question_answer

    In a right angle\[\Delta ABC,\text{ }\angle A=90{}^\circ \]and sides a, b, c are respectively, 5 cm, 4 cm and 3 cm. If a force F has moments 0, 9 and 16 in N cm unit respectively about vertices A, B and C, the magnitude of F is

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                 \[{{F}^{2}}={{3}^{2}}+{{4}^{2}}\]                 \[F=5\]