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The 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
A
$\frac{Wx}{d}$
B
$\frac{Wd}{x}$
C
$\frac{W(d-x)}{x}$
D
$\frac{W(d-x)}{d}$
Detailed Solution
For rotational equilibrium, net moment about any point is zero.
Moments about B: $R_A d - W(d - x) - R_B\cdot 0 = 0$
$R_A = \frac{W(d-x)}{d}$
(Similarly, moments about A give $R_B = \frac{Wx}{d}$.)
Moments about B: $R_A d - W(d - x) - R_B\cdot 0 = 0$
$R_A = \frac{W(d-x)}{d}$
(Similarly, moments about A give $R_B = \frac{Wx}{d}$.)