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4-1.Newton's Laws of Motion
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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
A${F_1}/m$
B${F_2}{F_3}/m{F_1}$
C$({F_2} - {F_3})/m$
D${F_2}/m$
(AIEEE-2002)
Solution
(a)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{{{\rm{Net force}}}}{{{\rm{Mass}}}}$
$ = \frac{{\sqrt {F_2^2 + F_3^2} }}{m} = \frac{{{F_1}}}{m}$
In the absence of force ${F_1}$, Acceleration$ = \frac{{{\rm{Net force}}}}{{{\rm{Mass}}}}$
$ = \frac{{\sqrt {F_2^2 + F_3^2} }}{m} = \frac{{{F_1}}}{m}$
Standard 11
Physics
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