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 \[\vec{F}\] has moments 0, 9 and 16 in N cm. units respectively about vertices A, B and C, then magnitude of \[\vec{F}\] is

A) 9

B) 4

C) 5

D) 3

Correct Answer: C

Solution :

[c] Since, the moment about A is zero, hence \[\vec{F}\] passes through A. Taking A as origin. Let the line of action of force \[\vec{F}\] be y = mx. (see figure) moment about \[B=\frac{3m}{\sqrt{1+{{m}^{2}}}}|\vec{F}|=9....(1)\] Moment about \[C=\frac{4}{\sqrt{1+{{m}^{2}}}}|\vec{F}|=16...(2)\]  Dividing (1) by (2), we get: \[m=\frac{3}{4}\Rightarrow |\vec{F}|=5N\].