System of Particles and Rotational Motion - Result Question 24
24. A rod of weight $W$ is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance $d$ from each other. The centre of mass of the rod is at distance $x$ from $A$. The normal reaction on $A$ is
[2015]
(a) $\frac{Wd}{x}$
(b) $\frac{W(d-x)}{x}$
(c) $\frac{W(d-x)}{d}$
(d) $\frac{Wx}{d}$
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Answer:
Correct Answer: 24. (c)
Solution:
(c) By torque balancing about $B$
$N_A(d)=W(d-x)$
$N_A=\frac{W(d-x)}{d}$
