Vector Algebra Question 40
In a right angle $ \Delta ABC,\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
Options:
A) 9
B) 4
C) 5
D) 3
Show Answer
Answer:
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 $ .
BETA
