A) 3
B) 4
C) 5
D) 9
Correct Answer: C
Solution :
Moment about A of force\[\overrightarrow{F}=0\] \[\Rightarrow \]Force\[\overrightarrow{F}\]passes through vertex A. Let\[\overrightarrow{F}\]makes angle\[\theta \]with AC and\[90{}^\circ -\theta \]with AB. Moment about B of force \[\overrightarrow{F}=9\] \[\Rightarrow \] \[F.3\cos \theta =9\Rightarrow F\,\cos \theta =3\] ...(i) Moment about C of force \[\overrightarrow{F}=16\] \[F.4\sin \theta =16\Rightarrow F.\sin \theta =4\] ...(ii) On squaring Eqs. (i) and (ii) and then adding \[{{F}^{2}}={{3}^{2}}+{{4}^{2}}\] \[F=5\]You need to login to perform this action.
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