _{3} $ are mutually perpendicular, then the particle remains stationary. If the force $ F _{1} $ is now removed then the acceleration of the particle is [AIEEE 2002]
Options:
A) $ F _{1}/m $
B) $ F _{2}F _{3}/mF _{1} $
C) $ (F _{2}-F _{3})/m $
D) $ F _{2}/m $
Show Answer
Answer:
Correct Answer: A
Solution:
For equilibrium of system, $ F _{1}=\sqrt{F _{2}^{2}+F _{3}^{2}} $ As $ \theta =90{}^\circ $ In the absence of force $ F _{1} $ ,
Acceleration $ =\frac{\text{Net force}}{m} $
$ =\frac{\sqrt{F _{2}^{2}+F _{3}^{2}}}{m}=\frac{F _{1}}{m} $
BETA
