A) \[\frac{{{F}_{1}}}{m}\]
B) \[\frac{{{F}_{2}}{{F}_{3}}}{m{{F}_{1}}}\]
C) \[\frac{({{F}_{2}}-{{F}_{3}})}{m}\]
D) \[\frac{{{F}_{3}}}{m}\]
Correct Answer: A
Solution :
The particle remains stationary under the action of three forces \[{{\vec{F}}_{1}}={{\vec{F}}_{2}}\]and\[{{\vec{F}}_{3}},\]it means resultant force is zero. \[{{\vec{F}}_{1}}=-({{\vec{F}}_{2}}+{{\vec{F}}_{3}})\] Since in second case \[{{F}_{1}}\]is removed (in terms of magnitude we are taking now), the forces acting are \[{{F}_{2}}\]and \[{{F}_{3}}\]the resultant of which has the magnitude as \[{{F}_{1}},\]so acceleration of particle is \[\frac{{{F}_{1}}}{m}\]in the direction opposite to that of\[{{\vec{F}}_{1}}.\] Hence, the correction option is [a].
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